Chapter 13 Probability

Given:

Probability that A is selected, $P(A) = 0.7$

Probability that exactly one of A or B is selected, $P(\text{exactly one}) = 0.6$

To Find:

Probability that B is selected, $P(B)$.

Solution:

Let A be the event that candidate A is selected.

Let B be the event that candidate B is selected.

We are given the probability of A being selected:

$P(A) = 0.7$

We are given the probability that exactly one of them is selected:

$P(\text{exactly one}) = 0.6$

The event "exactly one of A or B is selected" means that either A is selected and B is not selected ($A \cap B'$), or B is selected and A is not selected ($A' \cap B$). These two outcomes are mutually exclusive events.

Therefore, the probability of exactly one being selected can be written as:

$P(\text{exactly one}) = P(A \cap B') + P(A' \cap B)$

In many problems of this type, it is assumed that the selection of one candidate is independent of the selection of the other unless stated otherwise. Assuming independence of events A and B, we can express the probabilities of intersections as products of individual probabilities:

$P(A \cap B') = P(A) \times P(B')$

$P(A' \cap B) = P(A') \times P(B)$

We know that the probability of A not being selected, $P(A')$, is:

$P(A') = 1 - P(A) = 1 - 0.7 = 0.3$

Let $P(B)$ be the probability that B is selected, which we want to find. Let $P(B) = x$.

Then, the probability of B not being selected, $P(B')$, is:

$P(B') = 1 - P(B) = 1 - x$

Substitute these expressions back into the equation for the probability of exactly one selection:

$P(A) \times P(B') + P(A') \times P(B) = 0.6$

$0.7 \times (1 - x) + 0.3 \times x = 0.6$

Now, solve this linear equation for $x$:

$0.7 - 0.7x + 0.3x = 0.6$

Combine the terms with $x$:

$0.7 - 0.4x = 0.6$

Subtract 0.7 from both sides of the equation:

$-0.4x = 0.6 - 0.7$

$-0.4x = -0.1$

Multiply both sides by -1:

$0.4x = 0.1$

Divide both sides by 0.4:

$x = \frac{0.1}{0.4}$

$x = \frac{1}{4}$

$x = 0.25$

So, the probability that B is selected, $P(B)$, is 0.25.

Answer:

The probability that B is selected is 0.25.

Given:

Probability of simultaneous occurrence of at least one of two events A and B is $p$.

This can be written as $P(A \cup B) = p$.

Probability that exactly one of A, B occurs is $q$.

This can be written as $P(\text{exactly one}) = q$.

To Prove:

$P(A') + P(B') = 2 - 2p + q$

Proof:

Let A and B be two events.

We are given that $P(A \cup B) = p$.

We are also given that the probability that exactly one of A, B occurs is $q$.

The event "exactly one of A, B occurs" consists of two mutually exclusive events: (A occurs and B does not occur) or (B occurs and A does not occur).

So, $P(\text{exactly one}) = P(A \cap B') + P(A' \cap B)$.

Thus, we have $P(A \cap B') + P(A' \cap B) = q$.

We know the formula for the probability of the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Substituting the given value $P(A \cup B) = p$, we get:

$p = P(A) + P(B) - P(A \cap B)$

The probability of exactly one event occurring can also be expressed using the union and intersection:

$P(\text{exactly one}) = P(A \cup B) - P(A \cap B)$

This is because $P(A \cup B) = P(A \cap B') + P(A' \cap B) + P(A \cap B)$.

Substituting the given values $P(A \cup B) = p$ and $P(\text{exactly one}) = q$, we have:

$q = p - P(A \cap B)$

From this equation, we can find the probability of the intersection $P(A \cap B)$:

$P(A \cap B) = p - q$

Now, substitute this expression for $P(A \cap B)$ back into the union formula $p = P(A) + P(B) - P(A \cap B)$:

$p = P(A) + P(B) - (p - q)$

$p = P(A) + P(B) - p + q$

Rearrange the equation to find the sum of probabilities $P(A) + P(B)$:

$P(A) + P(B) = p + p - q$

$P(A) + P(B) = 2p - q$

We want to prove $P(A') + P(B') = 2 - 2p + q$.

Using the complement rule, $P(E') = 1 - P(E)$, we have:

$P(A') = 1 - P(A)$

$P(B') = 1 - P(B)$

Adding these two probabilities:

$P(A') + P(B') = (1 - P(A)) + (1 - P(B))$

$P(A') + P(B') = 2 - (P(A) + P(B))$

Now, substitute the expression we found for $P(A) + P(B)$, which is $2p - q$, into this equation:

$P(A') + P(B') = 2 - (2p - q)$

$P(A') + P(B') = 2 - 2p + q$

This is the desired result.

Hence, $P(A') + P(B') = 2 - 2p + q$ is proved.

Given:

Let R be the event that a randomly picked bulb is of red colour.

Let D be the event that a randomly picked bulb is defective.

The probability that a bulb is of red colour is $P(R) = 10\% = 0.10$.

The probability that a bulb is red and defective is $P(R \cap D) = 2\% = 0.02$.

To Find:

The probability that a bulb is defective if it is red. This is the conditional probability $P(D|R)$.

Solution:

We need to find the probability of event D occurring given that event R has already occurred. This is given by the formula for conditional probability:

We are given the values for $P(R)$ and $P(R \cap D)$. Note that $P(D \cap R)$ is the same as $P(R \cap D)$.

Substitute the given values into the formula:

$P(D|R) = \frac{0.02}{0.10}$

Simplify the fraction:

$P(D|R) = \frac{2/100}{10/100}$

$P(D|R) = \frac{2}{10}$

$P(D|R) = 0.2$

Answer:

The probability that the bulb is defective if it is red is $0.2$ or $20\%$.

Given:

Two dice are thrown together.

Event A: Getting 6 on the first die.

Event B: Getting 2 on the second die.

To Determine:

Whether events A and B are independent.

Solution:

When two dice are thrown, the total number of possible outcomes is $6 \times 6 = 36$. The sample space S is the set of all ordered pairs $(d_1, d_2)$ where $d_1, d_2 \in \{1, 2, 3, 4, 5, 6\}$.

$|S| = 36$

Event A is 'getting 6 on the first die'. The outcomes belonging to event A are:

$A = \{(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$

The number of outcomes in A is $|A| = 6$.

The probability of event A is:

$P(A) = \frac{\text{Number of outcomes in A}}{\text{Total number of outcomes}} = \frac{|A|}{|S|} = \frac{6}{36} = \frac{1}{6}$

Event B is 'getting 2 on the second die'. The outcomes belonging to event B are:

$B = \{(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)\}$

The number of outcomes in B is $|B| = 6$.

The probability of event B is:

$P(B) = \frac{\text{Number of outcomes in B}}{\text{Total number of outcomes}} = \frac{|B|}{|S|} = \frac{6}{36} = \frac{1}{6}$

Now, let's consider the intersection of events A and B, which is 'getting 6 on the first die AND getting 2 on the second die'.

$A \cap B = \{(6,2)\}$

The number of outcomes in $A \cap B$ is $|A \cap B| = 1$.

The probability of the intersection $A \cap B$ is:

$P(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{1}{36}$

Two events A and B are independent if and only if $P(A \cap B) = P(A) \times P(B)$.

Let's calculate the product of the individual probabilities $P(A) \times P(B)$:

$P(A) \times P(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

We compare $P(A \cap B)$ with $P(A) \times P(B)$:

$P(A \cap B) = \frac{1}{36}$

$P(A) \times P(B) = \frac{1}{36}$

Since $P(A \cap B) = P(A) \times P(B)$, the condition for independence is satisfied.

Answer:

Yes, the events A and B are independent.

Given:

Total number of boys = 8

Total number of girls = 4

Total number of students = $8 + 4 = 12$

A committee of 4 students is selected at random.

Event E: There is at least one girl on the committee.

Event F: There are exactly 2 girls on the committee.

To Find:

The probability that there are exactly 2 girls on the committee, given that there is at least one girl on the committee. This is $P(F|E)$.

Solution:

The total number of ways to select a committee of 4 students from 12 students is given by $\binom{12}{4}$.

Total number of outcomes = $\binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$.

Event E is that there is at least one girl on the committee. The complement of this event, E', is that there are no girls on the committee, meaning all 4 students selected are boys.

The number of ways to select a committee of 4 boys from 8 boys is $\binom{8}{4}$.

Number of outcomes in E' = $\binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.

The number of outcomes in E (at least one girl) is the total number of outcomes minus the number of outcomes with no girls.

Number of outcomes in E = Total outcomes - Number of outcomes in E'

$|E| = 495 - 70 = 425$.

The probability of event E is $P(E) = \frac{|E|}{\text{Total outcomes}} = \frac{425}{495}$.

Event F is that there are exactly 2 girls on the committee. For this, we need to select 2 girls from 4 girls AND 2 boys from 8 boys.

Number of ways to select 2 girls from 4 = $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$.

Number of ways to select 2 boys from 8 = $\binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28$.

The number of outcomes in F is the product of these two numbers:

$|F| = \binom{4}{2} \times \binom{8}{2} = 6 \times 28 = 168$.

The probability of event F is $P(F) = \frac{|F|}{\text{Total outcomes}} = \frac{168}{495}$.

We need to calculate $P(F|E)$, the probability of event F given event E. Using the formula for conditional probability:

The event $F \cap E$ is the event that there are exactly 2 girls on the committee AND there is at least one girl on the committee. If there are exactly 2 girls, it automatically implies there is at least one girl. Thus, the event $F \cap E$ is the same as event F.

$F \cap E = F$

So, $P(F \cap E) = P(F) = \frac{168}{495}$.

Substitute the probabilities $P(F \cap E)$ and $P(E)$ into the conditional probability formula (i):

$P(F|E) = \frac{168/495}{425/495}$

$P(F|E) = \frac{168}{425}$

We can check if the fraction $\frac{168}{425}$ can be simplified further. The prime factors of 168 are $2^3 \times 3 \times 7$. The prime factors of 425 are $5^2 \times 17$. There are no common factors, so the fraction is in its simplest form.

Answer:

The probability that there are exactly 2 girls on the committee, given that there is at least one girl on the committee, is $\frac{168}{425}$.

Given:

Let E₁, E₂, and E₃ be the events that a randomly picked tube was produced by machine E₁, E₂, and E₃ respectively.

Let D be the event that a randomly picked tube is defective.

The proportions of total output produced by each machine are:

$P(E_1) = 50\% = 0.50$

$P(E_2) = 25\% = 0.25$

$P(E_3) = 25\% = 0.25$

Note that $P(E_1) + P(E_2) + P(E_3) = 0.50 + 0.25 + 0.25 = 1$, confirming that these events form a partition of the sample space.

The probabilities of a tube being defective, given the machine it was produced by, are:

$P(D|E_1) = 4\% = 0.04$

$P(D|E_2) = 4\% = 0.04$

$P(D|E_3) = 5\% = 0.05$

To Find:

The probability that a randomly picked tube is defective, i.e., $P(D)$.

Solution:

We can use the Law of Total Probability to find the probability of a tube being defective. The law states that if E₁, E₂, ..., E_n are mutually exclusive and exhaustive events (forming a partition of the sample space), then for any event D:

$P(D) = \sum_{i=1}^{n} P(D|E_i)P(E_i)$

In this case, we have three events E₁, E₂, and E₃. So, the formula becomes:

Substitute the given values into the formula (i):

$P(D) = (0.04 \times 0.50) + (0.04 \times 0.25) + (0.05 \times 0.25)$

Perform the multiplications:

$P(D) = 0.0200 + 0.0100 + 0.0125$

Perform the addition:

$P(D) = 0.0425$

The probability that a randomly picked tube is defective is 0.0425.

Answer:

The probability that a randomly picked electric tube is defective is 0.0425 or 4.25%.

Given:

Number of throws of a fair die, $n = 10$.

Success is defined as getting a score which is a multiple of 3.

The scores on a fair die are $\{1, 2, 3, 4, 5, 6\}$. The scores which are multiples of 3 are $\{3, 6\}$.

Probability of getting a multiple of 3 in a single throw:

$p = P(\text{Success}) = \frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}$

Probability of not getting a multiple of 3 in a single throw:

$q = P(\text{Failure}) = 1 - p = 1 - \frac{1}{3} = \frac{2}{3}$

To Find:

The probability of obtaining a multiple of 3 in at least 8 of the 10 throws, i.e., $P(X \geq 8)$, where X is the number of successes in 10 trials.

Solution:

This is a binomial probability problem with parameters $n=10$, $p=\frac{1}{3}$, and $q=\frac{2}{3}$.

The probability of getting exactly $k$ successes in $n$ trials is given by the binomial probability formula:

We need to find the probability of getting at least 8 successes, which means the number of successes can be 8, 9, or 10.

$P(X \geq 8) = P(X=8) + P(X=9) + P(X=10)$

Let's calculate each term:

For $k=8$:

$$P(X=8) = \binom{10}{8} \left(\frac{1}{3}\right)^8 \left(\frac{2}{3}\right)^{10-8} = \binom{10}{8} \left(\frac{1}{3}\right)^8 \left(\frac{2}{3}\right)^2$$

Calculate $\binom{10}{8}$:

$$\binom{10}{8} = \binom{10}{10-8} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45$$

$$P(X=8) = 45 \times \frac{1}{3^8} \times \frac{2^2}{3^2} = 45 \times \frac{1}{3^8} \times \frac{4}{3^2} = \frac{45 \times 4}{3^{8+2}} = \frac{180}{3^{10}}$$

For $k=9$:

$$P(X=9) = \binom{10}{9} \left(\frac{1}{3}\right)^9 \left(\frac{2}{3}\right)^{10-9} = \binom{10}{9} \left(\frac{1}{3}\right)^9 \left(\frac{2}{3}\right)^1$$

Calculate $\binom{10}{9}$:

$$\binom{10}{9} = \binom{10}{10-9} = \binom{10}{1} = 10$$

$$P(X=9) = 10 \times \frac{1}{3^9} \times \frac{2}{3^1} = \frac{10 \times 2}{3^{9+1}} = \frac{20}{3^{10}}$$

For $k=10$:

$$P(X=10) = \binom{10}{10} \left(\frac{1}{3}\right)^{10} \left(\frac{2}{3}\right)^{10-10} = \binom{10}{10} \left(\frac{1}{3}\right)^{10} \left(\frac{2}{3}\right)^0$$

Calculate $\binom{10}{10}$:

$$\binom{10}{10} = 1$$

$$P(X=10) = 1 \times \frac{1}{3^{10}} \times 1 = \frac{1}{3^{10}}$$

Now, sum the probabilities:

$$P(X \geq 8) = \frac{180}{3^{10}} + \frac{20}{3^{10}} + \frac{1}{3^{10}} = \frac{180 + 20 + 1}{3^{10}} = \frac{201}{3^{10}}$$

Calculate $3^{10}$:

$3^{10} = 59049$

So, the probability is:

$$P(X \geq 8) = \frac{201}{59049}$$

Answer:

The probability that a score which is a multiple of 3 will be obtained in at least 8 of the 10 throws is $\frac{201}{59049}$.

Given:

The probability distribution of the discrete random variable X is given by the following table:

To Find:

The value of C.

The mean of the distribution, $E(X)$.

Solution:

For a discrete probability distribution, the sum of all probabilities must be equal to 1.

$$\sum P(X=x_i) = 1$$

Summing the probabilities from the given table:

$C + 2C + 2C + 3C + C^2 + 2C^2 + (7C^2 + C) = 1$

Combine the terms involving C and C²:

$(C + 2C + 2C + 3C + C) + (C^2 + 2C^2 + 7C^2) = 1$

$9C + 10C^2 = 1$

Rearrange the equation into a standard quadratic form:

$10C^2 + 9C - 1 = 0$

We can solve this quadratic equation for C using the quadratic formula $C = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=10$, $b=9$, and $c=-1$.

$$C = \frac{-9 \pm \sqrt{9^2 - 4(10)(-1)}}{2(10)}$$

$$C = \frac{-9 \pm \sqrt{81 + 40}}{20}$$

$$C = \frac{-9 \pm \sqrt{121}}{20}$$

$$C = \frac{-9 \pm 11}{20}$$

This gives two possible values for C:

$$C_1 = \frac{-9 + 11}{20} = \frac{2}{20} = \frac{1}{10} = 0.1$$

$$C_2 = \frac{-9 - 11}{20} = \frac{-20}{20} = -1$$

Since probabilities must be non-negative, we must check which value of C yields valid probabilities. If $C = -1$, $P(X=1) = -1$, which is not possible for a probability. If $C = 0.1$, all probabilities are non-negative:

$P(X=1) = 0.1$

$P(X=2) = 2(0.1) = 0.2$

$P(X=3) = 2(0.1) = 0.2$

$P(X=4) = 3(0.1) = 0.3$

$P(X=5) = (0.1)^2 = 0.01$

$P(X=6) = 2(0.1)^2 = 0.02$

$P(X=7) = 7(0.1)^2 + 0.1 = 7(0.01) + 0.1 = 0.07 + 0.1 = 0.17$

All these probabilities are between 0 and 1, inclusive. Thus, the correct value of C is 0.1.

Now, we find the mean (expected value) of the distribution, $E(X)$, using the formula:

$$E(X) = \sum_{i=1}^{7} x_i P(X=x_i)$$

$E(X) = (1 \times C) + (2 \times 2C) + (3 \times 2C) + (4 \times 3C) + (5 \times C^2) + (6 \times 2C^2) + (7 \times (7C^2 + C))$

$E(X) = C + 4C + 6C + 12C + 5C^2 + 12C^2 + 49C^2 + 7C$

Combine the terms:

$E(X) = (C + 4C + 6C + 12C + 7C) + (5C^2 + 12C^2 + 49C^2)$

$E(X) = 30C + 66C^2$

Substitute the value $C = 0.1$ into the expression for $E(X)$:

$E(X) = 30(0.1) + 66(0.1)^2$

$E(X) = 30(0.1) + 66(0.01)$

$E(X) = 3.0 + 0.66$

$E(X) = 3.66$

Answer:

The value of C is $0.1$.

The mean of the distribution is $3.66$.

Given:

Number of red balls in the box = 8

Number of white balls in the box = 4

Total number of balls = $8 + 4 = 12$

Number of balls drawn without replacement = 4

X denotes the number of red balls drawn.

To Find:

The probability distribution of X.

Solution:

X is the number of red balls drawn when 4 balls are selected without replacement from a group of 8 red and 4 white balls. This is a problem involving the hypergeometric distribution.

The total number of ways to draw 4 balls from 12 balls is $\binom{12}{4}$.

$$\binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$$

The possible values for the number of red balls drawn (X) can range from the minimum number of red balls required to make a group of 4 to the maximum number of red balls available (8) or the total number of balls drawn (4), whichever is smaller. Since we draw 4 balls and there are only 4 white balls, the number of red balls can be $4-4=0$ (if all 4 are white) up to $\text{min}(8, 4) = 4$. So, the possible values of X are $0, 1, 2, 3, 4$.

To find the probability $P(X=k)$ for $k \in \{0, 1, 2, 3, 4\}$, we need to find the number of ways to choose $k$ red balls from 8 and $(4-k)$ white balls from 4, and divide by the total number of ways to choose 4 balls from 12.

The number of ways to select $k$ red balls from 8 is $\binom{8}{k}$.

The number of ways to select $(4-k)$ white balls from 4 is $\binom{4}{4-k}$.

The number of ways to select $k$ red balls and $(4-k)$ white balls is $\binom{8}{k} \times \binom{4}{4-k}$.

So, the probability is:

$$P(X=k) = \frac{\binom{8}{k} \binom{4}{4-k}}{\binom{12}{4}}$$

Calculate $P(X=k)$ for each value of k:

For $k=0$ (0 red balls and 4 white balls):

$$P(X=0) = \frac{\binom{8}{0} \binom{4}{4}}{\binom{12}{4}} = \frac{1 \times 1}{495} = \frac{1}{495}$$

For $k=1$ (1 red ball and 3 white balls):

$$P(X=1) = \frac{\binom{8}{1} \binom{4}{3}}{\binom{12}{4}} = \frac{8 \times 4}{495} = \frac{32}{495}$$

For $k=2$ (2 red balls and 2 white balls):

$$P(X=2) = \frac{\binom{8}{2} \binom{4}{2}}{\binom{12}{4}} = \frac{\frac{8 \times 7}{2 \times 1} \times \frac{4 \times 3}{2 \times 1}}{495} = \frac{28 \times 6}{495} = \frac{168}{495}$$

For $k=3$ (3 red balls and 1 white ball):

$$P(X=3) = \frac{\binom{8}{3} \binom{4}{1}}{\binom{12}{4}} = \frac{\frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times 4}{495} = \frac{56 \times 4}{495} = \frac{224}{495}$$

For $k=4$ (4 red balls and 0 white balls):

$$P(X=4) = \frac{\binom{8}{4} \binom{4}{0}}{\binom{12}{4}} = \frac{\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \times 1}{495} = \frac{70 \times 1}{495} = \frac{70}{495}$$

The probability distribution of X can be presented in a table:

Let's verify that the sum of probabilities is 1:

$$\sum P(X=k) = \frac{1}{495} + \frac{32}{495} + \frac{168}{495} + \frac{224}{495} + \frac{70}{495} = \frac{1 + 32 + 168 + 224 + 70}{495} = \frac{495}{495} = 1$$

The probabilities sum to 1, so this is a valid probability distribution.

Answer:

The probability distribution of X is:

Given:

Number of tosses of a fair coin = 3.

The random variable X represents the number of heads in three tosses.

To Determine:

The variance of X, $\text{Var}(X)$.

The standard deviation of X, $\sigma_X$.

Solution:

When a fair coin is tossed, the probability of getting a head (H) is $p = 0.5$, and the probability of getting a tail (T) is $q = 1 - p = 0.5$.

We are tossing the coin 3 times. The possible number of heads (X) can be 0, 1, 2, or 3.

Let's find the probability distribution of X. The total number of possible outcomes for 3 tosses is $2^3 = 8$. The possible outcomes are:

HHH (3 heads)

HHT, HTH, THH (2 heads)

HTT, THT, TTH (1 head)

TTT (0 heads)

Assuming the coin is fair, each outcome is equally likely with probability $\frac{1}{8}$.

The probability distribution of X is:

$P(X=0) = P(\text{TTT}) = \frac{1}{8}$

$P(X=1) = P(\text{HTT, THT, TTH}) = \frac{3}{8}$

$P(X=2) = P(\text{HHT, HTH, THH}) = \frac{3}{8}$

$P(X=3) = P(\text{HHH}) = \frac{1}{8}$

We can represent the probability distribution in a table:

To find the variance, we first need to calculate the mean (Expected value) $E(X)$. The formula for the mean of a discrete random variable is $E(X) = \sum x_i P(x_i)$.

$$E(X) = (0 \times \frac{1}{8}) + (1 \times \frac{3}{8}) + (2 \times \frac{3}{8}) + (3 \times \frac{1}{8})$$

$$E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{3+6+3}{8} = \frac{12}{8} = \frac{3}{2}$$

So, the mean number of heads is $E(X) = 1.5$.

Now, we calculate the variance $\text{Var}(X)$. The formula is $\text{Var}(X) = E(X^2) - [E(X)]^2$.

First, calculate $E(X^2) = \sum x_i^2 P(x_i)$.

$$E(X^2) = (0^2 \times \frac{1}{8}) + (1^2 \times \frac{3}{8}) + (2^2 \times \frac{3}{8}) + (3^2 \times \frac{1}{8})$$

$$E(X^2) = (0 \times \frac{1}{8}) + (1 \times \frac{3}{8}) + (4 \times \frac{3}{8}) + (9 \times \frac{1}{8})$$

$$E(X^2) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{3+12+9}{8} = \frac{24}{8} = 3$$

Now, calculate the variance:

$$\text{Var}(X) = E(X^2) - [E(X)]^2 = 3 - \left(\frac{3}{2}\right)^2$$

$$\text{Var}(X) = 3 - \frac{9}{4} = \frac{12}{4} - \frac{9}{4} = \frac{3}{4}$$

The variance is $0.75$.

The standard deviation $\sigma_X$ is the square root of the variance.

$$\sigma_X = \sqrt{\text{Var}(X)} = \sqrt{\frac{3}{4}}$$

$$\sigma_X = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2}$$

In decimal form, $\sqrt{3} \approx 1.732$, so $\sigma_X \approx \frac{1.732}{2} = 0.866$.

Alternatively, for a binomial distribution $B(n, p)$, the mean is $np$ and the variance is $npq$.

$n=3$, $p=0.5$, $q=0.5$.

$E(X) = np = 3 \times 0.5 = 1.5$.

$\text{Var}(X) = npq = 3 \times 0.5 \times 0.5 = 3 \times 0.25 = 0.75 = \frac{3}{4}$.

$\sigma_X = \sqrt{npq} = \sqrt{0.75} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$.

Both methods yield the same result.

Answer:

The variance of the number of heads is $\frac{3}{4}$ or $0.75$.

The standard deviation of the number of heads is $\frac{\sqrt{3}}{2}$ or approximately $0.866$.

Given:

From Example 6, we have the following probabilities:

Probability that a tube is from machine E₁, $P(E_1) = 0.50$.

Probability that a tube is from machine E₂, $P(E_2) = 0.25$.

Probability that a tube is from machine E₃, $P(E_3) = 0.25$.

Probability that a tube is defective given it's from E₁, $P(D|E_1) = 0.04$.

Probability that a tube is defective given it's from E₂, $P(D|E_2) = 0.04$.

Probability that a tube is defective given it's from E₃, $P(D|E_3) = 0.05$.

Also from Example 6, the total probability of a tube being defective is $P(D) = 0.0425$.

To Find:

The probability that the defective tube was produced on machine E₁, i.e., $P(E_1|D)$.

Solution:

We need to calculate the conditional probability $P(E_1|D)$. This can be done using Bayes' Theorem, which states:

For machine E₁, we have:

Substitute the known values into formula (ii):

$P(E_1|D) = \frac{0.04 \times 0.50}{0.0425}$

Calculate the numerator:

$0.04 \times 0.50 = 0.0200$

Now substitute this back into the expression for $P(E_1|D)$:

$P(E_1|D) = \frac{0.0200}{0.0425}$

To simplify the fraction, multiply the numerator and denominator by 10000:

$P(E_1|D) = \frac{0.0200 \times 10000}{0.0425 \times 10000} = \frac{200}{425}$

Both 200 and 425 are divisible by 25. Divide both numerator and denominator by 25:

$$ \frac{\cancel{200}^{8}}{\cancel{425}_{17}} = \frac{8}{17} $$

So, $P(E_1|D) = \frac{8}{17}$.

Answer:

The probability that the defective tube was produced on machine E₁ is $\frac{8}{17}$.

Given:

Let A be the event that a car is manufactured at Plant X.

Let B be the event that a car is manufactured at Plant Y.

Let S be the event that a car is of standard quality.

Probability of a car originating from Plant X, $P(A) = 70\% = 0.7$.

Probability of a car originating from Plant Y, $P(B) = 30\% = 0.3$.

Note that $P(A) + P(B) = 0.7 + 0.3 = 1$, so A and B form a partition of the sample space.

Probability of a car being of standard quality given it is from Plant X, $P(S|A) = 80\% = 0.8$.

Probability of a car being of standard quality given it is from Plant Y, $P(S|B) = 90\% = 0.9$.

To Find:

The probability that a randomly chosen car, found to be of standard quality, has come from Plant X. This is the conditional probability $P(A|S)$.

Solution:

We need to calculate the probability $P(A|S)$. We can use Bayes' Theorem for this, which requires the probabilities $P(A)$, $P(S|A)$, and $P(S)$.

Bayes' Theorem is given by:

We are given $P(A) = 0.7$ and $P(S|A) = 0.8$. We need to find the total probability of a car being of standard quality, $P(S)$. Since any car must be produced by either Plant X or Plant Y, we can use the Law of Total Probability:

Using the definition of conditional probability $P(E \cap F) = P(E|F)P(F)$, we can write:

Substitute the given values into formula (ii):

$$P(S) = (0.8)(0.7) + (0.9)(0.3)$$

$$P(S) = 0.56 + 0.27$$

$$P(S) = 0.83$$

Now that we have $P(S)$, we can substitute the values of $P(S)$, $P(S|A)$, and $P(A)$ into Bayes' Theorem formula (i):

$$P(A|S) = \frac{0.8 \times 0.7}{0.83}$$

$$P(A|S) = \frac{0.56}{0.83}$$

To express this probability as a fraction, we can multiply the numerator and denominator by 100:

$$P(A|S) = \frac{0.56 \times 100}{0.83 \times 100} = \frac{56}{83}$$

The fraction $\frac{56}{83}$ cannot be simplified further since 83 is a prime number and 56 is not divisible by 83.

Answer:

The probability that the standard quality car has come from Plant X is $\frac{56}{83}$.

Given:

Probability of event A, $P(A) = 0.2$.

Probability of event B, $P(B) = 0.4$.

Probability of the union of events A and B, $P(A \cup B) = 0.6$.

To Find:

The conditional probability of A given B, $P(A | B)$.

Solution:

We use the formula for the probability of the union of two events:

Substitute the given values into formula (i) to find the probability of the intersection $P(A \cap B)$:

$$0.6 = 0.2 + 0.4 - P(A \cap B)$$

$$0.6 = 0.6 - P(A \cap B)$$

Rearrange the equation to solve for $P(A \cap B)$:

$$P(A \cap B) = 0.6 - 0.6$$

$$P(A \cap B) = 0$$

Now, we use the formula for conditional probability $P(A | B)$:

Substitute the value of $P(A \cap B)$ and the given value of $P(B)$ into formula (ii):

$$P(A | B) = \frac{0}{0.4}$$

$$P(A | B) = 0$$

The probability $P(A | B)$ is 0.

Answer:

The value of $P(A | B)$ is 0.

This matches option (D).

Given:

Probability of event A, $P(A) = 0.6$.

Probability of event B, $P(B) = 0.2$.

Conditional probability of A given B, $P(A | B) = 0.5$.

To Find:

The conditional probability $P(A' | B')$.

Solution:

We are asked to find $P(A' | B')$. Using the definition of conditional probability, we have:

First, let's find $P(A \cap B)$ using the given $P(A | B)$ and $P(B)$. The formula for conditional probability $P(A | B)$ is:

Substitute the given values into formula (ii):

$$0.5 = \frac{P(A \cap B)}{0.2}$$

Solve for $P(A \cap B)$:

$$P(A \cap B) = 0.5 \times 0.2 = 0.1$$

Next, we need to find $P(A' \cap B')$. Using De Morgan's Law, $A' \cap B' = (A \cup B)'$. Therefore, $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$.

We find $P(A \cup B)$ using the formula:

Substitute the known values into formula (iii):

$$P(A \cup B) = 0.6 + 0.2 - 0.1$$

$$P(A \cup B) = 0.8 - 0.1$$

$$P(A \cup B) = 0.7$$

Now, we can find $P(A' \cap B')$:

$$P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.7 = 0.3$$

We also need $P(B')$. Using the complement rule:

$$P(B') = 1 - P(B) = 1 - 0.2 = 0.8$$

Finally, substitute the values of $P(A' \cap B')$ and $P(B')$ into the conditional probability formula (i):

$$P(A' | B') = \frac{0.3}{0.8}$$

To simplify the fraction, we can write it as:

$$P(A' | B') = \frac{3/10}{8/10} = \frac{3}{8}$$

The probability $P(A' | B')$ is $\frac{3}{8}$.

Answer:

The value of $P(A' | B')$ is $\frac{3}{8}$.

This matches option (C).

Given:

A and B are independent events.

$0 < P(A) < 1$

$0 < P(B) < 1$

To Determine:

Which of the given statements is not correct.

Solution:

If events A and B are independent, then the definition of independence holds:

We are given that $0 < P(A) < 1$ and $0 < P(B) < 1$. This means that $P(A)$ and $P(B)$ are both positive, and neither is equal to 0 or 1.

Let's analyze each statement:

(A) A and B are mutually exclusive.

If A and B were mutually exclusive events, then their intersection would be empty, and the probability of their intersection would be 0.

$$P(A \cap B) = 0$$

From the independence condition (i), we have $P(A \cap B) = P(A) \times P(B)$.

So, if A and B were both independent and mutually exclusive (and $P(A) > 0, P(B) > 0$), then:

$$P(A) \times P(B) = 0$$

This would imply that either $P(A) = 0$ or $P(B) = 0$. However, we are given that $P(A) > 0$ and $P(B) > 0$.

Therefore, if A and B are independent and have non-zero probabilities, they cannot be mutually exclusive. Mutually exclusive events can only be independent if the probability of at least one of them is zero.

Since $P(A) > 0$ and $P(B) > 0$, the statement that A and B are mutually exclusive is incorrect.

Let's quickly check the other statements (which are known properties of independent events) to confirm they are correct given A and B are independent.

(B) A and B′ are independent.

If A and B are independent, then $P(A \cap B) = P(A)P(B)$.

We want to check if $P(A \cap B') = P(A)P(B')$.

We know that $P(A) = P(A \cap B) + P(A \cap B')$.

So, $P(A \cap B') = P(A) - P(A \cap B)$.

Since A and B are independent, substitute $P(A \cap B) = P(A)P(B)$:

$P(A \cap B') = P(A) - P(A)P(B) = P(A)(1 - P(B)) = P(A)P(B')$.

Thus, A and B′ are independent. This statement is correct.

(C) A′ and B are independent.

This is similar to (B). If A and B are independent, then $P(A \cap B) = P(A)P(B)$.

We want to check if $P(A' \cap B) = P(A')P(B)$.

We know that $P(B) = P(A \cap B) + P(A' \cap B)$.

So, $P(A' \cap B) = P(B) - P(A \cap B)$.

Since A and B are independent, substitute $P(A \cap B) = P(A)P(B)$:

$P(A' \cap B) = P(B) - P(A)P(B) = P(B)(1 - P(A)) = P(B)P(A')$.

Thus, A′ and B are independent. This statement is correct.

(D) A′ and B′ are independent.

If A and B are independent, then $P(A \cap B) = P(A)P(B)$.

We want to check if $P(A' \cap B') = P(A')P(B')$.

Using De Morgan's Law, $P(A' \cap B') = P((A \cup B)')$.

$P((A \cup B)') = 1 - P(A \cup B) = 1 - (P(A) + P(B) - P(A \cap B))$.

Since A and B are independent, substitute $P(A \cap B) = P(A)P(B)$:

$P(A' \cap B') = 1 - (P(A) + P(B) - P(A)P(B)) = 1 - P(A) - P(B) + P(A)P(B)$.

Also, $P(A')P(B') = (1 - P(A))(1 - P(B)) = 1 - P(B) - P(A) + P(A)P(B)$.

Since $P(A' \cap B') = P(A')P(B')$, A′ and B′ are independent. This statement is correct.

Based on the analysis, the statement that is not correct is (A).

Answer:

The statement that is not correct is (A) A and B are mutually exclusive.

Given:

The probability distribution of the discrete random variable X is given by the table:

To Find:

The expected value of X, $E(X)$.

Solution:

The expected value (mean) of a discrete random variable X with probability distribution $P(X=x_i)$ is given by the formula:

Substitute the values from the table into formula (i):

$$E(X) = (30 \times \frac{1}{5}) + (10 \times \frac{3}{10}) + (-10 \times \frac{1}{2})$$

Perform the multiplications:

$$E(X) = \frac{30}{5} + \frac{30}{10} - \frac{10}{2}$$

$$E(X) = 6 + 3 - 5$$

Perform the addition and subtraction:

$$E(X) = 9 - 5$$

$$E(X) = 4$$

The expected value of X is 4.

Answer:

The value of $E(X)$ is 4.

This matches option (B).

Given:

A discrete random variable X with values $x_1, x_2, ..., x_n$ and corresponding probabilities $p_1, p_2, ..., p_n$.

To Determine:

The correct formula for the variance of X.

Solution:

The variance of a discrete random variable X, denoted by $\text{Var}(X)$ or $\sigma_X^2$, is defined as the expected value of the squared deviation from the mean ($E(X)$).

Let $\mu = E(X)$ be the mean of X.

The definition of variance is:

$$\text{Var}(X) = E[(X - \mu)^2]$$

Expanding the term inside the expectation:

$$(X - \mu)^2 = X^2 - 2X\mu + \mu^2$$

Taking the expectation of both sides, and using the linearity of expectation $E(aY + bZ) = aE(Y) + bE(Z)$ and $E(c) = c$ for a constant c:

$$E[(X - \mu)^2] = E[X^2 - 2X\mu + \mu^2]$$

$$E[(X - \mu)^2] = E(X^2) - E(2X\mu) + E(\mu^2)$$

Since $\mu$ is the mean, it is a constant value. So $2\mu$ and $\mu^2$ are also constants.

$$E(2X\mu) = 2\mu E(X) = 2\mu(\mu) = 2\mu^2$$

$$E(\mu^2) = \mu^2$$

Substitute these back into the equation for variance:

$$\text{Var}(X) = E(X^2) - 2\mu^2 + \mu^2$$

$$\text{Var}(X) = E(X^2) - \mu^2$$

Since $\mu = E(X)$, we can write $\mu^2 = [E(X)]^2$.

Therefore, the variance of X is:

$$\text{Var}(X) = E(X^2) - [E(X)]^2$$

Compare this formula with the given options:

(A) $E (X^2)$ - Incorrect, this is the expected value of the square of X.

(B) $E (X^2) + E (X)$ - Incorrect.

(C) $E (X^2) – [E (X)]^2$ - Correct.

(D) $\sqrt{E (X^2) − [E(X)]^2}$ - Incorrect, this is the formula for the standard deviation, which is the square root of the variance.

Answer:

The correct formula for the variance of X is $E (X^2) – [E (X)]^2$.

This matches option (C).

Given:

Events A and B are independent.

$P(A) = p$

$P(B) = 2p$

$P(\text{Exactly one of A, B}) = \frac{5}{9}$

Also given $0 < P(A) < 1$ and $0 < P(B) < 1$.

To Find:

The value(s) of p.

Solution:

Since A and B are independent events, the probability of their intersection is the product of their probabilities:

$$P(A \cap B) = P(A) \times P(B) = p \times 2p = 2p^2$$

The probability that exactly one of the events A or B occurs is given by:

$$P(\text{Exactly one of A, B}) = P(A \cap B') + P(A' \cap B)$$

Since A and B are independent, A and B' are also independent, and A' and B are also independent. Thus:

$$P(A \cap B') = P(A)P(B')$$

$$P(A' \cap B) = P(A')P(B)$$

We know $P(A') = 1 - P(A) = 1 - p$ and $P(B') = 1 - P(B) = 1 - 2p$.

So, the equation for the probability of exactly one event occurring becomes:

$$\frac{5}{9} = P(A)P(B') + P(A')P(B)$$

$$\frac{5}{9} = p(1 - 2p) + (1 - p)(2p)$$

Expand and simplify the equation:

$$\frac{5}{9} = p - 2p^2 + 2p - 2p^2$$

$$\frac{5}{9} = 3p - 4p^2$$

Multiply both sides by 9 to remove the fraction:

$$5 = 9(3p - 4p^2)$$

$$5 = 27p - 36p^2$$

Rearrange the terms to form a quadratic equation:

$$36p^2 - 27p + 5 = 0$$

We can solve this quadratic equation using factoring or the quadratic formula. Let's try factoring. We look for two numbers that multiply to $36 \times 5 = 180$ and add up to -27. These numbers are -15 and -12.

$$36p^2 - 15p - 12p + 5 = 0$$

Group the terms and factor:

$$3p(12p - 5) - 1(12p - 5) = 0$$

$$(3p - 1)(12p - 5) = 0$$

This gives two possible values for p:

$$3p - 1 = 0 \implies 3p = 1 \implies p = \frac{1}{3}$$

$$12p - 5 = 0 \implies 12p = 5 \implies p = \frac{5}{12}$$

We need to check if these values of p satisfy the condition $0 < P(A) < 1$ and $0 < P(B) < 1$.

Case 1: $p = \frac{1}{3}$

$P(A) = p = \frac{1}{3}$. This satisfies $0 < \frac{1}{3} < 1$.

$P(B) = 2p = 2 \times \frac{1}{3} = \frac{2}{3}$. This satisfies $0 < \frac{2}{3} < 1$.

This value of p is valid.

Case 2: $p = \frac{5}{12}$

$P(A) = p = \frac{5}{12}$. This satisfies $0 < \frac{5}{12} < 1$.

$P(B) = 2p = 2 \times \frac{5}{12} = \frac{10}{12} = \frac{5}{6}$. This satisfies $0 < \frac{5}{6} < 1$.

This value of p is also valid.

Both values $p = \frac{1}{3}$ and $p = \frac{5}{12}$ are possible solutions.

Answer:

p = $\frac{1}{3}$ or $\frac{5}{12}$

The blank can be filled with either value.

Given:

A and B′ are independent events.

To Find:

The expression that completes the statement $P(A' \cup B) = 1 – \text{blank}$.

Solution:

We are given that A and B′ are independent events. This means that the probability of their intersection is the product of their individual probabilities:

We want to find an expression for $P(A' \cup B)$. We can use the complement rule for probability: $P(E) = 1 - P(E')$.

Let $E = A' \cup B$. Then $E' = (A' \cup B)'$.

Using De Morgan's Law, $(A' \cup B)' = (A')' \cap B' = A \cap B'$.

So, $P(A' \cup B) = 1 - P((A' \cup B)') = 1 - P(A \cap B')$.

Now, substitute the result from the independence condition (i) into this equation:

$$P(A' \cup B) = 1 - P(A)P(B')$$

Comparing this with the given statement $P(A' \cup B) = 1 – \text{blank}$, we see that the blank should be filled with the expression $P(A)P(B')$.

Answer:

The completed statement is: If A and B′ are independent events then P (A′ ∪ B) = 1 – P(A) P(B′)

The blank is filled by P(A) P(B′).

Given Statement:

Let A and B be two independent events. Then $P(A \cap B) = P(A) + P(B)$.

To Determine:

Whether the given statement is True or False.

Solution:

The definition of two independent events A and B is that the probability of their simultaneous occurrence is the product of their individual probabilities:

The given statement says $P(A \cap B) = P(A) + P(B)$.

This latter formula, $P(A \cap B) = P(A) + P(B)$, is generally not true for independent events, unless one of the probabilities is zero (in which case both sides would be zero). It is related to the formula for the probability of the union of two events, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. For mutually exclusive events, $P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B)$.

Let's consider a counterexample with independent events where $P(A) > 0$ and $P(B) > 0$.

Suppose we toss a fair coin twice.

Let A be the event 'getting a head on the first toss', so $P(A) = \frac{1}{2}$.

Let B be the event 'getting a head on the second toss', so $P(B) = \frac{1}{2}$.

Events A and B are independent.

The intersection $A \cap B$ is the event 'getting a head on the first toss AND getting a head on the second toss' (HH). The probability of this is $P(A \cap B) = \frac{1}{4}$.

According to the definition of independence (i):

$P(A) \times P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

This matches $P(A \cap B) = \frac{1}{4}$, so A and B are indeed independent.

Now let's check the given statement: $P(A \cap B) = P(A) + P(B)$.

Using the values from our example:

$\frac{1}{4} = \frac{1}{2} + \frac{1}{2}$

$\frac{1}{4} = 1$

This is clearly false ($\frac{1}{4} \neq 1$).

The formula $P(A \cap B) = P(A) + P(B)$ would only hold if $P(A) + P(B) = P(A) \times P(B)$. This is generally not true for positive probabilities.

Thus, the statement that for independent events, $P(A \cap B) = P(A) + P(B)$ is incorrect.

Answer:

False

Given Statement:

Three events A, B and C are said to be independent if $P(A \cap B \cap C) = P(A) P(B) P(C)$.

To Determine:

Whether the given statement is True or False.

Solution:

For three events A, B, and C to be considered independent, they must satisfy the following conditions:

1. Pairwise independence:

$P(A \cap B) = P(A) P(B)$

$P(A \cap C) = P(A) P(C)$

$P(B \cap C) = P(B) P(C)$

2. Independence of the intersection of all three events:

$P(A \cap B \cap C) = P(A) P(B) P(C)$

The given statement says that the events are independent *if* only the condition $P(A \cap B \cap C) = P(A) P(B) P(C)$ holds. This is incorrect because satisfying the condition $P(A \cap B \cap C) = P(A) P(B) P(C)$ does not necessarily imply pairwise independence.

It is possible for events to satisfy $P(A \cap B \cap C) = P(A) P(B) P(C)$ but fail one or more of the pairwise independence conditions. If pairwise independence does not hold, the events are not considered fully independent.

Therefore, the condition $P(A \cap B \cap C) = P(A) P(B) P(C)$ is a necessary condition for the independence of three events, but it is not sufficient on its own.

Answer:

False

Given Statement:

One of the condition of Bernoulli trials is that the trials are independent of each other.

To Determine:

Whether the given statement is True or False.

Solution:

Bernoulli trials are a sequence of independent experiments, each of which has only two possible outcomes, usually called success and failure.

The essential characteristics of Bernoulli trials are:

1. Each trial has exactly two possible outcomes (Success or Failure).

2. The probability of success (p) is constant for every trial. Consequently, the probability of failure (q = 1 - p) is also constant for every trial.

3. The trials are independent of each other. This means the outcome of one trial does not affect the outcome of any other trial.

The given statement explicitly mentions that the trials are independent of each other as one of the conditions. This is indeed a fundamental requirement for a sequence of trials to be considered Bernoulli trials.

Answer:

True

Given:

Probabilities of outcomes for a loaded die are:

$P(1) = 0.2$

$P(2) = 0.2$

$P(3) = 0.1$

$P(4) = 0.3$

$P(5) = 0.1$

$P(6) = 0.1$

The die is thrown two times. The outcomes are independent.

Event A: 'same number each time'.

Event B: 'a total score is 10 or more'.

To Determine:

Whether events A and B are independent.

Solution:

When the die is thrown two times, the outcome is an ordered pair $(x, y)$, where $x$ is the result of the first throw and $y$ is the result of the second throw.

Since the two throws are independent, the probability of the outcome $(x, y)$ is $P(x, y) = P(x) \times P(y)$.

Event A is 'same number each time'. The outcomes in event A are:

$A = \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$

The probability of event A is the sum of the probabilities of these outcomes:

$P(A) = P(1,1) + P(2,2) + P(3,3) + P(4,4) + P(5,5) + P(6,6)$

$P(A) = P(1)P(1) + P(2)P(2) + P(3)P(3) + P(4)P(4) + P(5)P(5) + P(6)P(6)$

$P(A) = (0.2)(0.2) + (0.2)(0.2) + (0.1)(0.1) + (0.3)(0.3) + (0.1)(0.1) + (0.1)(0.1)$

$P(A) = 0.04 + 0.04 + 0.01 + 0.09 + 0.01 + 0.01$

$P(A) = 0.20$

Event B is 'a total score is 10 or more'. This means the sum of the outcomes $x + y \geq 10$. The outcomes in event B are:

Possible sums $\geq 10$ are 10, 11, 12.

Sum = 10: $(4,6), (5,5), (6,4)$

Sum = 11: $(5,6), (6,5)$

Sum = 12: $(6,6)$

So, $B = \{(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$

The probability of event B is the sum of the probabilities of these outcomes:

$P(B) = P(4,6) + P(5,5) + P(6,4) + P(5,6) + P(6,5) + P(6,6)$

$P(B) = P(4)P(6) + P(5)P(5) + P(6)P(4) + P(5)P(6) + P(6)P(5) + P(6)P(6)$

$P(B) = (0.3)(0.1) + (0.1)(0.1) + (0.1)(0.3) + (0.1)(0.1) + (0.1)(0.1) + (0.1)(0.1)$

$P(B) = 0.03 + 0.01 + 0.03 + 0.01 + 0.01 + 0.01$

$P(B) = 0.10$

For events A and B to be independent, the condition $P(A \cap B) = P(A) \times P(B)$ must be satisfied.

First, let's find the intersection of events A and B, $A \cap B$. These are the outcomes that are in both A and B.

$A = \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$

$B = \{(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$

The common outcomes are $(5,5)$ and $(6,6)$.

$A \cap B = \{(5,5), (6,6)\}$

The probability of the intersection is:

$P(A \cap B) = P(5,5) + P(6,6)$

$P(A \cap B) = P(5)P(5) + P(6)P(6)$

$P(A \cap B) = (0.1)(0.1) + (0.1)(0.1)$

$P(A \cap B) = 0.01 + 0.01$

$P(A \cap B) = 0.02$

Now, let's calculate the product of the probabilities of A and B:

$P(A) \times P(B) = (0.20) \times (0.10)$

$P(A) \times P(B) = 0.02$

Comparing $P(A \cap B)$ and $P(A) \times P(B)$:

$P(A \cap B) = 0.02$

$P(A) \times P(B) = 0.02$

Since $P(A \cap B) = P(A) \times P(B)$, the events A and B are independent.

Conclusion:

Events A and B are independent.

Given:

A fair die is thrown two times.

For a fair die, the probability of each outcome (1, 2, 3, 4, 5, 6) is $\frac{1}{6}$.

Event A: 'same number each time'.

Event B: 'a total score is 10 or more'.

To Determine:

Whether events A and B are independent if the die is fair.

Solution:

When a fair die is thrown two times, there are $6 \times 6 = 36$ possible outcomes in the sample space. Each outcome $(x, y)$ has a probability of $P(x, y) = P(x) \times P(y) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$, since the throws are independent.

Event A is 'same number each time'. The outcomes in A are:

$A = \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$

There are 6 outcomes in event A.

The probability of event A is:

$P(A) = \text{Number of outcomes in A} \times P(\text{each outcome})$

$P(A) = 6 \times \frac{1}{36} = \frac{6}{36} = \frac{1}{6}$

Event B is 'a total score is 10 or more'. The outcomes $(x, y)$ such that $x+y \geq 10$ are:

Possible sums $\geq 10$ are 10, 11, 12.

Sum = 10: $(4,6), (5,5), (6,4)$

Sum = 11: $(5,6), (6,5)$

Sum = 12: $(6,6)$

So, $B = \{(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$

There are 6 outcomes in event B.

The probability of event B is:

$P(B) = \text{Number of outcomes in B} \times P(\text{each outcome})$

$P(B) = 6 \times \frac{1}{36} = \frac{6}{36} = \frac{1}{6}$

To check for independence, we need to find the probability of the intersection of A and B, $A \cap B$. The intersection consists of outcomes that are in both A and B.

$A = \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$

$B = \{(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$

The outcomes common to both sets are $(5,5)$ and $(6,6)$.

$A \cap B = \{(5,5), (6,6)\}$

There are 2 outcomes in $A \cap B$.

The probability of the intersection is:

$P(A \cap B) = \text{Number of outcomes in } A \cap B \times P(\text{each outcome})$

$P(A \cap B) = 2 \times \frac{1}{36} = \frac{2}{36} = \frac{1}{18}$

For events A and B to be independent, $P(A \cap B)$ must be equal to $P(A) \times P(B)$.

Let's calculate $P(A) \times P(B)$:

$P(A) \times P(B) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$

Comparing $P(A \cap B)$ and $P(A) \times P(B)$:

$P(A \cap B) = \frac{1}{18}$

$P(A) \times P(B) = \frac{1}{36}$

Since $\frac{1}{18} \neq \frac{1}{36}$, $P(A \cap B) \neq P(A) \times P(B)$.

Conclusion:

For a fair die, events A and B are not independent (they are dependent).

Given:

Probability that at least one of the two events A and B occurs is 0.6.

This means $P(A \cup B) = 0.6$.

Probability that A and B occur simultaneously is 0.3.

This means $P(A \cap B) = 0.3$.

To Evaluate:

$P( \overline{A} ) + P( \overline{B} )$

Solution:

We use the formula for the probability of the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the given values into the formula:

$0.6 = P(A) + P(B) - 0.3$

Add 0.3 to both sides of the equation to find the sum of $P(A)$ and $P(B)$:

$0.6 + 0.3 = P(A) + P(B)$

$P(A) + P(B) = 0.9$

We are asked to evaluate $P( \overline{A} ) + P( \overline{B} )$.

We know that the probability of the complement of an event E is given by $P(\overline{E}) = 1 - P(E)$.

Applying this to events A and B:

$P(\overline{A}) = 1 - P(A)$

$P(\overline{B}) = 1 - P(B)$

Now, sum these probabilities:

$P(\overline{A}) + P(\overline{B}) = (1 - P(A)) + (1 - P(B))$

$P(\overline{A}) + P(\overline{B}) = 1 - P(A) + 1 - P(B)$

$P(\overline{A}) + P(\overline{B}) = 2 - (P(A) + P(B))$

Substitute the value of $P(A) + P(B)$ that we found earlier:

$P(\overline{A}) + P(\overline{B}) = 2 - 0.9$

$P(\overline{A}) + P(\overline{B}) = 1.1$

Answer:

The value of $P( \overline{A} ) + P( \overline{B} )$ is $1.1$.

Given:

Number of red marbles = 5

Number of black marbles = 3

Total number of marbles in the bag initially = $5 + 3 = 8$

Three marbles are drawn one by one without replacement.

The first marble drawn is red.

To Find:

The probability that at least one of the three marbles drawn is black, given that the first marble is red.

Let R$_i$ denote the event that the $i$-th marble drawn is red.

Let B$_i$ denote the event that the $i$-th marble drawn is black.

Let E be the event that at least one of the three marbles drawn is black.

We are asked to find $P(E | R_1)$.

Solution:

The problem gives that the first marble drawn is red. This means we are working within the conditional sample space where the first draw was $R_1$.

After the first marble drawn is red, the contents of the bag are updated:

Number of remaining red marbles = $5 - 1 = 4$

Number of remaining black marbles = 3

Total number of remaining marbles = $4 + 3 = 7$

We draw two more marbles from these 7 marbles without replacement.

The event E is "at least one of the three marbles drawn is black". Given that the first marble is red, event E occurs if at least one of the second or third marbles drawn is black.

The complement of event E, given $R_1$, is that none of the three marbles is black. Since the first is already red ($R_1$), this means the second and third marbles drawn must also be red ($R_2$ and $R_3$).

So, $\overline{E}$ (given $R_1$) is the event that the second marble is red AND the third marble is red.

We calculate the probability of this event, given $R_1$. This is $P(R_2 \cap R_3 | R_1)$.

The probability of drawing a red marble on the second draw, given the first was red ($R_1$), is the probability of drawing a red from the remaining 4 red and 3 black marbles:

After drawing a second red marble (given $R_1$ and $R_2$), the bag contains 3 red and 3 black marbles (total 6).

The probability of drawing a red marble on the third draw, given the first two were red ($R_1$ and $R_2$), is the probability of drawing a red from these 3 red and 3 black marbles:

The probability that both the second and third marbles drawn are red, given the first was red, is the product of these conditional probabilities:

Substitute values from (ii) and (iv):

This probability, $P(R_2 \cap R_3 | R_1)$, is the probability that none of the three marbles are black, given the first is red. This is $P(\overline{E} | R_1)$.

$P(\overline{E} | R_1) = \frac{2}{7}$

The probability that at least one of the three marbles drawn is black, given the first is red, is the complement of the event that none of the remaining two are black (given the first was red).

$P(E | R_1) = 1 - P(\overline{E} | R_1)$

$P(E | R_1) = 1 - \frac{2}{7}$

$P(E | R_1) = \frac{7}{7} - \frac{2}{7} = \frac{5}{7}$

Answer:

The probability that at least one of the three marbles drawn be black, if the first marble is red, is $\frac{5}{7}$.

Given:

Two dice are thrown together.

Event E: 'a total of 4'

Event F: 'a total of 9 or more'

Event G: 'a total divisible by 5'

To Calculate and Decide:

Calculate P(E), P(F), and P(G).

Decide which pairs of events (E and F, E and G, F and G) are independent.

Solution:

When two fair dice are thrown, the total number of possible outcomes is $6 \times 6 = 36$. Each outcome is an ordered pair $(d_1, d_2)$ where $d_1$ is the result of the first die and $d_2$ is the result of the second die. Assuming the dice are fair, each outcome has a probability of $\frac{1}{36}$.

Event E: 'a total of 4'

The outcomes that result in a total of 4 are:

$E = \{(1,3), (2,2), (3,1)\}$

Number of outcomes in E is $|E| = 3$.

The probability of event E is:

$P(E) = \frac{|E|}{\text{Total outcomes}} = \frac{3}{36} = \frac{1}{12}$

Event F: 'a total of 9 or more'

The outcomes that result in a total of 9 or more (i.e., 9, 10, 11, or 12) are:

Sum 9: $(3,6), (4,5), (5,4), (6,3)$

Sum 10: $(4,6), (5,5), (6,4)$

Sum 11: $(5,6), (6,5)$

Sum 12: $(6,6)$

$F = \{(3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$

Number of outcomes in F is $|F| = 10$.

The probability of event F is:

$P(F) = \frac{|F|}{\text{Total outcomes}} = \frac{10}{36} = \frac{5}{18}$

Event G: 'a total divisible by 5'

The sums of two dice can range from $1+1=2$ to $6+6=12$. The totals divisible by 5 in this range are 5 and 10.

Sum 5: $(1,4), (2,3), (3,2), (4,1)$

Sum 10: $(4,6), (5,5), (6,4)$

$G = \{(1,4), (2,3), (3,2), (4,1), (4,6), (5,5), (6,4)\}$

Number of outcomes in G is $|G| = 7$.

The probability of event G is:

$P(G) = \frac{|G|}{\text{Total outcomes}} = \frac{7}{36}$

Checking for Independence:

Two events A and B are independent if and only if $P(A \cap B) = P(A) \times P(B)$.

Pair E and F:

Intersection $E \cap F$: Outcomes common to both E and F.

$E = \{(1,3), (2,2), (3,1)\}$

$F = \{(3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$

There are no common outcomes. $E \cap F = \emptyset$.

$P(E \cap F) = 0$

Calculate the product of probabilities:

$P(E) \times P(F) = \frac{1}{12} \times \frac{10}{36} = \frac{10}{432} = \frac{5}{216}$

Since $P(E \cap F) = 0$ and $P(E) \times P(F) = \frac{5}{216}$, we have $P(E \cap F) \neq P(E) \times P(F)$.

Therefore, events E and F are not independent.

Pair E and G:

Intersection $E \cap G$: Outcomes common to both E and G.

$E = \{(1,3), (2,2), (3,1)\}$

$G = \{(1,4), (2,3), (3,2), (4,1), (4,6), (5,5), (6,4)\}$

There are no common outcomes. $E \cap G = \emptyset$.

$P(E \cap G) = 0$

Calculate the product of probabilities:

$P(E) \times P(G) = \frac{1}{12} \times \frac{7}{36} = \frac{7}{432}$

Since $P(E \cap G) = 0$ and $P(E) \times P(G) = \frac{7}{432}$, we have $P(E \cap G) \neq P(E) \times P(G)$.

Therefore, events E and G are not independent.

Pair F and G:

Intersection $F \cap G$: Outcomes common to both F and G.

$F = \{(3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$

$G = \{(1,4), (2,3), (3,2), (4,1), (4,6), (5,5), (6,4)\}$

The common outcomes are $(4,6), (5,5), (6,4)$.

$F \cap G = \{(4,6), (5,5), (6,4)\}$

Number of outcomes in $F \cap G$ is $|F \cap G| = 3$.

$P(F \cap G) = \frac{|F \cap G|}{\text{Total outcomes}} = \frac{3}{36} = \frac{1}{12}$

Calculate the product of probabilities:

$P(F) \times P(G) = \frac{10}{36} \times \frac{7}{36} = \frac{70}{1296}$

Simplify the product: $\frac{\cancel{70}^{35}}{\cancel{1296}_{648}} = \frac{35}{648}$

We compare $P(F \cap G) = \frac{1}{12}$ with $P(F) \times P(G) = \frac{35}{648}$.

To compare, find a common denominator or cross-multiply. $12 \times 54 = 648$, so $\frac{1}{12} = \frac{1 \times 54}{12 \times 54} = \frac{54}{648}$.

Since $\frac{1}{12} = \frac{54}{648}$ and $P(F) \times P(G) = \frac{35}{648}$, we have $P(F \cap G) \neq P(F) \times P(G)$.

Therefore, events F and G are not independent.

Conclusion:

$P(E) = \frac{1}{12}$

$P(F) = \frac{5}{18}$

$P(G) = \frac{7}{36}$

Comparing the probabilities of intersections with the products of probabilities, we found that:

$P(E \cap F) \neq P(E) \times P(F)$

$P(E \cap G) \neq P(E) \times P(G)$

$P(F \cap G) \neq P(F) \times P(G)$

None of the pairs of events (E and F, E and G, F and G) are independent.

The experiment of tossing a coin three times is said to have a binomial distribution because it satisfies all the essential conditions required for a binomial experiment.

A random experiment follows a binomial distribution if it has the following properties:

1. Fixed Number of Trials: The experiment consists of a fixed number of trials, denoted by $n$.

In the case of tossing a coin three times, the number of trials is fixed at $n = 3$.

2. Independent Trials: Each trial is independent of the others. The outcome of one trial does not affect the outcome of any other trial.

When tossing a coin multiple times, the result of one toss does not influence the result of the subsequent tosses. Each toss is an independent event.

3. Only Two Possible Outcomes: Each trial has only two mutually exclusive outcomes, usually referred to as "success" and "failure".

For a coin toss, there are two possible outcomes: Heads or Tails. We can define one as "success" (e.g., getting a Head) and the other as "failure" (getting a Tail).

4. Constant Probability of Success: The probability of "success", denoted by $p$, is the same for each trial.

Assuming the coin is fair or has a consistent bias, the probability of getting a Head (or a Tail) remains constant for every toss.

5. Random Variable: The random variable, denoted by $X$, represents the number of "successes" in the $n$ trials.

In this experiment, we can define a random variable $X$ as the number of Heads (or the number of Tails) obtained in the three tosses. $X$ can take values $0, 1, 2, 3$.

Since the experiment of tossing a coin three times meets all these criteria (fixed trials, independent trials, two outcomes per trial, constant probability of success, counting successes), it is correctly described as an experiment with a binomial distribution. The number of successes (e.g., number of Heads) follows a binomial distribution with parameters $n=3$ (number of trials) and $p$ (probability of success on a single trial, e.g., $P(\text{Head})$).

Given:

$P(A) = \frac{1}{2}$

$P(B) = \frac{1}{3}$

$P(A \cap B) = \frac{1}{4}$

To Find:

(i) $P(A|B)$

(ii) $P(B|A)$

(iii) $P(A'|B)$

(iv) $P(A'|B')$

Solution:

We will use the formula for conditional probability: $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$, provided $P(Y) \neq 0$. We also use the property $P(E') = 1 - P(E)$.

(i) $P(A|B)$

Using the conditional probability formula:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

Substitute the given values:

$P(A|B) = \frac{\frac{1}{4}}{\frac{1}{3}} = \frac{1}{4} \times \frac{3}{1} = \frac{3}{4}$

(ii) $P(B|A)$

Using the conditional probability formula:

$P(B|A) = \frac{P(B \cap A)}{P(A)}$

Since $P(B \cap A) = P(A \cap B)$, we have:

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

Substitute the given values:

$P(B|A) = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{4} \times \frac{2}{1} = \frac{\cancel{2}^1}{\cancel{4}^2} = \frac{1}{2}$

(iii) $P(A'|B)$

Using the conditional probability formula:

$P(A'|B) = \frac{P(A' \cap B)}{P(B)}$

We need to find $P(A' \cap B)$. We know that $P(B) = P(A \cap B) + P(A' \cap B)$.

Therefore, $P(A' \cap B) = P(B) - P(A \cap B)$

Substitute the given values:

$P(A' \cap B) = \frac{1}{3} - \frac{1}{4}$

$P(A' \cap B) = \frac{4}{12} - \frac{3}{12} = \frac{4-3}{12} = \frac{1}{12}$

Now substitute this value back into the conditional probability formula:

$P(A'|B) = \frac{\frac{1}{12}}{\frac{1}{3}} = \frac{1}{12} \times \frac{3}{1} = \frac{\cancel{3}^1}{\cancel{12}^4} = \frac{1}{4}$

(iv) $P(A'|B')$

Using the conditional probability formula:

$P(A'|B') = \frac{P(A' \cap B')}{P(B')}$

We need to find $P(A' \cap B')$ and $P(B')$.

First, find $P(B')$:

$P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}$

Next, find $P(A' \cap B')$. Using De Morgan's law, $A' \cap B' = (A \cup B)'$.

So, $P(A' \cap B') = P((A \cup B)')$.

We know that $P((A \cup B)') = 1 - P(A \cup B)$.

First, find $P(A \cup B)$ using the formula:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the given values:

$P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4}$

Find a common denominator, which is 12:

$P(A \cup B) = \frac{6}{12} + \frac{4}{12} - \frac{3}{12} = \frac{6+4-3}{12} = \frac{7}{12}$

Now find $P(A' \cap B') = P((A \cup B)')$:

$P(A' \cap B') = 1 - P(A \cup B) = 1 - \frac{7}{12} = \frac{12}{12} - \frac{7}{12} = \frac{5}{12}$

Finally, substitute $P(A' \cap B')$ and $P(B')$ into the conditional probability formula:

$P(A'|B') = \frac{P(A' \cap B')}{P(B')} = \frac{\frac{5}{12}}{\frac{2}{3}} = \frac{5}{12} \times \frac{3}{2} = \frac{5 \times \cancel{3}^1}{\cancel{12}^4 \times 2} = \frac{5}{8}$

Answer:

(i) $P(A|B) = \frac{3}{4}$

(ii) $P(B|A) = \frac{1}{2}$

(iii) $P(A'|B) = \frac{1}{4}$

(iv) $P(A'|B') = \frac{5}{8}$

Given:

$P(A) = \frac{2}{5}$

$P(B) = \frac{1}{3}$

$P(C) = \frac{1}{2}$

$P(A \cap C) = \frac{1}{5}$

$P(B \cap C) = \frac{1}{4}$

To Find:

$P(C | B)$

$P(A' \cap C')$

Solution:

First, let's find $P(C | B)$. We use the formula for conditional probability:

$P(C | B) = \frac{P(C \cap B)}{P(B)}$

Substitute the given values:

$P(C | B) = \frac{\frac{1}{4}}{\frac{1}{3}}$

$P(C | B) = \frac{1}{4} \times \frac{3}{1}$

$P(C | B) = \frac{3}{4}$

Next, let's find $P(A' \cap C')$. We can use De Morgan's law which states that $A' \cap C' = (A \cup C)'$.

So, $P(A' \cap C') = P((A \cup C)')$.

The probability of the complement of an event is $P(E') = 1 - P(E)$. Thus,

$P((A \cup C)') = 1 - P(A \cup C)$

We need to find $P(A \cup C)$ using the formula for the probability of the union of two events:

$P(A \cup C) = P(A) + P(C) - P(A \cap C)$

Substitute the given values:

$P(A \cup C) = \frac{2}{5} + \frac{1}{2} - \frac{1}{5}$

Combine the fractions with the same denominator first:

$P(A \cup C) = \left(\frac{2}{5} - \frac{1}{5}\right) + \frac{1}{2}$

$P(A \cup C) = \frac{1}{5} + \frac{1}{2}$

To add these fractions, find a common denominator, which is 10:

$P(A \cup C) = \frac{1 \times 2}{5 \times 2} + \frac{1 \times 5}{2 \times 5} = \frac{2}{10} + \frac{5}{10}$

$P(A \cup C) = \frac{2+5}{10} = \frac{7}{10}$

Now substitute this value back into the expression for $P(A' \cap C')$:

$P(A' \cap C') = 1 - P(A \cup C)$

$P(A' \cap C') = 1 - \frac{7}{10}$

$P(A' \cap C') = \frac{10}{10} - \frac{7}{10} = \frac{10-7}{10} = \frac{3}{10}$

Answer:

$P(C | B) = \frac{3}{4}$

$P(A' \cap C') = \frac{3}{10}$

Given:

E$_1$ and E$_2$ are two independent events.

$P(E_1) = p_1$

$P(E_2) = p_2$

Properties of Independent Events:

If E$_1$ and E$_2$ are independent, then:

$P(E_1 \cap E_2) = P(E_1) \times P(E_2) = p_1 p_2$

The complements E$_1'$ and E$_2'$ are also independent of E$_1$ and E$_2$, and of each other.

$P(E_1') = 1 - P(E_1) = 1 - p_1$

$P(E_2') = 1 - P(E_2) = 1 - p_2$

$P(E_1' \cap E_2) = P(E_1') \times P(E_2) = (1 - p_1) p_2$

$P(E_1 \cap E_2') = P(E_1) \times P(E_2') = p_1 (1 - p_2)$

$P(E_1' \cap E_2') = P(E_1') \times P(E_2') = (1 - p_1) (1 - p_2)$

Descriptions of the Events:

(i) $p_1 p_2$

This probability is $P(E_1) \times P(E_2)$. Since E$_1$ and E$_2$ are independent, this is equal to $P(E_1 \cap E_2)$.

Description: The event where both E$_1$ and E$_2$ occur.

(ii) $(1 – p_1) p_2$

This probability is $(1 - P(E_1)) \times P(E_2) = P(E_1') \times P(E_2)$. Since E$_1'$ and E$_2$ are independent, this is equal to $P(E_1' \cap E_2)$.

Description: The event where E$_1$ does not occur and E$_2$ occurs (or only E$_2$ occurs).

(iii) $1 – (1 – p_1) (1 – p_2)$

This probability is $1 - (1 - p_1)(1 - p_2)$.

We know that $(1 - p_1)(1 - p_2) = P(E_1') P(E_2')$. Since E$_1'$ and E$_2'$ are independent, this is $P(E_1' \cap E_2')$.

Using De Morgan's law, $E_1' \cap E_2' = (E_1 \cup E_2)'$. So, $(1 - p_1)(1 - p_2) = P((E_1 \cup E_2)')$.

Therefore, $1 - (1 - p_1)(1 - p_2) = 1 - P((E_1 \cup E_2)') = P(E_1 \cup E_2)$.

Alternatively, expanding the expression: $1 - (1 - p_2 - p_1 + p_1 p_2) = 1 - 1 + p_2 + p_1 - p_1 p_2 = p_1 + p_2 - p_1 p_2$.

This is the formula for $P(E_1 \cup E_2)$ when E$_1$ and E$_2$ are independent.

Description: The event where at least one of the events E$_1$ or E$_2$ occurs.

(iv) $p_1 + p_2 – 2p_1p_2$

Let's rewrite the expression:

$p_1 + p_2 - 2p_1p_2 = p_1 + p_2 - p_1p_2 - p_1p_2$

Since $p_1p_2 = P(E_1 \cap E_2)$, and for independent events $P(E_1 \cup E_2) = p_1 + p_2 - p_1p_2$, we have:

$p_1 + p_2 - 2p_1p_2 = (p_1 + p_2 - p_1p_2) - p_1p_2 = P(E_1 \cup E_2) - P(E_1 \cap E_2)$.

The event $(E_1 \cup E_2) \setminus (E_1 \cap E_2)$ corresponds to the outcomes in the union but not in the intersection. This is the event that exactly one of the two events occurs.

Alternatively, $p_1 + p_2 - 2p_1p_2 = p_1(1 - p_2) + p_2(1 - p_1)$.

Since the events are independent, this is $P(E_1)P(E_2') + P(E_2)P(E_1') = P(E_1 \cap E_2') + P(E_1' \cap E_2)$.

The event $(E_1 \cap E_2')$ is that E$_1$ occurs and E$_2$ does not occur.

The event $(E_1' \cap E_2)$ is that E$_1$ does not occur and E$_2$ occurs.

These two events are mutually exclusive. Their union $(E_1 \cap E_2') \cup (E_1' \cap E_2)$ is the event that exactly one of E$_1$ or E$_2$ occurs.

Description: The event where exactly one of the events E$_1$ or E$_2$ occurs.

Given:

A discrete random variable X with the following probability distribution:

To Find:

(i) The value of k.

(ii) The mean of the distribution ($E(X)$).

Solution:

(i) Finding the value of k:

For a valid probability distribution, the sum of all probabilities must be equal to 1.

Substitute the given probabilities into equation (i):

$k + k^2 + 2k^2 + k = 1$

Combine like terms:

$3k^2 + 2k - 1 = 0$

This is a quadratic equation. We can solve it by factoring:

$(3k - 1)(k + 1) = 0$

This gives two possible values for k:

$3k - 1 = 0 \Rightarrow 3k = 1 \Rightarrow k = \frac{1}{3}$

or

$k + 1 = 0 \Rightarrow k = -1$

For a probability distribution, all probabilities must be non-negative ($P(X) \geq 0$). Let's check both values of k.

If $k = -1$:

$P(X=0.5) = -1$

$P(X=1) = (-1)^2 = 1$

$P(X=1.5) = 2(-1)^2 = 2$

$P(X=2) = -1$

Since $P(X=0.5)$ and $P(X=2)$ are negative, $k=-1$ is not a valid value for a probability distribution.

If $k = \frac{1}{3}$:

$P(X=0.5) = \frac{1}{3}$

$P(X=1) = (\frac{1}{3})^2 = \frac{1}{9}$

$P(X=1.5) = 2(\frac{1}{3})^2 = 2 \times \frac{1}{9} = \frac{2}{9}$

$P(X=2) = \frac{1}{3}$

All these probabilities are non-negative, and their sum is $\frac{1}{3} + \frac{1}{9} + \frac{2}{9} + \frac{1}{3} = \frac{3}{9} + \frac{1}{9} + \frac{2}{9} + \frac{3}{9} = \frac{3+1+2+3}{9} = \frac{9}{9} = 1$. This is a valid probability distribution.

Thus, the value of k is $\frac{1}{3}$.

(ii) Determining the mean of the distribution:

The mean (or expected value) of a discrete random variable X is given by the formula:

$E(X) = \sum X \cdot P(X)$

In this case, the sum is over the given values of X:

$E(X) = (0.5) \times P(X=0.5) + (1) \times P(X=1) + (1.5) \times P(X=1.5) + (2) \times P(X=2)$

Substitute the probabilities in terms of k and the value $k = \frac{1}{3}$:

$E(X) = (0.5)(k) + (1)(k^2) + (1.5)(2k^2) + (2)(k)$

$E(X) = (\frac{1}{2})(\frac{1}{3}) + (1)((\frac{1}{3})^2) + (\frac{3}{2})(2(\frac{1}{3})^2) + (2)(\frac{1}{3})$

$E(X) = \frac{1}{6} + 1 \times \frac{1}{9} + \frac{3}{2} \times 2 \times \frac{1}{9} + \frac{2}{3}$

$E(X) = \frac{1}{6} + \frac{1}{9} + \frac{6}{18} + \frac{2}{3}$

Combine the last two terms and find a common denominator for all terms (18):

$E(X) = \frac{1}{6} + \frac{1}{9} + \frac{3}{3} = \frac{1}{6} + \frac{1}{9} + 1$ (This approach is incorrect; stick to common denominator for all)

$E(X) = \frac{1}{6} + \frac{1}{9} + \frac{1}{3} + \frac{2}{3}$

The least common multiple of 6, 9, and 3 is 18.

$E(X) = \frac{1 \times 3}{6 \times 3} + \frac{1 \times 2}{9 \times 2} + \frac{1 \times 6}{3 \times 6} + \frac{2 \times 6}{3 \times 6}$

$E(X) = \frac{3}{18} + \frac{2}{18} + \frac{6}{18} + \frac{12}{18}$

$E(X) = \frac{3 + 2 + 6 + 12}{18} = \frac{23}{18}$

Answer:

(i) The value of $k = \frac{1}{3}$.

(ii) The mean of the distribution is $E(X) = \frac{23}{18}$.

Given:

A and B are two events associated with a random experiment.

To Prove:

(i) $P(A) = P(A ∩ B) + P(A ∩ \overline{B} )$

(ii) $P(A ∪ B) = P(A ∩ B) + P(A ∩ \overline{B} ) + P( \overline{A} ∩ B)$

Proof:

We use the properties of sets and probability axioms.

(i) Proof of $P(A) = P(A ∩ B) + P(A ∩ \overline{B} )$

Consider event A. The sample space $\Omega$ can be partitioned by event B and its complement $\overline{B}$. Any event A can be written as the union of its intersection with B and its intersection with $\overline{B}$.

In terms of sets, event A can be expressed as:

The event $(A \cap B)$ represents the outcomes where both A and B occur.

The event $(A \cap \overline{B})$ represents the outcomes where A occurs and B does not occur.

These two events, $(A \cap B)$ and $(A \cap \overline{B})$, are mutually exclusive because an outcome cannot be in both B and $\overline{B}$ simultaneously.

Since $(A \cap B)$ and $(A \cap \overline{B})$ are mutually exclusive, the probability of their union is the sum of their probabilities, according to the addition rule for mutually exclusive events:

From (1), the union of $(A \cap B)$ and $(A \cap \overline{B})$ is event A. Therefore, the probability of their union is $P(A)$.

Equating (2) and (3), we get:

$P(A) = P(A \cap B) + P(A \cap \overline{B} )$

This proves the first identity.

(ii) Proof of $P(A ∪ B) = P(A ∩ B) + P(A ∩ \overline{B} ) + P( \overline{A} ∩ B)$

Consider the union of events A and B, $A \cup B$. This union can be partitioned into three mutually exclusive events:

1. The event where both A and B occur: $A \cap B$

2. The event where A occurs but B does not occur: $A \cap \overline{B}$

3. The event where B occurs but A does not occur: $\overline{A} \cap B$

In terms of sets, the union of A and B can be expressed as the union of these three sets:

These three events $(A \cap B)$, $(A \cap \overline{B})$, and $(\overline{A} \cap B)$ are pairwise mutually exclusive. For example, an outcome in $(A \cap B)$ cannot be in $(A \cap \overline{B})$ (because it must be in B and not in B simultaneously, which is impossible), and similarly for other pairs.

Since these three events are mutually exclusive, the probability of their union is the sum of their probabilities:

From (4), the union of these three events is $A \cup B$. Therefore, the probability of their union is $P(A \cup B)$.

Equating (5) and (6), we get:

$P(A ∪ B) = P(A ∩ B) + P(A ∩ \overline{B} ) + P( \overline{A} ∩ B)$

This proves the second identity.

Given:

X is the number of tails in three tosses of a coin.

We assume the coin is fair, so the probability of getting a tail (T) on a single toss is $p = P(T) = \frac{1}{2}$, and the probability of getting a head (H) is $1-p = P(H) = \frac{1}{2}$.

The number of trials (tosses) is $n = 3$.

This experiment follows a binomial distribution with parameters $n=3$ and $p=\frac{1}{2}$. The random variable X, the number of tails, follows the distribution $B(3, \frac{1}{2})$.

To Determine:

The standard deviation of X, denoted by $\sigma_X$.

Solution:

The standard deviation of a binomial distribution $B(n, p)$ is given by the formula $\sigma = \sqrt{np(1-p)}$.

In this case, $n=3$, $p=\frac{1}{2}$, and $1-p = 1 - \frac{1}{2} = \frac{1}{2}$.

The variance of X, $\sigma_X^2$, is:

Substitute the values of n, p, and (1-p) into equation (i):

$\sigma_X^2 = 3 \times \frac{1}{2} \times \frac{1}{2}$

$\sigma_X^2 = 3 \times \frac{1}{4} = \frac{3}{4}$

The standard deviation is the square root of the variance:

$\sigma_X = \sqrt{\sigma_X^2} = \sqrt{\frac{3}{4}}$

$\sigma_X = \frac{\sqrt{3}}{\sqrt{4}} = \frac{\sqrt{3}}{2}$

Alternate Solution (using probability distribution):

We can list the possible outcomes and the number of tails (X):

TTT: X=3

TTH: X=2

THT: X=2

HTT: X=2

THH: X=1

HTH: X=1

HHT: X=1

HHH: X=0

Assuming a fair coin, each outcome has a probability of $(\frac{1}{2})^3 = \frac{1}{8}$.

The probability distribution of X is:

$P(X=0) = P(HHH) = \frac{1}{8}$

$P(X=1) = P(THH, HTH, HHT) = 3 \times \frac{1}{8} = \frac{3}{8}$

$P(X=2) = P(TTH, THT, HTT) = 3 \times \frac{1}{8} = \frac{3}{8}$

$P(X=3) = P(TTT) = \frac{1}{8}$

The mean $E(X)$ is:

$E(X) = \sum X \cdot P(X) = 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 2 \times \frac{3}{8} + 3 \times \frac{1}{8}$

$E(X) = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{3+6+3}{8} = \frac{12}{8} = \frac{3}{2} = 1.5$

The variance $Var(X)$ is $E(X^2) - (E(X))^2$.

First, find $E(X^2) = \sum X^2 \cdot P(X)$:

$E(X^2) = 0^2 \times \frac{1}{8} + 1^2 \times \frac{3}{8} + 2^2 \times \frac{3}{8} + 3^2 \times \frac{1}{8}$

$E(X^2) = 0 \times \frac{1}{8} + 1 \times \frac{3}{8} + 4 \times \frac{3}{8} + 9 \times \frac{1}{8}$

$E(X^2) = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{3+12+9}{8} = \frac{24}{8} = 3$

Now calculate the variance:

$Var(X) = E(X^2) - (E(X))^2 = 3 - (\frac{3}{2})^2$

$Var(X) = 3 - \frac{9}{4} = \frac{12}{4} - \frac{9}{4} = \frac{3}{4}$

The standard deviation is the square root of the variance:

$\sigma_X = \sqrt{Var(X)} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

Both methods yield the same result.

Answer:

The standard deviation of X is $\frac{\sqrt{3}}{2}$.

Given:

Stake paid per throw = $\textsf{₹}1$

Amount received if die shows 3 = $\textsf{₹}5$

Amount received if die shows 1 or 6 = $\textsf{₹}2$

Amount received otherwise (if die shows 2, 4, or 5) = $\textsf{₹}0$

Assuming the die is fair, the probability of each outcome (1, 2, 3, 4, 5, 6) is $\frac{1}{6}$.

To Find:

The player's expected profit per throw.

Solution:

Let X be the random variable representing the player's net profit per throw. The net profit is the amount received minus the stake paid ($\textsf{₹}1$).

We identify the possible values of X and their probabilities based on the outcomes of the die roll:

1. If the die shows 3:

Net profit = Amount received - Stake = $\textsf{₹}5 - \textsf{₹}1 = \textsf{₹}4$.

The probability of this outcome is $P(\text{Die shows 3}) = \frac{1}{6}$. So, $P(X=4) = \frac{1}{6}$.

2. If the die shows 1 or 6:

Net profit = Amount received - Stake = $\textsf{₹}2 - \textsf{₹}1 = \textsf{₹}1$.

The probability of this event is $P(\text{Die shows 1 or 6}) = P(1) + P(6) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$. So, $P(X=1) = \frac{1}{3}$.

3. If the die shows 2, 4, or 5:

Net profit = Amount received - Stake = $\textsf{₹}0 - \textsf{₹}1 = \textsf{₹}-1$.

The probability of this event is $P(\text{Die shows 2, 4, or 5}) = P(2) + P(4) + P(5) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$. So, $P(X=-1) = \frac{1}{2}$.

The probability distribution of the random variable X is:

The expected profit is the expected value of the random variable X, $E(X)$.

The formula for the expected value of a discrete random variable is $E(X) = \sum x \cdot P(X=x)$.

$E(X) = (4 \times P(X=4)) + (1 \times P(X=1)) + (-1 \times P(X=-1))$

$E(X) = (4 \times \frac{1}{6}) + (1 \times \frac{1}{3}) + (-1 \times \frac{1}{2})$

$E(X) = \frac{4}{6} + \frac{1}{3} - \frac{1}{2}$

Simplify the fractions:

$E(X) = \frac{2}{3} + \frac{1}{3} - \frac{1}{2}$

Combine the first two fractions:

$E(X) = \frac{2+1}{3} - \frac{1}{2} = \frac{3}{3} - \frac{1}{2} = 1 - \frac{1}{2}$

$E(X) = \frac{1}{2}$

The expected profit per throw is $\textsf{₹}\frac{1}{2}$ or $\textsf{₹}0.50$.

Answer:

The player's expected profit per throw is $\textsf{₹}\frac{1}{2}$.

Given:

Three fair dice are thrown at the same time.

To Find:

The probability of getting three two’s, given that the sum of the numbers on the dice is six.

Solution:

When three fair dice are thrown, the total number of possible outcomes is $6 \times 6 \times 6 = 216$. Each outcome is an ordered triplet $(d_1, d_2, d_3)$, where $d_i \in \{1, 2, 3, 4, 5, 6\}$. Each specific outcome has a probability of $\frac{1}{216}$.

Let A be the event of getting three two’s. The outcome for this event is (2, 2, 2).

$A = \{(2,2,2)\}$

The number of outcomes in A is $|A| = 1$.

$P(A) = \frac{1}{216}$

Let B be the event that the sum of the numbers on the dice is six.

We need to find all possible ordered triplets $(d_1, d_2, d_3)$ such that $d_1 + d_2 + d_3 = 6$, where $d_i \in \{1, 2, 3, 4, 5, 6\}$.

The possible combinations of three numbers from $\{1, 2, 3, 4, 5, 6\}$ that sum to 6 are:

- (1, 1, 4)

- (1, 2, 3)

- (2, 2, 2)

Now, we list the permutations for each combination:

- For (1, 1, 4): The permutations are (1,1,4), (1,4,1), (4,1,1). There are $\frac{3!}{2!1!} = 3$ unique permutations.

- For (1, 2, 3): The permutations are (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1). There are $3! = 6$ unique permutations.

- For (2, 2, 2): The only permutation is (2,2,2). There is $\frac{3!}{3!} = 1$ unique permutation.

The outcomes in event B are:

$B = \{(1,1,4), (1,4,1), (4,1,1), (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1), (2,2,2)\}$

The number of outcomes in B is $|B| = 3 + 6 + 1 = 10$.

The probability of event B is:

$P(B) = \frac{|B|}{\text{Total outcomes}} = \frac{10}{216}$

We need to find the probability of event A occurring given that event B has occurred. This is the conditional probability $P(A|B)$.

The formula for conditional probability is:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

First, find the intersection of events A and B, $A \cap B$. These are the outcomes common to both A and B.

$A = \{(2,2,2)\}$

$B = \{(1,1,4), (1,4,1), (4,1,1), (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1), (2,2,2)\}$

The only common outcome is (2,2,2).

$A \cap B = \{(2,2,2)\}$

The number of outcomes in $A \cap B$ is $|A \cap B| = 1$.

The probability of the intersection is:

$P(A \cap B) = \frac{|A \cap B|}{\text{Total outcomes}} = \frac{1}{216}$

Now, substitute the probabilities $P(A \cap B)$ and $P(B)$ into the conditional probability formula:

$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{216}}{\frac{10}{216}}$

$P(A|B) = \frac{1}{216} \times \frac{216}{10} = \frac{1}{10}$

Answer:

The probability of getting three two’s, given that the sum of the numbers on the dice was six, is $\frac{1}{10}$.

Given:

Total number of tickets sold = 10,000

Cost of each ticket = $\textsf{₹}1$

First prize amount = $\textsf{₹}3000$ (1 prize)

Second prize amount = $\textsf{₹}2000$ (1 prize)

Third prize amount = $\textsf{₹}500$ (3 prizes)

You buy one ticket.

To Find:

Your expectation (expected net gain or profit) when buying one ticket.

Solution:

Let X be the random variable representing your net gain (or profit) from buying one ticket. The net gain is the amount you receive minus the cost of the ticket.

We identify the possible outcomes and the corresponding values of X and their probabilities:

1. Winning the first prize:

Amount received = $\textsf{₹}3000$.

Net gain X = Amount received - Cost = $\textsf{₹}3000 - \textsf{₹}1 = \textsf{₹}2999$.

There is 1 first prize out of 10,000 tickets.

Probability $P(X=2999) = \frac{1}{10000}$.

2. Winning the second prize:

Amount received = $\textsf{₹}2000$.

Net gain X = Amount received - Cost = $\textsf{₹}2000 - \textsf{₹}1 = \textsf{₹}1999$.

There is 1 second prize out of 10,000 tickets.

Probability $P(X=1999) = \frac{1}{10000}$.

3. Winning a third prize:

Amount received = $\textsf{₹}500$.

Net gain X = Amount received - Cost = $\textsf{₹}500 - \textsf{₹}1 = \textsf{₹}499$.

There are 3 third prizes out of 10,000 tickets.

Probability $P(X=499) = \frac{3}{10000}$.

4. Not winning any prize:

Amount received = $\textsf{₹}0$.

Net gain X = Amount received - Cost = $\textsf{₹}0 - \textsf{₹}1 = \textsf{₹}-1$.

The total number of prize-winning tickets is $1 + 1 + 3 = 5$.

The number of tickets that do not win a prize is $10000 - 5 = 9995$.

Probability $P(X=-1) = \frac{9995}{10000}$.

The probability distribution of the random variable X is:

The expectation (expected net gain) is the expected value of X, $E(X)$.

$E(X) = \sum x \cdot P(X=x)$

$E(X) = (2999 \times \frac{1}{10000}) + (1999 \times \frac{1}{10000}) + (499 \times \frac{3}{10000}) + (-1 \times \frac{9995}{10000})$

$E(X) = \frac{2999}{10000} + \frac{1999}{10000} + \frac{499 \times 3}{10000} + \frac{-9995}{10000}$

$E(X) = \frac{2999 + 1999 + (499 \times 3) - 9995}{10000}$

Calculate $499 \times 3$:

$499 \times 3 = 1497$

$E(X) = \frac{2999 + 1999 + 1497 - 9995}{10000}$

$E(X) = \frac{(2999 + 1999 + 1497) - 9995}{10000}$

$2999 + 1999 + 1497 = 6495$

$E(X) = \frac{6495 - 9995}{10000}$

$6495 - 9995 = -3500$

$E(X) = \frac{-3500}{10000} = -\frac{35}{100} = -0.35$

The expected net gain is $\textsf{₹}-0.35$. This means on average, the player is expected to lose $\textsf{₹}0.35$ per ticket over a long series of purchases.

Answer:

Your expectation (expected profit) is $\textsf{₹}-0.35$ (or a loss of $\textsf{₹}0.35$).

Given:

Bag 1: 4 white balls (W$_1$), 5 black balls (B$_1$). Total = 9 balls.

Bag 2: 9 white balls (W$_2$), 7 black balls (B$_2$). Total = 16 balls.

A ball is transferred from Bag 1 to Bag 2.

Then, a ball is drawn from Bag 2.

To Find:

The probability that the ball drawn from the second bag is white.

Solution:

Let T$_W$ be the event that a white ball is transferred from Bag 1 to Bag 2.

Let T$_B$ be the event that a black ball is transferred from Bag 1 to Bag 2.

Let D$_W$ be the event that a white ball is drawn from Bag 2 after the transfer.

Calculate the probabilities of the transfer events:

The probability of transferring a white ball from Bag 1 is $P(T_W) = \frac{\text{Number of white balls in Bag 1}}{\text{Total balls in Bag 1}} = \frac{4}{9}$.

The probability of transferring a black ball from Bag 1 is $P(T_B) = \frac{\text{Number of black balls in Bag 1}}{\text{Total balls in Bag 1}} = \frac{5}{9}$.

Now, consider the composition of Bag 2 after the transfer for each case:

Case 1: A white ball is transferred from Bag 1 to Bag 2 (Event T$_W$ occurs).

Bag 2 now contains: $(9+1)$ white balls and 7 black balls = 10 white, 7 black.

Total balls in Bag 2 = $10 + 7 = 17$.

The probability of drawing a white ball from Bag 2, given that a white ball was transferred, is $P(D_W | T_W) = \frac{\text{Number of white balls in Bag 2}}{\text{Total balls in Bag 2}} = \frac{10}{17}$.

Case 2: A black ball is transferred from Bag 1 to Bag 2 (Event T$_B$ occurs).

Bag 2 now contains: 9 white balls and $(7+1)$ black balls = 9 white, 8 black.

Total balls in Bag 2 = $9 + 8 = 17$.

The probability of drawing a white ball from Bag 2, given that a black ball was transferred, is $P(D_W | T_B) = \frac{\text{Number of white balls in Bag 2}}{\text{Total balls in Bag 2}} = \frac{9}{17}$.

To find the overall probability that the ball drawn from the second bag is white, $P(D_W)$, we use the Law of Total Probability. The events T$_W$ and T$_B$ form a partition of the sample space of the transfer.

$P(D_W) = P(D_W \cap T_W) + P(D_W \cap T_B)$

Using the formula for the probability of intersection of two events: $P(A \cap B) = P(A|B) \times P(B)$, we get:

$P(D_W) = P(D_W | T_W) \times P(T_W) + P(D_W | T_B) \times P(T_B)$

Substitute the probabilities we calculated:

$P(D_W) = \left(\frac{10}{17} \times \frac{4}{9}\right) + \left(\frac{9}{17} \times \frac{5}{9}\right)$

$P(D_W) = \frac{10 \times 4}{17 \times 9} + \frac{9 \times 5}{17 \times 9}$

$P(D_W) = \frac{40}{153} + \frac{45}{153}$

$P(D_W) = \frac{40 + 45}{153} = \frac{85}{153}$

We can check if the fraction can be simplified. The sum of digits of 85 is 13 (not divisible by 3 or 9). It ends in 5, so potentially divisible by 5. $85 = 5 \times 17$. The sum of digits of 153 is $1+5+3=9$ (divisible by 3 and 9). $153 = 9 \times 17 = 3^2 \times 17$.

$P(D_W) = \frac{5 \times 17}{9 \times 17} = \frac{5}{9}$

Answer:

The probability that the ball drawn from the second bag is white is $\frac{5}{9}$.

Given:

Bag I contains: 3 black balls, 2 white balls. Total = $3 + 2 = 5$ balls.

Bag II contains: 2 black balls, 4 white balls. Total = $2 + 4 = 6$ balls.

A bag is selected at random, and then a ball is selected at random from the selected bag.

To Determine:

The probability of selecting a black ball.

Solution:

Let B$_1$ be the event that Bag I is selected.

Let B$_2$ be the event that Bag II is selected.

Since a bag is selected at random, the probability of selecting either bag is equal:

Let D$_B$ be the event that a black ball is drawn.

We need to find the probability of drawing a black ball, $P(D_B)$. We can use the Law of Total Probability, considering the two cases: drawing a black ball given Bag I was selected, or drawing a black ball given Bag II was selected.

If Bag I was selected, the probability of drawing a black ball is the number of black balls in Bag I divided by the total number of balls in Bag I:

If Bag II was selected, the probability of drawing a black ball is the number of black balls in Bag II divided by the total number of balls in Bag II:

The Law of Total Probability states:

Substitute the probabilities from equations (i), (ii), (iii), and (iv) into equation (v):

$P(D_B) = \left(\frac{3}{5} \times \frac{1}{2}\right) + \left(\frac{1}{3} \times \frac{1}{2}\right)$

$P(D_B) = \frac{3}{10} + \frac{1}{6}$

To add these fractions, we find a common denominator. The least common multiple of 10 and 6 is 30.

$P(D_B) = \frac{3 \times 3}{10 \times 3} + \frac{1 \times 5}{6 \times 5}$

$P(D_B) = \frac{9}{30} + \frac{5}{30}$

$P(D_B) = \frac{9+5}{30} = \frac{14}{30}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2:

$P(D_B) = \frac{\cancel{14}^{7}}{\cancel{30}_{15}} = \frac{7}{15}$

Answer:

The probability of selecting a black ball is $\frac{7}{15}$.

Given:

A box contains: 5 blue balls (B), 4 red balls (R). Total = $5 + 4 = 9$ balls.

One ball is drawn at random and not replaced.

The colour of the first ball is not noted.

Then, a second ball is drawn at random.

To Find:

The probability that the second ball drawn is blue.

Solution:

Let F$_B$ be the event that the first ball drawn is blue.

Let F$_R$ be the event that the first ball drawn is red.

Let S$_B$ be the event that the second ball drawn is blue.

We need to find $P(S_B)$. We can use the Law of Total Probability by considering the colour of the first ball drawn (even though it's not noted). The first ball drawn must be either blue or red.

The probability of drawing a blue ball first is $P(F_B) = \frac{\text{Number of blue balls initially}}{\text{Total number of balls initially}} = \frac{5}{9}$.

The probability of drawing a red ball first is $P(F_R) = \frac{\text{Number of red balls initially}}{\text{Total number of balls initially}} = \frac{4}{9}$.

Note that $P(F_B) + P(F_R) = \frac{5}{9} + \frac{4}{9} = \frac{9}{9} = 1$.

Now, consider the state of the box after the first draw and before the second draw, based on the colour of the first ball (which is unknown to us, but determines the probabilities for the second draw):

Case 1: The first ball drawn was blue (Event F$_B$ occurred).

The box now contains: $(5-1) = 4$ blue balls and 4 red balls. Total = $4 + 4 = 8$ balls.

The probability of drawing a blue ball on the second draw, given the first was blue, is $P(S_B | F_B) = \frac{\text{Number of blue balls remaining}}{\text{Total balls remaining}} = \frac{4}{8} = \frac{1}{2}$.

Case 2: The first ball drawn was red (Event F$_R$ occurred).

The box now contains: 5 blue balls and $(4-1) = 3$ red balls. Total = $5 + 3 = 8$ balls.

The probability of drawing a blue ball on the second draw, given the first was red, is $P(S_B | F_R) = \frac{\text{Number of blue balls remaining}}{\text{Total balls remaining}} = \frac{5}{8}$.

Using the Law of Total Probability, the probability of the second ball being blue is:

$P(S_B) = P(S_B \cap F_B) + P(S_B \cap F_R)$

Using the conditional probability formula $P(A \cap B) = P(A|B) P(B)$, we get:

$P(S_B) = P(S_B | F_B) P(F_B) + P(S_B | F_R) P(F_R)$

Substitute the probabilities we calculated:

$P(S_B) = \left(\frac{1}{2} \times \frac{5}{9}\right) + \left(\frac{5}{8} \times \frac{4}{9}\right)$

$P(S_B) = \frac{5}{18} + \frac{20}{72}$

Simplify the second fraction: $\frac{\cancel{20}^5}{\cancel{72}_{18}} = \frac{5}{18}$.

$P(S_B) = \frac{5}{18} + \frac{5}{18}$

$P(S_B) = \frac{5 + 5}{18} = \frac{10}{18}$

Simplify the result:

$P(S_B) = \frac{\cancel{10}^5}{\cancel{18}_9} = \frac{5}{9}$

Answer:

The probability that the second ball drawn is blue is $\frac{5}{9}$.

Given:

A standard deck of 52 playing cards.

Number of kings in a deck = 4.

Four cards are drawn successively without replacement.

To Find:

The probability that all four cards drawn are kings.

Solution:

Let K$_1$ be the event that the first card drawn is a king.

Let K$_2$ be the event that the second card drawn is a king.

Let K$_3$ be the event that the third card drawn is a king.

Let K$_4$ be the event that the fourth card drawn is a king.

We want to find the probability of the event that all four cards are kings, which is the intersection of these events: $P(K_1 \cap K_2 \cap K_3 \cap K_4)$.

Using the multiplication rule of probability for dependent events (since drawing is without replacement):

$P(K_1 \cap K_2 \cap K_3 \cap K_4) = P(K_1) \times P(K_2 | K_1) \times P(K_3 | K_1 \cap K_2) \times P(K_4 | K_1 \cap K_2 \cap K_3)$

Step 1: Probability of the first card being a king ($P(K_1)$)

Initially, there are 52 cards in the deck, and 4 of them are kings.

$P(K_1) = \frac{\text{Number of kings}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13}$

Step 2: Probability of the second card being a king, given the first was a king ($P(K_2 | K_1)$)

After drawing one king without replacement, there are 51 cards left in the deck, and 3 of them are kings.

$P(K_2 | K_1) = \frac{\text{Number of kings remaining}}{\text{Total number of cards remaining}} = \frac{3}{51} = \frac{1}{17}$

Step 3: Probability of the third card being a king, given the first two were kings ($P(K_3 | K_1 \cap K_2)$)

After drawing two kings without replacement, there are 50 cards left in the deck, and 2 of them are kings.

$P(K_3 | K_1 \cap K_2) = \frac{\text{Number of kings remaining}}{\text{Total number of cards remaining}} = \frac{2}{50} = \frac{1}{25}$

Step 4: Probability of the fourth card being a king, given the first three were kings ($P(K_4 | K_1 \cap K_2 \cap K_3)$)

After drawing three kings without replacement, there are 49 cards left in the deck, and 1 of them is a king.

$P(K_4 | K_1 \cap K_2 \cap K_3) = \frac{\text{Number of kings remaining}}{\text{Total number of cards remaining}} = \frac{1}{49}$

Now, multiply these probabilities together:

$P(\text{All four cards are kings}) = P(K_1) \times P(K_2 | K_1) \times P(K_3 | K_1 \cap K_2) \times P(K_4 | K_1 \cap K_2 \cap K_3)$

$P(\text{All four cards are kings}) = \frac{4}{52} \times \frac{3}{51} \times \frac{2}{50} \times \frac{1}{49}$

$P(\text{All four cards are kings}) = \frac{\cancel{4}^1}{\cancel{52}_{13}} \times \frac{\cancel{3}^1}{\cancel{51}_{17}} \times \frac{\cancel{2}^1}{\cancel{50}_{25}} \times \frac{1}{49}$

$P(\text{All four cards are kings}) = \frac{1}{13} \times \frac{1}{17} \times \frac{1}{25} \times \frac{1}{49}$

$P(\text{All four cards are kings}) = \frac{1}{13 \times 17 \times 25 \times 49}$

Calculate the denominator:

$13 \times 17 = 221$

$25 \times 49 = 1225$

$221 \times 1225 = 270725$

Alternatively, using combinations:

The total number of ways to draw 4 cards from 52 is $\binom{52}{4}$.

$\binom{52}{4} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = \frac{52}{4 \times 1} \times \frac{51}{3 \times 1} \times \frac{50}{2 \times 1} \times \frac{49}{1} = 13 \times 17 \times 25 \times 49 = 270725$

The number of ways to draw 4 kings from the 4 kings available is $\binom{4}{4}$.

$\binom{4}{4} = 1$

The probability of drawing 4 kings is the number of ways to draw 4 kings divided by the total number of ways to draw 4 cards.

$P(\text{All four cards are kings}) = \frac{\binom{4}{4}}{\binom{52}{4}} = \frac{1}{270725}$

Answer:

The probability that all the four cards drawn are kings is $\frac{1}{270725}$.

Given:

A die is thrown 5 times.

We assume it is a fair die.

Number of trials, $n = 5$.

Success is defined as getting an odd number on a single throw. The odd numbers on a die are {1, 3, 5}.

Probability of success on a single trial, $p = P(\text{odd}) = P(1) + P(3) + P(5) = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.

Probability of failure on a single trial, $1-p = P(\text{even}) = 1 - \frac{1}{2} = \frac{1}{2}$.

We want to find the probability of getting exactly 3 successes.

To Find:

The probability that an odd number will come up exactly three times in 5 throws.

Solution:

This experiment is a sequence of independent Bernoulli trials with a fixed number of trials, two possible outcomes (odd/even), and a constant probability of success. Therefore, the number of odd numbers in 5 throws follows a binomial distribution.

Let X be the random variable representing the number of odd numbers in 5 throws.

X follows a binomial distribution with parameters $n=5$ and $p=\frac{1}{2}$.

The probability of getting exactly $k$ successes in $n$ trials in a binomial distribution is given by the formula:

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$

We want to find the probability of getting exactly 3 odd numbers, so we set $k=3$.

$P(X=3) = \binom{5}{3} (\frac{1}{2})^3 (1-\frac{1}{2})^{5-3}$

$P(X=3) = \binom{5}{3} (\frac{1}{2})^3 (\frac{1}{2})^2$

$P(X=3) = \binom{5}{3} (\frac{1}{2})^{3+2}$

$P(X=3) = \binom{5}{3} (\frac{1}{2})^5$

First, calculate the binomial coefficient $\binom{5}{3}$:

$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(2 \times 1)} = \frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10$

Next, calculate $(\frac{1}{2})^5$:

$(\frac{1}{2})^5 = \frac{1^5}{2^5} = \frac{1}{32}$

Now, substitute these values back into the probability formula:

$P(X=3) = 10 \times \frac{1}{32}$

$P(X=3) = \frac{10}{32}$

Simplify the fraction:

$P(X=3) = \frac{\cancel{10}^{5}}{\cancel{32}_{16}} = \frac{5}{16}$

Answer:

The probability that an odd number will come up exactly three times is $\frac{5}{16}$.

Given:

Ten coins are tossed.

We assume the coins are fair.

Number of trials (tosses), $n = 10$.

Success is defined as getting a head (H) on a single toss.

Probability of success on a single trial, $p = P(H) = \frac{1}{2}$.

Probability of failure on a single trial, $1-p = P(T) = 1 - \frac{1}{2} = \frac{1}{2}$.

Let X be the random variable representing the number of heads in 10 tosses.

X follows a binomial distribution with parameters $n=10$ and $p=\frac{1}{2}$.

The probability of getting exactly $k$ successes in $n$ trials is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.

To Find:

The probability of getting at least 8 heads, which means $P(X \geq 8)$.

Solution:

The event "at least 8 heads" corresponds to getting exactly 8 heads, exactly 9 heads, or exactly 10 heads.

$P(X \geq 8) = P(X=8) + P(X=9) + P(X=10)$

We calculate each term using the binomial probability formula with $n=10$ and $p = 1-p = \frac{1}{2}$. Note that $(\frac{1}{2})^k (\frac{1}{2})^{n-k} = (\frac{1}{2})^n = (\frac{1}{2})^{10} = \frac{1}{1024}$ for any $k$.

$P(X=k) = \binom{10}{k} (\frac{1}{2})^{10}$

Calculate $P(X=8)$:

$P(X=8) = \binom{10}{8} (\frac{1}{2})^{10}$

$\binom{10}{8} = \binom{10}{10-8} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = \frac{90}{2} = 45$

$P(X=8) = 45 \times \frac{1}{1024} = \frac{45}{1024}$

Calculate $P(X=9)$:

$P(X=9) = \binom{10}{9} (\frac{1}{2})^{10}$

$\binom{10}{9} = \binom{10}{10-9} = \binom{10}{1} = 10$

$P(X=9) = 10 \times \frac{1}{1024} = \frac{10}{1024}$

Calculate $P(X=10)$:

$P(X=10) = \binom{10}{10} (\frac{1}{2})^{10}$

$\binom{10}{10} = 1$

$P(X=10) = 1 \times \frac{1}{1024} = \frac{1}{1024}$

Now, sum these probabilities:

$P(X \geq 8) = P(X=8) + P(X=9) + P(X=10)$

$P(X \geq 8) = \frac{45}{1024} + \frac{10}{1024} + \frac{1}{1024}$

$P(X \geq 8) = \frac{45 + 10 + 1}{1024} = \frac{56}{1024}$

We can simplify the fraction by dividing the numerator and denominator by their greatest common divisor. Both are divisible by 8.

$56 \div 8 = 7$

$1024 \div 8 = 128$

$P(X \geq 8) = \frac{7}{128}$

Answer:

The probability of getting at least 8 heads is $\frac{7}{128}$.

Given:

Probability of hitting the target on a single shot, $p = 0.25 = \frac{1}{4}$.

Number of shots (trials), $n = 7$.

Success is defined as hitting the target on a single shot.

Probability of failure on a single shot, $1-p = 1 - 0.25 = 0.75 = \frac{3}{4}$.

Let X be the random variable representing the number of times the man hits the target in 7 shots.

X follows a binomial distribution with parameters $n=7$ and $p=\frac{1}{4}$.

The probability of getting exactly $k$ successes in $n$ trials is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.

To Find:

The probability of hitting at least twice, which means $P(X \geq 2)$.

Solution:

The event "at least twice" corresponds to hitting the target 2, 3, 4, 5, 6, or 7 times. Calculating each of these probabilities and summing them would be lengthy.

It is easier to calculate the probability of the complementary event, which is hitting the target less than twice (i.e., 0 or 1 time), and subtract this from 1.

$P(X \geq 2) = 1 - P(X < 2)$

$P(X < 2) = P(X=0) + P(X=1)$

We calculate $P(X=0)$ and $P(X=1)$ using the binomial probability formula with $n=7$, $p=\frac{1}{4}$, and $1-p=\frac{3}{4}$.

Calculate $P(X=0)$:

$P(X=0) = \binom{7}{0} (\frac{1}{4})^0 (\frac{3}{4})^{7-0}$

$\binom{7}{0} = 1$

$(\frac{1}{4})^0 = 1$

$(\frac{3}{4})^7 = \frac{3^7}{4^7} = \frac{2187}{16384}$

$P(X=0) = 1 \times 1 \times \frac{2187}{16384} = \frac{2187}{16384}$

Calculate $P(X=1)$:

$P(X=1) = \binom{7}{1} (\frac{1}{4})^1 (\frac{3}{4})^{7-1}$

$\binom{7}{1} = 7$

$(\frac{1}{4})^1 = \frac{1}{4}$

$(\frac{3}{4})^6 = \frac{3^6}{4^6} = \frac{729}{4096}$

$P(X=1) = 7 \times \frac{1}{4} \times \frac{729}{4096}$

$P(X=1) = \frac{7 \times 729}{4 \times 4096} = \frac{5103}{16384}$

Now, calculate $P(X < 2) = P(X=0) + P(X=1)$:

$P(X < 2) = \frac{2187}{16384} + \frac{5103}{16384}$

$P(X < 2) = \frac{2187 + 5103}{16384} = \frac{7290}{16384}$

Simplify the fraction by dividing numerator and denominator by 2:

$P(X < 2) = \frac{\cancel{7290}^{3645}}{\cancel{16384}_{8192}} = \frac{3645}{8192}$

Finally, calculate $P(X \geq 2)$:

$P(X \geq 2) = 1 - P(X < 2) = 1 - \frac{7290}{16384}$

$P(X \geq 2) = \frac{16384}{16384} - \frac{7290}{16384} = \frac{16384 - 7290}{16384}$

$16384 - 7290 = 9094$

$P(X \geq 2) = \frac{9094}{16384}$

Simplify the fraction by dividing numerator and denominator by 2:

$P(X \geq 2) = \frac{\cancel{9094}^{4547}}{\cancel{16384}_{8192}} = \frac{4547}{8192}$

Answer:

The probability of the man hitting the target at least twice is $\frac{9094}{16384}$ or $\frac{4547}{8192}$.

Given:

Total number of watches in the lot = 100.

Number of defective watches = 10.

Number of non-defective (good) watches = $100 - 10 = 90$.

Number of watches selected = 8.

The watches are selected one by one with replacement.

To Find:

The probability that there will be at least one defective watch among the 8 selected watches.

Solution:

Since the selection is done with replacement, each selection is an independent trial.

This scenario can be modeled using a binomial distribution.

Define a "success" on a single trial (selecting one watch) as selecting a defective watch.

The probability of success on a single trial is $p = P(\text{defective}) = \frac{\text{Number of defective watches}}{\text{Total number of watches}} = \frac{10}{100} = 0.1$.

The probability of failure on a single trial (selecting a non-defective watch) is $1-p = P(\text{non-defective}) = \frac{90}{100} = 0.9$.

The number of trials is $n = 8$ (since 8 watches are selected).

Let X be the random variable representing the number of defective watches among the 8 selected watches.

X follows a binomial distribution with parameters $n=8$ and $p=0.1$.

The probability of getting exactly $k$ successes in $n$ trials is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.

We want to find the probability of getting at least one defective watch, which means $P(X \geq 1)$.

The event "at least one defective watch" is the complement of the event "no defective watches".

$P(X \geq 1) = 1 - P(X = 0)$

We calculate the probability of getting exactly 0 defective watches, $P(X=0)$, using the binomial probability formula with $k=0$, $n=8$, $p=0.1$, and $1-p=0.9$.

$P(X=0) = \binom{8}{0} (0.1)^0 (0.9)^{8-0}$

$\binom{8}{0} = 1$

$(0.1)^0 = 1$

$(0.9)^8$

$P(X=0) = 1 \times 1 \times (0.9)^8 = (0.9)^8$

Now, calculate $(0.9)^8$: $(0.9)^8 = 0.43046721$

So, $P(X=0) \approx 0.4305$

Now calculate $P(X \geq 1)$:

$P(X \geq 1) = 1 - P(X=0) = 1 - (0.9)^8$

$P(X \geq 1) = 1 - 0.43046721$

$P(X \geq 1) = 0.56953279$

Answer:

The probability that there will be at least one defective watch is $1 - (0.9)^8$, which is approximately $0.5695$.

Given:

The probability distribution of a discrete random variable X is:

Check if the sum of probabilities is 1: $0.10 + 0.25 + 0.30 + 0.20 + 0.15 = 1.00$. The distribution is valid.

To Calculate:

(i) $V \left( \frac{X}{2} \right)$

(ii) Variance of X, $Var(X)$ or $\sigma_X^2$

Solution:

We know that for a random variable X and a constant 'a', $Var(aX) = a^2 Var(X)$.

In part (i), we need to calculate $V(\frac{X}{2})$. Here, $a = \frac{1}{2}$.

$V(\frac{X}{2}) = V(\frac{1}{2} X) = (\frac{1}{2})^2 Var(X) = \frac{1}{4} Var(X)$.

To find $V(\frac{X}{2})$, we first need to calculate the variance of X, $Var(X)$. This is part (ii).

(ii) Calculating the Variance of X:

The variance of a discrete random variable X is given by the formula:

$Var(X) = E(X^2) - (E(X))^2$

First, we need to find the expected value (mean) of X, $E(X)$.

$E(X) = \sum X \cdot P(X)$

$E(X) = (0 \times 0.1) + (1 \times 0.25) + (2 \times 0.3) + (3 \times 0.2) + (4 \times 0.15)$

$E(X) = 0 + 0.25 + 0.60 + 0.60 + 0.60$

$E(X) = 0.25 + 0.60 + 0.60 + 0.60 = 2.05$

Next, we need to find the expected value of $X^2$, $E(X^2)$.

$E(X^2) = \sum X^2 \cdot P(X)$

$E(X^2) = (0^2 \times 0.1) + (1^2 \times 0.25) + (2^2 \times 0.3) + (3^2 \times 0.2) + (4^2 \times 0.15)$

$E(X^2) = (0 \times 0.1) + (1 \times 0.25) + (4 \times 0.3) + (9 \times 0.2) + (16 \times 0.15)$

$E(X^2) = 0 + 0.25 + 1.20 + 1.80 + 2.40$

$E(X^2) = 0.25 + 1.20 + 1.80 + 2.40 = 5.65$

Now, calculate the variance of X:

$Var(X) = E(X^2) - (E(X))^2$

$Var(X) = 5.65 - (2.05)^2$

Calculate $(2.05)^2$: $2.05 \times 2.05$

$2.05 \times 2.05 = 4.2025$

$Var(X) = 5.65 - 4.2025$

$Var(X) = 1.4475$

(i) Calculating $V(\frac{X}{2})$:

Now that we have $Var(X)$, we can calculate $V(\frac{X}{2})$:

$V(\frac{X}{2}) = \frac{1}{4} Var(X)$

$V(\frac{X}{2}) = \frac{1}{4} \times 1.4475$

$V(\frac{X}{2}) = 0.25 \times 1.4475$

Calculate $0.25 \times 1.4475$:

$0.25 \times 1.4475 = 0.361875$

Answer:

(i) $V \left( \frac{X}{2} \right) = 0.361875$

(ii) Variance of X = $1.4475$

Given:

The probability distribution of a discrete random variable X is:

To Determine/Find:

(i) The value of k.

(ii) $P(X \leq 2)$ and $P(X > 2)$.

(iii) $P(X \leq 2) + P(X > 2)$.

Solution:

(i) Determining the value of k:

For a valid probability distribution, the sum of all probabilities must be equal to 1.

Substitute the given probabilities into equation (i):

$k + \frac{k}{2} + \frac{k}{4} + \frac{k}{8} = 1$

Find a common denominator for the fractions, which is 8:

$\frac{8k}{8} + \frac{4k}{8} + \frac{2k}{8} + \frac{k}{8} = 1$

Combine the terms on the left side:

$\frac{8k + 4k + 2k + k}{8} = 1$

$\frac{15k}{8} = 1$

Multiply both sides by 8 and divide by 15 to solve for k:

$15k = 8$

$k = \frac{8}{15}$

We must also check that all probabilities are non-negative. Since $k = \frac{8}{15}$ is positive, all probabilities $k, \frac{k}{2}, \frac{k}{4}, \frac{k}{8}$ will be positive. So, $k = \frac{8}{15}$ is the correct value.

(ii) Determining $P(X \leq 2)$ and $P(X > 2)$:

$P(X \leq 2)$ is the probability that X takes a value less than or equal to 2. This includes the values 0, 1, and 2.

$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$

Substitute the probabilities in terms of k and the value $k = \frac{8}{15}$:

$P(X \leq 2) = k + \frac{k}{2} + \frac{k}{4}$

$P(X \leq 2) = \frac{8}{15} + \frac{8/15}{2} + \frac{8/15}{4}$

$P(X \leq 2) = \frac{8}{15} + \frac{8}{15 \times 2} + \frac{8}{15 \times 4}$

$P(X \leq 2) = \frac{8}{15} + \frac{8}{30} + \frac{8}{60}$

Find a common denominator, which is 60:

$P(X \leq 2) = \frac{8 \times 4}{15 \times 4} + \frac{8 \times 2}{30 \times 2} + \frac{8}{60}$

$P(X \leq 2) = \frac{32}{60} + \frac{16}{60} + \frac{8}{60}$

$P(X \leq 2) = \frac{32 + 16 + 8}{60} = \frac{56}{60}$

Simplify the fraction:

$P(X \leq 2) = \frac{\cancel{56}^{14}}{\cancel{60}_{15}} = \frac{14}{15}$

Alternatively, using the sum in terms of k:

$k + \frac{k}{2} + \frac{k}{4} = \frac{4k}{4} + \frac{2k}{4} + \frac{k}{4} = \frac{4k+2k+k}{4} = \frac{7k}{4}$

Substitute $k = \frac{8}{15}$:

$P(X \leq 2) = \frac{7}{4} \times \frac{8}{15} = \frac{7 \times \cancel{8}^2}{\cancel{4}^1 \times 15} = \frac{7 \times 2}{15} = \frac{14}{15}$

$P(X > 2)$ is the probability that X takes a value strictly greater than 2. This includes only the value 3.

$P(X > 2) = P(X=3)$

Substitute the probability in terms of k and the value $k = \frac{8}{15}$:

$P(X > 2) = \frac{k}{8} = \frac{8/15}{8} = \frac{8}{15 \times 8} = \frac{\cancel{8}^1}{15 \times \cancel{8}^1} = \frac{1}{15}$

(iii) Finding $P(X \leq 2) + P(X > 2)$:

Add the probabilities found in part (ii):

$P(X \leq 2) + P(X > 2) = \frac{14}{15} + \frac{1}{15}$

$P(X \leq 2) + P(X > 2) = \frac{14 + 1}{15} = \frac{15}{15} = 1$

This result is expected because the events $(X \leq 2)$ and $(X > 2)$ are complementary events, covering all possible outcomes of X without overlap. The sum of the probabilities of complementary events is always 1.

Answer:

(i) The value of $k = \frac{8}{15}$.

(ii) $P(X \leq 2) = \frac{14}{15}$ and $P(X > 2) = \frac{1}{15}$.

(iii) $P(X \leq 2) + P(X > 2) = 1$.

Given:

The probability distribution of a discrete random variable X is:

Check if the sum of probabilities is 1: $0.2 + 0.5 + 0.3 = 1.0$. The distribution is valid.

To Determine:

The standard deviation of the random variable X, denoted by $\sigma_X$.

Solution:

The standard deviation $\sigma_X$ is the square root of the variance $Var(X)$.

$Var(X) = E(X^2) - (E(X))^2$

First, calculate the expected value (mean) of X, $E(X)$.

$E(X) = \sum X \cdot P(X)$

$E(X) = (2 \times 0.2) + (3 \times 0.5) + (4 \times 0.3)$

$E(X) = 0.4 + 1.5 + 1.2$

$E(X) = 3.1$

Next, calculate the expected value of $X^2$, $E(X^2)$.

$E(X^2) = \sum X^2 \cdot P(X)$

$E(X^2) = (2^2 \times 0.2) + (3^2 \times 0.5) + (4^2 \times 0.3)$

$E(X^2) = (4 \times 0.2) + (9 \times 0.5) + (16 \times 0.3)$

$E(X^2) = 0.8 + 4.5 + 4.8$

$E(X^2) = 10.1$

Now, calculate the variance of X:

$Var(X) = E(X^2) - (E(X))^2$

$Var(X) = 10.1 - (3.1)^2$

Calculate $(3.1)^2$: $3.1 \times 3.1 = 9.61$

$Var(X) = 10.1 - 9.61$

$Var(X) = 0.49$

Finally, calculate the standard deviation, which is the square root of the variance:

$\sigma_X = \sqrt{Var(X)} = \sqrt{0.49}$

$\sigma_X = 0.7$

Answer:

The standard deviation of the random variable X is $0.7$.

Given:

A biased die.

$P(4) = \frac{1}{10}$.

Other scores (1, 2, 3, 5, 6) are equally likely.

There are 5 other scores. Let the probability of each of these scores be $p_0$.

The sum of probabilities for all possible outcomes must be 1.

$P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1$

$p_0 + p_0 + p_0 + \frac{1}{10} + p_0 + p_0 = 1$

$5 p_0 + \frac{1}{10} = 1$

$5 p_0 = 1 - \frac{1}{10} = \frac{9}{10}$

$p_0 = \frac{9}{10 \times 5} = \frac{9}{50}$

So, $P(1) = P(2) = P(3) = P(5) = P(6) = \frac{9}{50}$.

The probability of getting a 4 is $P(4) = \frac{1}{10} = \frac{5}{50}$.

Check sum: $5 \times \frac{9}{50} + \frac{5}{50} = \frac{45}{50} + \frac{5}{50} = \frac{50}{50} = 1$. The probabilities are valid.

The die is tossed twice.

X is the 'number of fours seen'.

To Find:

The variance of the random variable X.

Solution:

The experiment consists of two independent trials (tosses). For each toss, there are two outcomes relevant to X: getting a 4 (success) or not getting a 4 (failure).

Probability of success (getting a 4), $p = P(4) = \frac{1}{10}$.

Probability of failure (not getting a 4), $1-p = P(\text{not 4}) = 1 - \frac{1}{10} = \frac{9}{10}$.

The number of trials is $n=2$.

The random variable X, the number of fours in 2 tosses, follows a binomial distribution with parameters $n=2$ and $p=\frac{1}{10}$.

For a binomial distribution $B(n, p)$, the variance is given by the formula:

$Var(X) = np(1-p)$

Substitute the values $n=2$, $p=\frac{1}{10}$, and $1-p=\frac{9}{10}$:

$Var(X) = 2 \times \frac{1}{10} \times \frac{9}{10}$

$Var(X) = \frac{2 \times 1 \times 9}{10 \times 10} = \frac{18}{100}$

$Var(X) = 0.18$

Alternate Solution (using probability distribution):

The possible values for X (number of fours in 2 tosses) are 0, 1, or 2.

Let F denote getting a 4, and N denote not getting a 4.

$P(F) = \frac{1}{10}$

$P(N) = \frac{9}{10}$

The possible outcomes of two tosses are (N, N), (F, N), (N, F), (F, F).

Calculate the probability of each outcome (since tosses are independent):

$P(N, N) = P(N) \times P(N) = \frac{9}{10} \times \frac{9}{10} = \frac{81}{100}$ (X=0 fours)

$P(F, N) = P(F) \times P(N) = \frac{1}{10} \times \frac{9}{10} = \frac{9}{100}$ (X=1 four)

$P(N, F) = P(N) \times P(F) = \frac{9}{10} \times \frac{1}{10} = \frac{9}{100}$ (X=1 four)

$P(F, F) = P(F) \times P(F) = \frac{1}{10} \times \frac{1}{10} = \frac{1}{100}$ (X=2 fours)

The probability distribution of X is:

$P(X=0) = \frac{81}{100}$

$P(X=1) = P(F,N) + P(N,F) = \frac{9}{100} + \frac{9}{100} = \frac{18}{100}$

$P(X=2) = \frac{1}{100}$

Check sum: $\frac{81}{100} + \frac{18}{100} + \frac{1}{100} = \frac{100}{100} = 1$. Valid.

Calculate the mean $E(X)$:

$E(X) = \sum X \cdot P(X) = (0 \times \frac{81}{100}) + (1 \times \frac{18}{100}) + (2 \times \frac{1}{100})$

$E(X) = 0 + \frac{18}{100} + \frac{2}{100} = \frac{20}{100} = 0.2$

Note: For a binomial distribution $B(n,p)$, $E(X) = np = 2 \times \frac{1}{10} = 0.2$. Matches.

Calculate $E(X^2)$:

$E(X^2) = \sum X^2 \cdot P(X) = (0^2 \times \frac{81}{100}) + (1^2 \times \frac{18}{100}) + (2^2 \times \frac{1}{100})$

$E(X^2) = (0 \times \frac{81}{100}) + (1 \times \frac{18}{100}) + (4 \times \frac{1}{100})$

$E(X^2) = 0 + \frac{18}{100} + \frac{4}{100} = \frac{22}{100} = 0.22$

Calculate the variance:

$Var(X) = E(X^2) - (E(X))^2$

$Var(X) = 0.22 - (0.2)^2$

$Var(X) = 0.22 - 0.04$

$Var(X) = 0.18$

Both methods yield the same result.

Answer:

The variance of the random variable X is $0.18$ or $\frac{18}{100} = \frac{9}{50}$.

Given:

A fair die is thrown three times.

X is the random variable representing the number of twos seen.

To Find:

The expectation of X, $E(X)$.

Solution:

The experiment consists of a fixed number of independent trials ($n=3$ die throws).

For each trial, there are two possible outcomes relevant to the random variable X: getting a '2' (defined as a success) or not getting a '2' (defined as a failure).

Assuming the die is fair, the probability of getting a '2' on a single throw is $p = P(\text{getting a 2}) = \frac{1}{6}$.

The probability of not getting a '2' on a single throw is $1-p = P(\text{not getting a 2}) = 1 - \frac{1}{6} = \frac{5}{6}$.

The random variable X, representing the number of successes (twos) in $n=3$ trials, follows a binomial distribution with parameters $n=3$ and $p=\frac{1}{6}$. We denote this as $X \sim B(3, \frac{1}{6})$.

For a binomial distribution $B(n, p)$, the expected value (mean) is given by the formula:

$E(X) = np$

Substitute the values of $n$ and $p$ into the formula:

$E(X) = 3 \times \frac{1}{6}$

$E(X) = \frac{3}{6}$

$E(X) = \frac{1}{2}$

Answer:

The expectation of the random variable X is $\frac{1}{2}$.

Given:

Two biased dice are thrown together. The throws are independent.

For the first die (Die 1):

$P_1(6) = \frac{1}{2}$.

The other 5 scores (1, 2, 3, 4, 5) are equally likely. Let the probability of each of these scores be $p_1$.

Sum of probabilities for Die 1 is 1:

$5 p_1 + P_1(6) = 1$

$5 p_1 + \frac{1}{2} = 1$

$5 p_1 = 1 - \frac{1}{2} = \frac{1}{2}$

$p_1 = \frac{1}{2 \times 5} = \frac{1}{10}$

So, for Die 1: $P_1(1) = P_1(2) = P_1(3) = P_1(4) = P_1(5) = \frac{1}{10}$. $P_1(6) = \frac{5}{10} = \frac{1}{2}$.

For the second die (Die 2):

$P_2(1) = \frac{2}{5}$.

The other 5 scores (2, 3, 4, 5, 6) are equally likely. Let the probability of each of these scores be $p_2$.

Sum of probabilities for Die 2 is 1:

$P_2(1) + 5 p_2 = 1$

$\frac{2}{5} + 5 p_2 = 1$

$5 p_2 = 1 - \frac{2}{5} = \frac{3}{5}$

$p_2 = \frac{3}{5 \times 5} = \frac{3}{25}$

So, for Die 2: $P_2(2) = P_2(3) = P_2(4) = P_2(5) = P_2(6) = \frac{3}{25}$. $P_2(1) = \frac{10}{25} = \frac{2}{5}$.

Let X be the random variable representing the number of ones seen in the two throws.

To Find:

The probability distribution of X.

Solution:

The possible values for X (number of ones) are 0, 1, or 2.

Let $1_1$ denote getting a 1 on Die 1, and $\overline{1}_1$ denote not getting a 1 on Die 1.

Let $1_2$ denote getting a 1 on Die 2, and $\overline{1}_2$ denote not getting a 1 on Die 2.

From the probabilities calculated above:

$P(1_1) = P_1(1) = \frac{1}{10}$

$P(\overline{1}_1) = 1 - P(1_1) = 1 - \frac{1}{10} = \frac{9}{10}$ (This is the probability of getting any score other than 1 on Die 1).

$P(1_2) = P_2(1) = \frac{2}{5}$

$P(\overline{1}_2) = 1 - P(1_2) = 1 - \frac{2}{5} = \frac{3}{5}$ (This is the probability of getting any score other than 1 on Die 2).

We find the probability for each possible value of X:

Case X=0: No ones are seen.

This means Die 1 is not a 1 AND Die 2 is not a 1. Since the dice throws are independent, the probability is:

$P(X=0) = P(\overline{1}_1 \cap \overline{1}_2) = P(\overline{1}_1) \times P(\overline{1}_2)$

$P(X=0) = \frac{9}{10} \times \frac{3}{5} = \frac{27}{50}$

Case X=1: Exactly one one is seen.

This can happen in two mutually exclusive ways:

(a) Die 1 is a 1 AND Die 2 is not a 1: $1_1 \cap \overline{1}_2$

(b) Die 1 is not a 1 AND Die 2 is a 1: $\overline{1}_1 \cap 1_2$

The probability is the sum of the probabilities of these two events (since they are mutually exclusive):

$P(X=1) = P(1_1 \cap \overline{1}_2) + P(\overline{1}_1 \cap 1_2)$

Due to independence:

$P(X=1) = P(1_1) \times P(\overline{1}_2) + P(\overline{1}_1) \times P(1_2)$

$P(X=1) = (\frac{1}{10} \times \frac{3}{5}) + (\frac{9}{10} \times \frac{2}{5})$

$P(X=1) = \frac{3}{50} + \frac{18}{50} = \frac{3+18}{50} = \frac{21}{50}$

Case X=2: Exactly two ones are seen.

This means Die 1 is a 1 AND Die 2 is a 1. Since the dice throws are independent, the probability is:

$P(X=2) = P(1_1 \cap 1_2) = P(1_1) \times P(1_2)$

$P(X=2) = \frac{1}{10} \times \frac{2}{5} = \frac{2}{50} = \frac{1}{25}$

The probability distribution of X is:

Check if the sum of probabilities is 1:

$\frac{27}{50} + \frac{21}{50} + \frac{2}{50} = \frac{27 + 21 + 2}{50} = \frac{50}{50} = 1$. The distribution is valid.

Answer:

The probability distribution of the random variable X (the number of ones seen) is:

Given:

Probability distribution for random variable X:

Probability distribution for random variable Y:

To Prove:

$E(Y^2) = 2 E(X)$

Proof:

First, calculate the expected value of X, $E(X)$.

The formula for the expected value of a discrete random variable X is $E(X) = \sum x P(X=x)$.

$E(X) = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2)) + (3 \times P(X=3))$

$E(X) = (0 \times \frac{1}{5}) + (1 \times \frac{2}{5}) + (2 \times \frac{1}{5}) + (3 \times \frac{1}{5})$

$E(X) = 0 + \frac{2}{5} + \frac{2}{5} + \frac{3}{5}$

$E(X) = \frac{2 + 2 + 3}{5} = \frac{7}{5}$

Next, calculate the expected value of $Y^2$, $E(Y^2)$.

The formula for the expected value of a function of a discrete random variable Y, $g(Y)$, is $E(g(Y)) = \sum g(y) P(Y=y)$. Here, $g(Y) = Y^2$.

$E(Y^2) = \sum y^2 P(Y=y)$

$E(Y^2) = (0^2 \times P(Y=0)) + (1^2 \times P(Y=1)) + (2^2 \times P(Y=2)) + (3^2 \times P(Y=3))$

$E(Y^2) = (0 \times \frac{1}{5}) + (1 \times \frac{3}{10}) + (4 \times \frac{2}{5}) + (9 \times \frac{1}{10})$

$E(Y^2) = 0 + \frac{3}{10} + \frac{8}{5} + \frac{9}{10}$

To sum these fractions, find a common denominator, which is 10.

$E(Y^2) = \frac{3}{10} + \frac{8 \times 2}{5 \times 2} + \frac{9}{10}$

$E(Y^2) = \frac{3}{10} + \frac{16}{10} + \frac{9}{10}$

$E(Y^2) = \frac{3 + 16 + 9}{10} = \frac{28}{10}$

Simplify the fraction:

$E(Y^2) = \frac{\cancel{28}^{14}}{\cancel{10}_5} = \frac{14}{5}$

Now, we compare $E(Y^2)$ with $2 E(X)$.

We calculated $E(X) = \frac{7}{5}$.

So, $2 E(X) = 2 \times \frac{7}{5} = \frac{14}{5}$.

We calculated $E(Y^2) = \frac{14}{5}$.

Thus, we have $E(Y^2) = \frac{14}{5}$ and $2 E(X) = \frac{14}{5}$.

Since $E(Y^2)$ is equal to $2 E(X)$, the relationship $E(Y^2) = 2 E(X)$ is proven.

Conclusion:

$E(Y^2) = \frac{14}{5}$ and $2 E(X) = \frac{14}{5}$.

Therefore, $E(Y^2) = 2 E(X)$.

Given:

Probability that a single bulb is defective, $p = \frac{1}{50}$.

The bulbs are packed in boxes of 10.

Number of bulbs in a box (number of trials), $n = 10$.

We assume the quality of each bulb is independent of others.

Let X be the random variable representing the number of defective bulbs in a box of 10.

X follows a binomial distribution with parameters $n=10$ and $p=\frac{1}{50}$.

The probability of success (a bulb being defective) is $p = \frac{1}{50}$.

The probability of failure (a bulb working properly) is $1-p = 1 - \frac{1}{50} = \frac{49}{50}$.

The probability of getting exactly $k$ defective bulbs in $n$ trials is $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.

To Find:

(i) Probability that none of the bulbs is defective ($P(X=0)$).

(ii) Probability that exactly two bulbs are defective ($P(X=2)$).

(iii) Probability that more than 8 bulbs work properly ($P(\text{more than 8 good bulbs})$).

Solution:

(i) Probability that none of the bulbs is defective:

This means the number of defective bulbs X is 0. We use the binomial formula with $n=10$, $p=\frac{1}{50}$, $1-p=\frac{49}{50}$, and $k=0$.

$P(X=0) = \binom{10}{0} (\frac{1}{50})^0 (\frac{49}{50})^{10-0}$

$\binom{10}{0} = 1$

$(\frac{1}{50})^0 = 1$

$P(X=0) = 1 \times 1 \times (\frac{49}{50})^{10} = (\frac{49}{50})^{10}$

(ii) Probability that exactly two bulbs are defective:

This means the number of defective bulbs X is 2. We use the binomial formula with $n=10$, $p=\frac{1}{50}$, $1-p=\frac{49}{50}$, and $k=2$.

$P(X=2) = \binom{10}{2} (\frac{1}{50})^2 (\frac{49}{50})^{10-2}$

$P(X=2) = \binom{10}{2} (\frac{1}{50})^2 (\frac{49}{50})^8$

Calculate $\binom{10}{2}$:

$\binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = \frac{90}{2} = 45$

Substitute this value:

$P(X=2) = 45 \times (\frac{1}{50})^2 \times (\frac{49}{50})^8$

$P(X=2) = 45 \times \frac{1}{50^2} \times \frac{49^8}{50^8}$

$P(X=2) = 45 \times \frac{1}{2500} \times \frac{49^8}{3906250000}$

$P(X=2) = \frac{45 \times 49^8}{50^{10}}$

We can express this without further calculation unless specified:

$P(X=2) = 45 \left(\frac{1}{50}\right)^2 \left(\frac{49}{50}\right)^8$

(iii) Probability that more than 8 bulbs work properly:

Let Y be the random variable representing the number of bulbs that work properly in a box of 10.

The probability that a single bulb works properly is $1-p = \frac{49}{50}$.

The number of trials is $n=10$.

Y follows a binomial distribution with parameters $n=10$ and $p' = \frac{49}{50}$.

We want to find the probability that more than 8 bulbs work properly, which means $P(Y > 8)$.

$P(Y > 8) = P(Y=9) + P(Y=10)$

We use the binomial formula for Y with $n=10$, $p'=\frac{49}{50}$, and $1-p'=\frac{1}{50}$.

Calculate $P(Y=9)$:

$P(Y=9) = \binom{10}{9} (\frac{49}{50})^9 (\frac{1}{50})^{10-9}$

$P(Y=9) = \binom{10}{9} (\frac{49}{50})^9 (\frac{1}{50})^1$

$\binom{10}{9} = \binom{10}{1} = 10$

$P(Y=9) = 10 \times \frac{49^9}{50^9} \times \frac{1}{50} = \frac{10 \times 49^9}{50^{10}}$

Calculate $P(Y=10)$:

$P(Y=10) = \binom{10}{10} (\frac{49}{50})^{10} (\frac{1}{50})^{10-10}$

$P(Y=10) = \binom{10}{10} (\frac{49}{50})^{10} (\frac{1}{50})^0$

$\binom{10}{10} = 1$

$(\frac{1}{50})^0 = 1$

$P(Y=10) = 1 \times (\frac{49}{50})^{10} \times 1 = (\frac{49}{50})^{10}$

Now, sum these probabilities:

$P(Y > 8) = P(Y=9) + P(Y=10)$

$P(Y > 8) = \frac{10 \times 49^9}{50^{10}} + \frac{49^{10}}{50^{10}}$

$P(Y > 8) = \frac{10 \times 49^9 + 49^{10}}{50^{10}}$

Factor out $49^9$ from the numerator:

$P(Y > 8) = \frac{49^9 (10 + 49^1)}{50^{10}}$

$P(Y > 8) = \frac{49^9 (10 + 49)}{50^{10}} = \frac{49^9 \times 59}{50^{10}}$

Alternatively, relate Y to X. The number of bulbs that work properly (Y) plus the number of defective bulbs (X) is equal to the total number of bulbs (10). So, $Y = 10 - X$.

The event "more than 8 bulbs work properly" means $Y > 8$.

$Y > 8 \Leftrightarrow 10 - X > 8$

$10 - 8 > X$

$2 > X$, or $X < 2$.

So, $P(Y > 8) = P(X < 2) = P(X=0) + P(X=1)$.

We already calculated $P(X=0) = (\frac{49}{50})^{10}$.

Now calculate $P(X=1)$ using the binomial formula for X with $n=10$, $p=\frac{1}{50}$, $1-p=\frac{49}{50}$, and $k=1$.

$P(X=1) = \binom{10}{1} (\frac{1}{50})^1 (\frac{49}{50})^{10-1}$

$P(X=1) = 10 \times \frac{1}{50} \times (\frac{49}{50})^9$

$P(X=1) = \frac{10}{50} \times (\frac{49}{50})^9 = \frac{1}{5} \times (\frac{49}{50})^9 = \frac{49^9}{5 \times 50^9} = \frac{49^9}{5 \times 50^9} \times \frac{10/10}{1} = \frac{10 \times 49^9}{50 \times 50^9} = \frac{10 \times 49^9}{50^{10}}$

$P(X < 2) = P(X=0) + P(X=1) = (\frac{49}{50})^{10} + \frac{10 \times 49^9}{50^{10}}$

$P(X < 2) = \frac{49^{10}}{50^{10}} + \frac{10 \times 49^9}{50^{10}} = \frac{49^{10} + 10 \times 49^9}{50^{10}}$

Factor out $49^9$ from the numerator:

$P(X < 2) = \frac{49^9 (49^1 + 10)}{50^{10}} = \frac{49^9 \times 59}{50^{10}}$

Both approaches yield the same result.

Answer:

(i) The probability that none of the bulbs is defective is $(\frac{49}{50})^{10}$.

(ii) The probability that exactly two bulbs are defective is $45 \left(\frac{1}{50}\right)^2 \left(\frac{49}{50}\right)^8$ or $\frac{45 \times 49^8}{50^{10}}$.

(iii) The probability that more than 8 bulbs work properly is $\frac{59 \times 49^9}{50^{10}}$.

Given:

There are two coins: one fair and one 2-headed.

One coin is selected at random.

The selected coin is tossed, and the outcome is a Head (H).

To Find:

The probability that the selected coin was a fair coin, given that the toss resulted in a Head.

Solution:

Let's define the events:

F: The selected coin is the fair coin.

T: The selected coin is the 2-headed coin.

H: The toss of the selected coin results in a Head.

Since one of the two coins is selected at random, the initial probabilities of selecting each type of coin are:

Now consider the probability of getting a Head for each type of coin:

If the coin is fair (event F), the probability of getting a Head is:

If the coin is 2-headed (event T), the probability of getting a Head is:

We want to find $P(F | H)$, the probability that the coin was fair given that a Head was obtained. We can use Bayes' Theorem:

To use Bayes' Theorem, we first need to find the overall probability of getting a Head, $P(H)$. We can find this using the Law of Total Probability. The event H occurs if either a fair coin is chosen AND a Head is obtained, OR if a 2-headed coin is chosen AND a Head is obtained. Events F and T are mutually exclusive and cover all possibilities for the chosen coin, so they form a partition.

Using the formula for the probability of intersection, $P(A \cap B) = P(A|B) P(B)$:

Substitute the probabilities from (1), (2), (3), and (4) into (7):

$P(H) = \left(\frac{1}{2} \times \frac{1}{2}\right) + \left(1 \times \frac{1}{2}\right)$

$P(H) = \frac{1}{4} + \frac{1}{2}$

$P(H) = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}$

Now substitute the values of $P(H | F)$, $P(F)$, and $P(H)$ into Bayes' Theorem (5):

$P(F | H) = \frac{P(H | F) P(F)}{P(H)}$

$P(F | H) = \frac{\frac{1}{2} \times \frac{1}{2}}{\frac{3}{4}}$

$P(F | H) = \frac{\frac{1}{4}}{\frac{3}{4}}$

$P(F | H) = \frac{1}{4} \times \frac{4}{3} = \frac{1}{3}$

Answer:

The probability that it was a fair coin, given that a head was obtained, is $\frac{1}{3}$.

Given:

Let O be the event that a person has blood group O.

Let O' be the event that a person has another blood group.

Let L be the event that a person is left-handed.

We are given the following probabilities:

$P(L | O) = 6\% = 0.06$

$P(L | O') = 10\% = 0.10$

$P(O) = 30\% = 0.30$

Inferred:

Since 30% of people have blood group O, the probability of having another blood group is:

$P(O') = 1 - P(O) = 1 - 0.30 = 0.70$

To Find:

The probability that a left-handed person selected at random has blood group O. This is the conditional probability $P(O | L)$.

Solution:

We can use Bayes' Theorem to find $P(O | L)$. Bayes' Theorem states:

To use this formula, we need to find the overall probability of a randomly selected person being left-handed, $P(L)$. We can find $P(L)$ using the Law of Total Probability. A person is left-handed if they have blood group O AND are left-handed, OR if they have another blood group AND are left-handed.

Using the definition of conditional probability, $P(A \cap B) = P(A|B) P(B)$, we can rewrite equation (ii):

Substitute the given and inferred probabilities into equation (iii):

$P(L) = (0.06 \times 0.30) + (0.10 \times 0.70)$

$P(L) = 0.018 + 0.070$

$P(L) = 0.088$

Now, substitute the values of $P(L | O)$, $P(O)$, and the calculated $P(L)$ into Bayes' Theorem formula (i):

$P(O | L) = \frac{0.06 \times 0.30}{0.088}$

$P(O | L) = \frac{0.018}{0.088}$

To simplify the fraction, multiply the numerator and denominator by 1000:

$P(O | L) = \frac{0.018 \times 1000}{0.088 \times 1000} = \frac{18}{88}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2:

$P(O | L) = \frac{\cancel{18}^{9}}{\cancel{88}_{44}} = \frac{9}{44}$

Answer:

The probability that a left-handed person has blood group O is $\frac{9}{44}$.

Given:

The set $S = \{1, 2, 3, \dots, n\}$, where $n$ is a natural number.

Two natural numbers $r$ and $s$ are drawn one at a time, without replacement, from $S$. This implies $r \in S$, $s \in S$, and $r \neq s$. Also, since two distinct numbers are drawn from $S$, the size of $S$ must be at least 2, so $n \geq 2$.

The number $p$ is an element of $S$, so $p \in \{1, 2, \dots, n\}$.

To Find:

The conditional probability $P[r \leq p | s \leq p]$.

Solution:

Let A be the event that $r \leq p$.

Let B be the event that $s \leq p$.

We want to find $P(A | B) = \frac{P(A \cap B)}{P(B)}$.

The sample space $\Omega$ consists of all ordered pairs $(r, s)$ such that $r, s \in S$ and $r \neq s$.

The total number of outcomes in the sample space is $|\Omega| = n \times (n-1)$. Each outcome is equally likely with probability $\frac{1}{n(n-1)}$.

Event B: $s \leq p$

This event consists of all pairs $(r, s)$ such that $r \neq s$ and $s \in \{1, 2, \dots, p\}$.

To find the number of outcomes in B, we can choose $s$ from the set $\{1, 2, \dots, p\}$ (there are $p$ choices). Then, $r$ can be any number in $S$ except the chosen value of $s$ (there are $n-1$ choices).

The number of outcomes in B is $|B| = p \times (n-1)$.

The probability of event B is $P(B) = \frac{|B|}{|\Omega|} = \frac{p(n-1)}{n(n-1)}$.

Since $n \geq 2$, $n-1 \neq 0$, so we can cancel $n-1$:

(Since $p \in S$, $p \geq 1$, so $P(B) > 0$ for $n \geq 1$. As $n \geq 2$, $P(B) > 0$).

Event $A \cap B$: $r \leq p$ and $s \leq p$

This event consists of all pairs $(r, s)$ such that $r \in \{1, 2, \dots, p\}$, $s \in \{1, 2, \dots, p\}$, and $r \neq s$.

To find the number of outcomes in $A \cap B$, we need to choose two distinct numbers from the set $\{1, 2, \dots, p\}$ and assign them as the first number ($r$) and the second number ($s$).

The number of ways to choose the first number $r$ from $\{1, 2, \dots, p\}$ is $p$.

The number of ways to choose the second number $s$ from the remaining $p-1$ numbers in $\{1, 2, \dots, p\}$ (since $s \neq r$) is $p-1$.

The number of outcomes in $A \cap B$ is $|A \cap B| = p \times (p-1)$.

Note that if $p=1$, this number is $1(1-1)=0$, which is correct since you cannot choose two distinct numbers from a set of size 1.

The probability of event $A \cap B$ is $P(A \cap B) = \frac{|A \cap B|}{|\Omega|} = \frac{p(p-1)}{n(n-1)}$.

Conditional Probability $P(A | B)$

Using the formula $P(A | B) = \frac{P(A \cap B)}{P(B)}$:

$P(r \leq p | s \leq p) = \frac{\frac{p(p-1)}{n(n-1)}}{\frac{p(n-1)}{n(n-1)}}$

$P(r \leq p | s \leq p) = \frac{p(p-1)}{n(n-1)} \times \frac{n(n-1)}{p(n-1)}$

Since $p \in \{1, \dots, n\}$ and $n \geq 2$, we have $p \geq 1$ and $n-1 \geq 1$. Thus $p \neq 0$ and $n-1 \neq 0$. We can cancel the terms $p$ and $(n-1)$ from the numerator and denominator:

$P(r \leq p | s \leq p) = \frac{\cancel{p}(p-1)}{\cancel{p}(n-1)}$

$P(r \leq p | s \leq p) = \frac{p-1}{n-1}$

Answer:

The probability $P [r \leq p | s \leq p]$ is $\frac{p-1}{n-1}$.

Given:

A fair die is thrown twice.

Let $D_1$ be the outcome of the first throw and $D_2$ be the outcome of the second throw. The outcomes are pairs $(d_1, d_2)$ where $d_1, d_2 \in \{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes is $6 \times 6 = 36$. Since the die is fair and the throws are independent, each outcome $(d_1, d_2)$ has a probability of $\frac{1}{36}$.

Let X be the random variable representing the maximum of the two scores, i.e., $X = \max(D_1, D_2)$.

To Find:

(i) The probability distribution of X.

(ii) The mean of the distribution, $E(X)$.

Solution:

The possible values for the maximum score X are the integers from 1 to 6.

We find the probability for each value $x \in \{1, 2, 3, 4, 5, 6\}$.

The event $X \leq x$ (the maximum of the two scores is less than or equal to x) occurs if and only if both scores are less than or equal to x, i.e., $D_1 \leq x$ and $D_2 \leq x$.

The possible outcomes for $D_1 \leq x$ are $\{1, 2, \dots, x\}$, which are $x$ outcomes. The probability is $\frac{x}{6}$.

The possible outcomes for $D_2 \leq x$ are $\{1, 2, \dots, x\}$, which are $x$ outcomes. The probability is $\frac{x}{6}$.

Since the throws are independent, $P(D_1 \leq x \text{ and } D_2 \leq x) = P(D_1 \leq x) \times P(D_2 \leq x) = \frac{x}{6} \times \frac{x}{6} = \frac{x^2}{36}$.

So, $P(X \leq x) = \frac{x^2}{36}$ for $x \in \{1, 2, 3, 4, 5, 6\}$.

Now we can find the probability mass function $P(X=x)$ for each value of x.

$P(X=x) = P(X \leq x) - P(X \leq x-1)$ for $x \in \{2, 3, 4, 5, 6\}$.

$P(X=1) = P(X \leq 1)$

Calculate the probabilities $P(X=x)$:

$P(X=1) = P(X \leq 1) = \frac{1^2}{36} = \frac{1}{36}$ (Outcome (1,1))

$P(X=2) = P(X \leq 2) - P(X \leq 1) = \frac{2^2}{36} - \frac{1^2}{36} = \frac{4}{36} - \frac{1}{36} = \frac{3}{36}$ (Outcomes (1,2), (2,1), (2,2))

$P(X=3) = P(X \leq 3) - P(X \leq 2) = \frac{3^2}{36} - \frac{2^2}{36} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36}$

$P(X=4) = P(X \leq 4) - P(X \leq 3) = \frac{4^2}{36} - \frac{3^2}{36} = \frac{16}{36} - \frac{9}{36} = \frac{7}{36}$

$P(X=5) = P(X \leq 5) - P(X \leq 4) = \frac{5^2}{36} - \frac{4^2}{36} = \frac{25}{36} - \frac{16}{36} = \frac{9}{36}$

$P(X=6) = P(X \leq 6) - P(X \leq 5) = \frac{6^2}{36} - \frac{5^2}{36} = \frac{36}{36} - \frac{25}{36} = \frac{11}{36}$

(i) Probability distribution of X:

Check sum of probabilities: $\frac{1+3+5+7+9+11}{36} = \frac{36}{36} = 1$. The distribution is valid.

(ii) Mean of the distribution:

The mean (expected value) of X is $E(X) = \sum x P(X=x)$.

$E(X) = (1 \times \frac{1}{36}) + (2 \times \frac{3}{36}) + (3 \times \frac{5}{36}) + (4 \times \frac{7}{36}) + (5 \times \frac{9}{36}) + (6 \times \frac{11}{36})$

$E(X) = \frac{1}{36} + \frac{6}{36} + \frac{15}{36} + \frac{28}{36} + \frac{45}{36} + \frac{66}{36}$

$E(X) = \frac{1 + 6 + 15 + 28 + 45 + 66}{36}$

$1 + 6 = 7$

$7 + 15 = 22$

$22 + 28 = 50$

$50 + 45 = 95$

$95 + 66 = 161$

$E(X) = \frac{161}{36}$

Answer:

(i) The probability distribution of X is given in the table above.

(ii) The mean of the distribution is $E(X) = \frac{161}{36}$.

Given:

The random variable X can take values 0, 1, and 2.

$P(X=0) = p$

$P(X=1) = p$

$E(X^2) = E(X)$

To Find:

The value of p.

Solution:

For a valid probability distribution, the sum of probabilities for all possible values of the random variable must equal 1.

Let $P(X=2)$ be denoted by $P_2$.

Substitute the given probabilities into equation (i):

$p + p + P_2 = 1$

$2p + P_2 = 1$

$P_2 = 1 - 2p$

For the probabilities to be valid, they must be non-negative.

$P(X=0) = p \geq 0$

$P(X=1) = p \geq 0$

$P(X=2) = 1 - 2p \geq 0 \Rightarrow 1 \geq 2p \Rightarrow p \leq \frac{1}{2}$.

So, the possible range for p is $0 \leq p \leq \frac{1}{2}$.

Now, we calculate the expected value of X, $E(X)$.

$E(X) = \sum x P(X=x)$

$E(X) = (0 \times P(X=0)) + (1 \times P(X=1)) + (2 \times P(X=2))$

$E(X) = (0 \times p) + (1 \times p) + (2 \times (1 - 2p))$

$E(X) = 0 + p + 2 - 4p$

$E(X) = 2 - 3p$

Next, we calculate the expected value of $X^2$, $E(X^2)$.

$E(X^2) = \sum x^2 P(X=x)$

$E(X^2) = (0^2 \times P(X=0)) + (1^2 \times P(X=1)) + (2^2 \times P(X=2))$

$E(X^2) = (0 \times p) + (1 \times p) + (4 \times (1 - 2p))$

$E(X^2) = 0 + p + 4 - 8p$

$E(X^2) = 4 - 7p$

We are given that $E(X^2) = E(X)$.

Set the expressions for $E(X^2)$ and $E(X)$ equal to each other:

Solve equation (ii) for p:

Add 7p to both sides:

$4 = 2 - 3p + 7p$

$4 = 2 + 4p$

Subtract 2 from both sides:

$4 - 2 = 4p$

$2 = 4p$

Divide by 4:

$p = \frac{2}{4} = \frac{1}{2}$

Check if this value of p is within the valid range $0 \leq p \leq \frac{1}{2}$.

If $p = \frac{1}{2}$, then $P(X=0) = \frac{1}{2}$, $P(X=1) = \frac{1}{2}$, and $P(X=2) = 1 - 2(\frac{1}{2}) = 1 - 1 = 0$.

The probabilities are $\frac{1}{2}, \frac{1}{2}, 0$, which are all non-negative and sum to 1. This is a valid probability distribution.

Answer:

The value of p is $\frac{1}{2}$.

Given:

The probability distribution of a discrete random variable X is:

Let's write the probabilities with a common denominator (18):

$P(X=0) = \frac{1}{6} = \frac{3}{18}$

$P(X=1) = \frac{5}{18}$

$P(X=2) = \frac{2}{9} = \frac{4}{18}$

$P(X=3) = \frac{1}{6} = \frac{3}{18}$

$P(X=4) = \frac{1}{9} = \frac{2}{18}$

$P(X=5) = \frac{1}{18}$

The sum of probabilities is $\frac{3+5+4+3+2+1}{18} = \frac{18}{18} = 1$. The distribution is valid.

To Find:

The variance of the distribution, $Var(X)$.

Solution:

The variance of a discrete random variable X is given by the formula:

$Var(X) = E(X^2) - (E(X))^2$

First, we need to calculate the expected value (mean) of X, $E(X)$.

$E(X) = \sum x P(X=x)$

$E(X) = (0 \times \frac{3}{18}) + (1 \times \frac{5}{18}) + (2 \times \frac{4}{18}) + (3 \times \frac{3}{18}) + (4 \times \frac{2}{18}) + (5 \times \frac{1}{18})$

$E(X) = \frac{0}{18} + \frac{5}{18} + \frac{8}{18} + \frac{9}{18} + \frac{8}{18} + \frac{5}{18}$

$E(X) = \frac{0 + 5 + 8 + 9 + 8 + 5}{18} = \frac{35}{18}$

Next, we calculate the expected value of $X^2$, $E(X^2)$.

$E(X^2) = \sum x^2 P(X=x)$

$E(X^2) = (0^2 \times \frac{3}{18}) + (1^2 \times \frac{5}{18}) + (2^2 \times \frac{4}{18}) + (3^2 \times \frac{3}{18}) + (4^2 \times \frac{2}{18}) + (5^2 \times \frac{1}{18})$

$E(X^2) = (0 \times \frac{3}{18}) + (1 \times \frac{5}{18}) + (4 \times \frac{4}{18}) + (9 \times \frac{3}{18}) + (16 \times \frac{2}{18}) + (25 \times \frac{1}{18})$

$E(X^2) = \frac{0}{18} + \frac{5}{18} + \frac{16}{18} + \frac{27}{18} + \frac{32}{18} + \frac{25}{18}$

$E(X^2) = \frac{0 + 5 + 16 + 27 + 32 + 25}{18} = \frac{105}{18}$

Simplify the fraction: $\frac{\cancel{105}^{35}}{\cancel{18}_{6}} = \frac{35}{6}$

Now, calculate the variance of X:

$Var(X) = E(X^2) - (E(X))^2$

$Var(X) = \frac{35}{6} - \left(\frac{35}{18}\right)^2$

$Var(X) = \frac{35}{6} - \frac{35^2}{18^2}$

$35^2 = 1225$

$18^2 = 324$

$Var(X) = \frac{35}{6} - \frac{1225}{324}$

Find a common denominator, which is 324. $324 = 6 \times 54$.

$Var(X) = \frac{35 \times 54}{6 \times 54} - \frac{1225}{324}$

$Var(X) = \frac{1890}{324} - \frac{1225}{324}$

$Var(X) = \frac{1890 - 1225}{324} = \frac{665}{324}$

Answer:

The variance of the distribution is $\frac{665}{324}$.

Given:

A and B throw a pair of dice alternately.

A wins if they get a total of 6.

B wins if they get a total of 7.

A starts the game.

To Find:

The probability that A wins the game on their third throw of the pair of dice.

Solution:

When a pair of fair dice is thrown, there are $6 \times 6 = 36$ possible outcomes. Each outcome is equally likely with probability $\frac{1}{36}$.

Let $W_A$ be the event that A gets a total of 6.

The outcomes that sum to 6 are: (1,5), (2,4), (3,3), (4,2), (5,1).

Number of outcomes for $W_A$ is 5.

$P(W_A) = \frac{5}{36}$.

Let $L_A$ be the event that A does not get a total of 6. $P(L_A) = 1 - P(W_A) = 1 - \frac{5}{36} = \frac{31}{36}$.

Let $W_B$ be the event that B gets a total of 7.

The outcomes that sum to 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

Number of outcomes for $W_B$ is 6.

$P(W_B) = \frac{6}{36} = \frac{1}{6}$.

Let $L_B$ be the event that B does not get a total of 7. $P(L_B) = 1 - P(W_B) = 1 - \frac{1}{6} = \frac{5}{6}$.

The throws of the dice are independent events.

A starts the game. The throws alternate between A and B.

A's turns are 1st, 3rd, 5th, ...

B's turns are 2nd, 4th, 6th, ...

The game ends when either A gets a 6 or B gets a 7. This means if a player achieves their winning total, the game stops, and the other player does not get their turn.

We want the probability that A wins on their third throw.

For A to win on their third throw, the following sequence of events must occur:

1. A does not win on their first throw (A gets a total other than 6). Probability is $P(L_A)$.

2. B does not win on their first throw (B gets a total other than 7). Probability is $P(L_B)$.

3. A does not win on their second throw (A gets a total other than 6). Probability is $P(L_A)$.

4. B does not win on their second throw (B gets a total other than 7). Probability is $P(L_B)$.

5. A wins on their third throw (A gets a total of 6). Probability is $P(W_A)$.

A's turns: 1st, 3rd, 5th

B's turns: 2nd, 4th

Sequence of events for A winning on 3rd throw:

Throw 1 (A): Not 6 ($L_A$)

Throw 2 (B): Not 7 ($L_B$)

Throw 3 (A): Not 6 ($L_A$)

Throw 4 (B): Not 7 ($L_B$)

Throw 5 (A): Get 6 ($W_A$)

Since the throws are independent, the probability of this sequence is the product of the individual probabilities:

$P(\text{A wins on 3rd throw}) = P(L_A) \times P(L_B) \times P(L_A) \times P(L_B) \times P(W_A)$

$P(\text{A wins on 3rd throw}) = P(L_A)^2 \times P(L_B)^2 \times P(W_A)$

Substitute the probabilities:

$P(\text{A wins on 3rd throw}) = \left(\frac{31}{36}\right)^2 \times \left(\frac{5}{6}\right)^2 \times \frac{5}{36}$

$P(\text{A wins on 3rd throw}) = \frac{31^2}{36^2} \times \frac{5^2}{6^2} \times \frac{5}{36}$

$P(\text{A wins on 3rd throw}) = \frac{961}{1296} \times \frac{25}{36} \times \frac{5}{36}$

$P(\text{A wins on 3rd throw}) = \frac{961 \times 25 \times 5}{1296 \times 36 \times 36}$

$1296 = 36^2$

$P(\text{A wins on 3rd throw}) = \frac{961 \times 125}{1296 \times 1296} = \frac{120125}{1679616}$

Alternatively, keep the denominators $36$ and $6$ separated initially:

$P(\text{A wins on 3rd throw}) = \left(\frac{31}{36}\right) \times \left(\frac{5}{6}\right) \times \left(\frac{31}{36}\right) \times \left(\frac{5}{6}\right) \times \frac{5}{36}$

$P(\text{A wins on 3rd throw}) = \frac{31 \times 5 \times 31 \times 5 \times 5}{36 \times 6 \times 36 \times 6 \times 36}$

$P(\text{A wins on 3rd throw}) = \frac{(31 \times 31) \times (5 \times 5) \times 5}{(36 \times 36 \times 36) \times (6 \times 6)}$

$P(\text{A wins on 3rd throw}) = \frac{961 \times 25 \times 5}{36^3 \times 6^2} = \frac{961 \times 125}{46656 \times 36}$

$46656 \times 36 = 1679616$

$961 \times 125 = 120125$

$P(\text{A wins on 3rd throw}) = \frac{120125}{1679616}$

Answer:

The probability of winning the game by A in the third throw of the pair of dice is $\frac{120125}{1679616}$.

Given:

Two fair dice are tossed. The sample space $\Omega$ consists of $6 \times 6 = 36$ ordered pairs $(x, y)$ where $x, y \in \{1, 2, 3, 4, 5, 6\}$. Each outcome has a probability of $\frac{1}{36}$.

Event A: $A = \{(x, y) : x + y = 1\}$

Event B: $B = \{(x, y) : x \neq 5\}$

To Find:

Whether events A and B are independent.

Solution:

Two events A and B are independent if and only if $P(A \cap B) = P(A) \times P(B)$.

Calculate P(A):

Event A is the set of outcomes $(x, y)$ such that the sum of the scores is 1. The minimum possible sum when rolling two standard dice is $1 + 1 = 2$. There are no outcomes where the sum is 1.

$A = \emptyset$ (the empty set)

The number of outcomes in A is $|A| = 0$.

$P(A) = \frac{|A|}{\text{Total outcomes}} = \frac{0}{36} = 0$

Calculate P(B):

Event B is the set of outcomes $(x, y)$ where the result of the first die (x) is not 5.

The possible values for x are $\{1, 2, 3, 4, 6\}$. There are 5 choices for x.

The possible values for y are $\{1, 2, 3, 4, 5, 6\}$. There are 6 choices for y.

The number of outcomes in B is $|B| = 5 \times 6 = 30$.

Alternatively, consider the complement event $B'$, where $x = 5$. The outcomes are $\{(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)\}$. $|B'| = 6$. Then $|B| = 36 - 6 = 30$.

$P(B) = \frac{|B|}{\text{Total outcomes}} = \frac{30}{36} = \frac{5}{6}$

Calculate $P(A \cap B)$:

The intersection $A \cap B$ is the set of outcomes that are in both A and B.

$A = \emptyset$.

The intersection of the empty set with any set is the empty set.

$A \cap B = \emptyset \cap B = \emptyset$

The number of outcomes in $A \cap B$ is $|A \cap B| = 0$.

$P(A \cap B) = \frac{|A \cap B|}{\text{Total outcomes}} = \frac{0}{36} = 0$

Check for Independence:

We check if $P(A \cap B) = P(A) \times P(B)$.

Left side: $P(A \cap B) = 0$

Right side: $P(A) \times P(B) = 0 \times \frac{5}{6} = 0$

Since $P(A \cap B) = P(A) \times P(B)$ (both are 0), the events A and B are independent.

Conclusion:

Events A and B are independent.

Given:

Initial contents of the urn: $m$ white balls (W) and $n$ black balls (B).

Total initial number of balls = $m+n$.

Step 1: A ball is drawn at random.

Step 2: The drawn ball is replaced, and $k$ additional balls of the same colour are added.

Step 3: A ball is again drawn at random from the urn.

To Show:

The probability of drawing a white ball on the second draw does not depend on $k$.

Solution:

Let $F_W$ be the event that the first ball drawn is white.

Let $F_B$ be the event that the first ball drawn is black.

Let $S_W$ be the event that the second ball drawn is white.

The probability of drawing a white ball on the first draw is:

The probability of drawing a black ball on the first draw is:

Events $F_W$ and $F_B$ form a partition of the sample space for the first draw.

Now, we consider the composition of the urn after the first draw and the addition of $k$ balls, and find the conditional probability of drawing a white ball on the second draw based on the colour of the first ball drawn.

Case 1: The first ball drawn was white ($F_W$ occurred).

The white ball drawn is replaced, and $k$ additional white balls are added.

Number of white balls in the urn = $m - 1 + 1 + k = m + k$.

Number of black balls in the urn = $n$.

Total number of balls in the urn = $(m+k) + n = m+n+k$.

The probability of drawing a white ball on the second draw, given the first was white, is:

Case 2: The first ball drawn was black ($F_B$ occurred).

The black ball drawn is replaced, and $k$ additional black balls are added.

Number of white balls in the urn = $m$.

Number of black balls in the urn = $n - 1 + 1 + k = n + k$.

Total number of balls in the urn = $m + (n+k) = m+n+k$.

The probability of drawing a white ball on the second draw, given the first was black, is:

To find the overall probability of drawing a white ball on the second draw, $P(S_W)$, we use the Law of Total Probability, conditioned on the outcome of the first draw:

Using the multiplication rule for probability, $P(A \cap B) = P(A|B) P(B)$, we can rewrite equation (5):

Substitute the probabilities from equations (1), (2), (3), and (4) into equation (6):

$P(S_W) = \left(\frac{m+k}{m+n+k}\right) \times \left(\frac{m}{m+n}\right) + \left(\frac{m}{m+n+k}\right) \times \left(\frac{n}{m+n}\right)$

Combine the terms over the common denominator $(m+n+k)(m+n)$:

$P(S_W) = \frac{(m+k)m + m n}{(m+n+k)(m+n)}$

$P(S_W) = \frac{m^2 + mk + mn}{(m+n+k)(m+n)}$

Factor out $m$ from the numerator:

$P(S_W) = \frac{m(m + k + n)}{(m+n+k)(m+n)}$

Since $m+n+k$ appears in both the numerator and the denominator (and $m+n+k > 0$ as long as $m, n, k$ are non-negative and $m+n > 0$), we can cancel this term:

$P(S_W) = \frac{m \cancel{(m + n + k)}}{\cancel{(m+n+k)}(m+n)}$

$P(S_W) = \frac{m}{m+n}$

The final expression for the probability of drawing a white ball on the second draw is $\frac{m}{m+n}$. This expression depends only on the initial number of white balls ($m$) and the initial total number of balls ($m+n$). It does not contain the variable $k$.

Therefore, the probability of drawing a white ball on the second draw does not depend on $k$.

Conclusion:

The probability of drawing a white ball on the second draw is $\frac{m}{m+n}$, which is independent of $k$. This completes the proof.

Given:

Bag 1 contains 3 red balls and 0 white balls.

Bag 2 contains 2 red balls and 1 white ball.

Bag 3 contains 0 red balls and 3 white balls.

The probability that bag $i$ is chosen is $P(\text{Bag } i) = \frac{i}{6}$, for $i=1, 2, 3$.

To Find:

(i) The probability that a red ball will be selected ($P(\text{Red})$).

(ii) The probability that a white ball will be selected ($P(\text{White})$).

Solution:

First, let's find the conditional probabilities of drawing a red or white ball from each bag.

For Bag 1:

$P(\text{Red | Bag 1}) = \frac{\text{Number of red balls in Bag 1}}{\text{Total number of balls in Bag 1}} = \frac{3}{3} = 1$

$P(\text{White | Bag 1}) = \frac{\text{Number of white balls in Bag 1}}{\text{Total number of balls in Bag 1}} = \frac{0}{3} = 0$

For Bag 2:

$P(\text{Red | Bag 2}) = \frac{\text{Number of red balls in Bag 2}}{\text{Total number of balls in Bag 2}} = \frac{2}{3}$

$P(\text{White | Bag 2}) = \frac{\text{Number of white balls in Bag 2}}{\text{Total number of balls in Bag 2}} = \frac{1}{3}$

For Bag 3:

$P(\text{Red | Bag 3}) = \frac{\text{Number of red balls in Bag 3}}{\text{Total number of balls in Bag 3}} = \frac{0}{3} = 0$

$P(\text{White | Bag 3}) = \frac{\text{Number of white balls in Bag 3}}{\text{Total number of balls in Bag 3}} = \frac{3}{3} = 1$

The probabilities of choosing each bag are given as:

$P(\text{Bag 1}) = \frac{1}{6}$

$P(\text{Bag 2}) = \frac{2}{6} = \frac{1}{3}$

$P(\text{Bag 3}) = \frac{3}{6} = \frac{1}{2}$

(i) Probability of selecting a red ball:

We use the Law of Total Probability.

$P(\text{Red}) = P(\text{Red | Bag 1}) P(\text{Bag 1}) + P(\text{Red | Bag 2}) P(\text{Bag 2}) + P(\text{Red | Bag 3}) P(\text{Bag 3})$

$P(\text{Red}) = (1) \cdot \left(\frac{1}{6}\right) + \left(\frac{2}{3}\right) \cdot \left(\frac{1}{3}\right) + (0) \cdot \left(\frac{1}{2}\right)$

$P(\text{Red}) = \frac{1}{6} + \frac{2}{9} + 0$

To sum these fractions, we find a common denominator, which is 18.

$P(\text{Red}) = \frac{1 \cdot 3}{6 \cdot 3} + \frac{2 \cdot 2}{9 \cdot 2}$

$P(\text{Red}) = \frac{3}{18} + \frac{4}{18}$

$P(\text{Red}) = \frac{3 + 4}{18}$

$P(\text{Red}) = \frac{7}{18}$

(ii) Probability of selecting a white ball:

We again use the Law of Total Probability.

$P(\text{White}) = P(\text{White | Bag 1}) P(\text{Bag 1}) + P(\text{White | Bag 2}) P(\text{Bag 2}) + P(\text{White | Bag 3}) P(\text{Bag 3})$

$P(\text{White}) = (0) \cdot \left(\frac{1}{6}\right) + \left(\frac{1}{3}\right) \cdot \left(\frac{1}{3}\right) + (1) \cdot \left(\frac{1}{2}\right)$

$P(\text{White}) = 0 + \frac{1}{9} + \frac{1}{2}$

To sum these fractions, we find a common denominator, which is 18.

$P(\text{White}) = \frac{1 \cdot 2}{9 \cdot 2} + \frac{1 \cdot 9}{2 \cdot 9}$

$P(\text{White}) = \frac{2}{18} + \frac{9}{18}$

$P(\text{White}) = \frac{2 + 9}{18}$

$P(\text{White}) = \frac{11}{18}$

Final Answers:

(i) The probability that a red ball will be selected is $\frac{7}{18}$.

(ii) The probability that a white ball will be selected is $\frac{11}{18}$.

Given:

Based on Question 41:

Probabilities of choosing the bags:

$P(\text{Bag 1}) = \frac{1}{6}$, $P(\text{Bag 2}) = \frac{2}{6} = \frac{1}{3}$, $P(\text{Bag 3}) = \frac{3}{6} = \frac{1}{2}$

Conditional probabilities of drawing a white ball given the bag chosen:

$P(\text{White | Bag 1}) = 0$

$P(\text{White | Bag 2}) = \frac{1}{3}$

$P(\text{White | Bag 3}) = 1$

From the solution to Question 41 (part ii), the total probability of selecting a white ball is $P(\text{White}) = \frac{11}{18}$.

To Find:

(i) The probability that the white ball came from Bag 2, given that a white ball was selected ($P(\text{Bag 2 | White})$).

(ii) The probability that the white ball came from Bag 3, given that a white ball was selected ($P(\text{Bag 3 | White})$).

Solution:

We can use Bayes' Theorem to find the required conditional probabilities.

Bayes' Theorem states: $P(B_i | W) = \frac{P(W | B_i) P(B_i)}{P(W)}$

(i) Probability that the white ball came from Bag 2:

Using Bayes' Theorem for Bag 2 and event White:

$P(\text{Bag 2 | White}) = \frac{P(\text{White | Bag 2}) P(\text{Bag 2})}{P(\text{White})}$

Substitute the known values:

$P(\text{Bag 2 | White}) = \frac{\left(\frac{1}{3}\right) \cdot \left(\frac{1}{3}\right)}{\frac{11}{18}}$

$P(\text{Bag 2 | White}) = \frac{\frac{1}{9}}{\frac{11}{18}}$

To simplify the fraction, multiply the numerator by the reciprocal of the denominator:

$P(\text{Bag 2 | White}) = \frac{1}{9} \times \frac{18}{11}$

$P(\text{Bag 2 | White}) = \frac{1 \times 18}{9 \times 11}$

$P(\text{Bag 2 | White}) = \frac{18}{99}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, 9:

$P(\text{Bag 2 | White}) = \frac{\cancel{18}^{2}}{\cancel{99}_{11}}$

$P(\text{Bag 2 | White}) = \frac{2}{11}$

(ii) Probability that the white ball came from Bag 3:

Using Bayes' Theorem for Bag 3 and event White:

$P(\text{Bag 3 | White}) = \frac{P(\text{White | Bag 3}) P(\text{Bag 3})}{P(\text{White})}$

Substitute the known values:

$P(\text{Bag 3 | White}) = \frac{(1) \cdot \left(\frac{1}{2}\right)}{\frac{11}{18}}$

$P(\text{Bag 3 | White}) = \frac{\frac{1}{2}}{\frac{11}{18}}$

To simplify the fraction, multiply the numerator by the reciprocal of the denominator:

$P(\text{Bag 3 | White}) = \frac{1}{2} \times \frac{18}{11}$

$P(\text{Bag 3 | White}) = \frac{1 \times 18}{2 \times 11}$

$P(\text{Bag 3 | White}) = \frac{18}{22}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, 2:

$P(\text{Bag 3 | White}) = \frac{\cancel{18}^{9}}{\cancel{22}_{11}}$

$P(\text{Bag 3 | White}) = \frac{9}{11}$

Final Answers:

(i) The probability that the white ball came from Bag 2 is $\frac{2}{11}$.

(ii) The probability that the white ball came from Bag 3 is $\frac{9}{11}$.

Given:

Three types of flower seeds: A$_1$, A$_2$, and A$_3$.

The proportions of seeds in the mixture are 4 : 4 : 2.

Let the total number of parts be $4 + 4 + 2 = 10$.

The probability of selecting each type of seed is:

$P(A_1) = \frac{4}{10} = \frac{2}{5} = 0.4$

$P(A_2) = \frac{4}{10} = \frac{2}{5} = 0.4$

$P(A_3) = \frac{2}{10} = \frac{1}{5} = 0.2$

The germination rates are:

$P(\text{Germinate} | A_1) = 45\% = 0.45$

$P(\text{Germinate} | A_2) = 60\% = 0.60$

$P(\text{Germinate} | A_3) = 35\% = 0.35$

To Find:

(i) The probability that a randomly chosen seed germinates.

(ii) The probability that a seed does not germinate given that it is of type A$_3$.

(iii) The probability that the seed is of type A$_2$ given that a randomly chosen seed does not germinate.

Solution:

Let $G$ be the event that a randomly chosen seed germinates, and $G'$ be the event that it does not germinate.

We are given the conditional probabilities of germination:

$P(G | A_1) = 0.45$

$P(G | A_2) = 0.60$

$P(G | A_3) = 0.35$

The probabilities of a seed not germinating given the type are:

$P(G' | A_1) = 1 - P(G | A_1) = 1 - 0.45 = 0.55$

$P(G' | A_2) = 1 - P(G | A_2) = 1 - 0.60 = 0.40$

$P(G' | A_3) = 1 - P(G | A_3) = 1 - 0.35 = 0.65$

(i) Probability of a randomly chosen seed to germinate ($P(G)$):

Using the Law of Total Probability:

$P(G) = P(G | A_1)P(A_1) + P(G | A_2)P(A_2) + P(G | A_3)P(A_3)$

$P(G) = (0.45)(0.4) + (0.60)(0.4) + (0.35)(0.2)$

$P(G) = 0.18 + 0.24 + 0.07$

$P(G) = 0.49$

(ii) Probability that it will not germinate given that the seed is of type A$_3$ ($P(G' | A_3)$):

This is directly calculated using the complement rule from the given germination rate for A$_3$ seeds.

$P(G' | A_3) = 1 - P(G | A_3)$

$P(G' | A_3) = 1 - 0.35$

$P(G' | A_3) = 0.65$

(iii) Probability that it is of the type A$_2$ given that a randomly chosen seed does not germinate ($P(A_2 | G')$):

We use Bayes' Theorem:

$P(A_2 | G') = \frac{P(G' | A_2) P(A_2)}{P(G')}$

We already have $P(G' | A_2) = 0.40$ and $P(A_2) = 0.4$.

We need to find the total probability of a randomly chosen seed not germinating, $P(G')$. We can find this using the Law of Total Probability for $G'$.

$P(G') = P(G' | A_1)P(A_1) + P(G' | A_2)P(A_2) + P(G' | A_3)P(A_3)$

$P(G') = (0.55)(0.4) + (0.40)(0.4) + (0.65)(0.2)$

$P(G') = 0.22 + 0.16 + 0.13$

$P(G') = 0.51$

Alternatively, since we found $P(G)$ in part (i), we can use the complement rule for $P(G')$:

$P(G') = 1 - P(G) = 1 - 0.49 = 0.51$

Now substitute the values into the Bayes' Theorem formula:

$P(A_2 | G') = \frac{(0.40)(0.4)}{0.51}$

$P(A_2 | G') = \frac{0.16}{0.51}$

To express this as a fraction, multiply the numerator and denominator by 100:

$P(A_2 | G') = \frac{0.16 \times 100}{0.51 \times 100} = \frac{16}{51}$

Final Answers:

(i) The probability of a randomly chosen seed to germinate is $0.49$ or $\frac{49}{100}$.

(ii) The probability that it will not germinate given that the seed is of type A$_3$ is $0.65$ or $\frac{65}{100} = \frac{13}{20}$.

(iii) The probability that it is of the type A$_2$ given that a randomly chosen seed does not germinate is $\frac{16}{51}$.

Given:

The letter is known to have come from either TATA NAGAR or CALCUTTA.

Two consecutive letters visible on the envelope are "TA".

We assume, in the absence of any other information, that the letter is equally likely to have come from either place.

To Find:

The probability that the letter came from TATA NAGAR given that the visible consecutive letters are "TA".

Solution:

Let A be the event that the letter came from TATA NAGAR.

Let B be the event that the letter came from CALCUTTA.

Let E be the event that the two consecutive visible letters are "TA".

Based on our assumption of equal likelihood:

$P(A) = \frac{1}{2}$

$P(B) = \frac{1}{2}$

We need to find the conditional probabilities of observing "TA" in each word.

For TATA NAGAR:

The word has 10 letters. The number of consecutive pairs of letters is $10 - 1 = 9$.

The consecutive pairs are: (T,A), (A,T), (T,A), (A, ), ( ,N), (N,A), (A,G), (G,A), (A,R).

The pairs which are "TA" are the 1st and 3rd pairs. There are 2 such pairs.

The probability of observing "TA" given the letter is from TATA NAGAR is:

$P(E | A) = \frac{\text{Number of "TA" pairs}}{\text{Total number of consecutive pairs}} = \frac{2}{9}$

For CALCUTTA:

The word has 8 letters. The number of consecutive pairs of letters is $8 - 1 = 7$.

The consecutive pairs are: (C,A), (A,L), (L,C), (C,U), (U,T), (T,T), (T,A).

The pair which is "TA" is the 7th pair. There is 1 such pair.

The probability of observing "TA" given the letter is from CALCUTTA is:

$P(E | B) = \frac{\text{Number of "TA" pairs}}{\text{Total number of consecutive pairs}} = \frac{1}{7}$

Now, we calculate the total probability of observing "TA" using the Law of Total Probability:

$P(E) = P(E | A)P(A) + P(E | B)P(B)$

$P(E) = \left(\frac{2}{9}\right) \cdot \left(\frac{1}{2}\right) + \left(\frac{1}{7}\right) \cdot \left(\frac{1}{2}\right)$

$P(E) = \frac{2}{18} + \frac{1}{14} = \frac{1}{9} + \frac{1}{14}$

Finding a common denominator (LCM of 9 and 14 is 126):

$P(E) = \frac{1 \cdot 14}{9 \cdot 14} + \frac{1 \cdot 9}{14 \cdot 9} = \frac{14}{126} + \frac{9}{126} = \frac{14 + 9}{126} = \frac{23}{126}$

We want to find the probability that the letter came from TATA NAGAR given that "TA" is visible, which is $P(A | E)$. We use Bayes' Theorem:

$P(A | E) = \frac{P(E | A) P(A)}{P(E)}$

Substitute the calculated values:

$P(A | E) = \frac{\left(\frac{2}{9}\right) \cdot \left(\frac{1}{2}\right)}{\frac{23}{126}}$

$P(A | E) = \frac{\frac{\cancel{2}^{1}}{9 \cdot \cancel{2}_{1}}}{\frac{23}{126}} = \frac{\frac{1}{9}}{\frac{23}{126}}$

To divide the fractions, multiply the numerator by the reciprocal of the denominator:

$P(A | E) = \frac{1}{9} \times \frac{126}{23}$

Cancel out common factors (126 divided by 9 is 14):

$P(A | E) = \frac{1}{\cancel{9}_{1}} \times \frac{\cancel{126}^{14}}{23}$

$P(A | E) = \frac{1 \times 14}{1 \times 23} = \frac{14}{23}$

Final Answer:

The probability that the letter came from TATA NAGAR given that the visible consecutive letters are "TA" is $\frac{14}{23}$.

Given:

There are two bags.

Bag 1 contains 3 black balls and 4 white balls. Total balls in Bag 1 = $3 + 4 = 7$.

Bag 2 contains 4 black balls and 3 white balls. Total balls in Bag 2 = $4 + 3 = 7$.

A standard die is thrown.

If the die shows 1 or 3, Bag 1 is chosen.

If the die shows any other number (2, 4, 5, or 6), Bag 2 is chosen.

To Find:

The probability of choosing a black ball.

Solution:

Let $D_1$ be the event that the die shows 1 or 3.

Let $D_2$ be the event that the die shows any other number (2, 4, 5, or 6).

The possible outcomes when a die is thrown are {1, 2, 3, 4, 5, 6}. Total number of outcomes = 6.

Number of outcomes for $D_1$ (1 or 3) = 2.

$P(D_1) = \frac{\text{Number of favorable outcomes for } D_1}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}$

Number of outcomes for $D_2$ (2, 4, 5, or 6) = 4.

$P(D_2) = \frac{\text{Number of favorable outcomes for } D_2}{\text{Total number of outcomes}} = \frac{4}{6} = \frac{2}{3}$

Note that $P(D_1) + P(D_2) = \frac{1}{3} + \frac{2}{3} = 1$.

Let $B$ be the event that a black ball is chosen.

Let $W$ be the event that a white ball is chosen.

The probability of choosing a black ball given that Bag 1 was chosen is:

$P(B | D_1) = \frac{\text{Number of black balls in Bag 1}}{\text{Total balls in Bag 1}} = \frac{3}{7}$

The probability of choosing a black ball given that Bag 2 was chosen is:

$P(B | D_2) = \frac{\text{Number of black balls in Bag 2}}{\text{Total balls in Bag 2}} = \frac{4}{7}$

To find the overall probability of choosing a black ball, we use the Law of Total Probability:

$P(B) = P(B | D_1)P(D_1) + P(B | D_2)P(D_2)$

Substitute the probabilities we calculated:

$P(B) = \left(\frac{3}{7}\right) \cdot \left(\frac{1}{3}\right) + \left(\frac{4}{7}\right) \cdot \left(\frac{2}{3}\right)$

$P(B) = \frac{3}{21} + \frac{8}{21}$

$P(B) = \frac{3 + 8}{21}$

$P(B) = \frac{11}{21}$

Final Answer:

The probability of choosing a black ball is $\frac{11}{21}$.

Given:

Three urns with the following contents:

Urn 1: 2 white balls, 3 black balls. Total = 5 balls.

Urn 2: 3 white balls, 2 black balls. Total = 5 balls.

Urn 3: 4 white balls, 1 black ball. Total = 5 balls.

There is an equal probability of choosing each urn.

A ball is drawn at random from the chosen urn and it is white.

To Find:

The probability that the ball drawn was from the second urn, given that it was white.

Solution:

Let $U_1$, $U_2$, and $U_3$ be the events that Urn 1, Urn 2, and Urn 3 are chosen, respectively.

Let $W$ be the event that a white ball is drawn.

Since there is an equal probability of choosing each urn, we have:

$P(U_1) = P(U_2) = P(U_3) = \frac{1}{3}$

The conditional probabilities of drawing a white ball from each urn are:

$P(W | U_1) = \frac{\text{Number of white balls in Urn 1}}{\text{Total balls in Urn 1}} = \frac{2}{5}$

$P(W | U_2) = \frac{\text{Number of white balls in Urn 2}}{\text{Total balls in Urn 2}} = \frac{3}{5}$

$P(W | U_3) = \frac{\text{Number of white balls in Urn 3}}{\text{Total balls in Urn 3}} = \frac{4}{5}$

We need to find $P(U_2 | W)$, the probability that the ball was drawn from Urn 2 given that it was white. We can use Bayes' Theorem for this:

$P(U_2 | W) = \frac{P(W | U_2) P(U_2)}{P(W)}$

First, we need to find the total probability of drawing a white ball, $P(W)$, using the Law of Total Probability:

$P(W) = P(W | U_1)P(U_1) + P(W | U_2)P(U_2) + P(W | U_3)P(U_3)$

Substitute the values:

$P(W) = \left(\frac{2}{5}\right)\left(\frac{1}{3}\right) + \left(\frac{3}{5}\right)\left(\frac{1}{3}\right) + \left(\frac{4}{5}\right)\left(\frac{1}{3}\right)$

$P(W) = \frac{2}{15} + \frac{3}{15} + \frac{4}{15}$

$P(W) = \frac{2+3+4}{15} = \frac{9}{15} = \frac{3}{5}$

Now, substitute $P(W | U_2)$, $P(U_2)$, and $P(W)$ into Bayes' Theorem:

$P(U_2 | W) = \frac{\left(\frac{3}{5}\right) \cdot \left(\frac{1}{3}\right)}{\frac{3}{5}}$

$P(U_2 | W) = \frac{\frac{\cancel{3}^{1}}{5 \cdot \cancel{3}_{1}}}{\frac{3}{5}} = \frac{\frac{1}{5}}{\frac{3}{5}}$

To divide the fractions, multiply the numerator by the reciprocal of the denominator:

$P(U_2 | W) = \frac{1}{5} \times \frac{5}{3}$

$P(U_2 | W) = \frac{1 \times \cancel{5}^{1}}{\cancel{5}_{1} \times 3} = \frac{1}{3}$

Final Answer:

The probability that the white ball drawn was from the second urn is $\frac{1}{3}$.

Given:

Let T be the event that a person actually has TB.

Let H be the event that a person is healthy (does not have TB).

Let D be the event that a person is diagnosed to have TB by chest X-ray.

The probability that TB is detected when a person is actually suffering (True Positive Rate or Sensitivity):

$P(D | T) = 0.99$

The probability of a healthy person being diagnosed with TB (False Positive Rate):

$P(D | H) = 0.001$

The prevalence of TB in the city:

$P(T) = \frac{1}{1000} = 0.001$

To Find:

The probability that the person actually has TB, given that they are diagnosed with TB ($P(T | D)$).

Solution:

We can use Bayes' Theorem to find the required probability $P(T | D)$. The formula is:

$P(T | D) = \frac{P(D | T) P(T)}{P(D)}$

First, we need to find the total probability of being diagnosed with TB, $P(D)$. We can use the Law of Total Probability:

$P(D) = P(D | T) P(T) + P(D | H) P(H)$

We know $P(T) = 0.001$. The probability of being healthy is the complement of having TB:

$P(H) = 1 - P(T) = 1 - 0.001 = 0.999$

Now, substitute the values into the Law of Total Probability formula for $P(D)$:

$P(D) = (0.99)(0.001) + (0.001)(0.999)$

$P(D) = 0.00099 + 0.000999$

$P(D) = 0.001989$

Now, substitute the values into Bayes' Theorem to find $P(T | D)$:

$P(T | D) = \frac{P(D | T) P(T)}{P(D)}$

$P(T | D) = \frac{(0.99)(0.001)}{0.001989}$

$P(T | D) = \frac{0.00099}{0.001989}$

To remove decimals, we can multiply the numerator and denominator by 1,000,000:

$P(T | D) = \frac{0.00099 \times 1,000,000}{0.001989 \times 1,000,000} = \frac{990}{1989}$

We can simplify this fraction. Both 990 and 1989 are divisible by 9.

$990 \div 9 = 110$

$1989 \div 9 = 221$

So,

$P(T | D) = \frac{110}{221}$

Final Answer:

The probability that the person actually has TB given that they are diagnosed with TB is $\frac{110}{221}$.

Given:

Items are manufactured by three machines A, B, and C.

Proportion of items manufactured by each machine:

$P(A) = 50\% = 0.50$

$P(B) = 30\% = 0.30$

$P(C) = 20\% = 0.20$

Note that $P(A) + P(B) + P(C) = 0.50 + 0.30 + 0.20 = 1.00$.

Defective rates for items produced on each machine:

Probability of an item being defective given it was produced on machine A:

$P(\text{Defective} | A) = 2\% = 0.02$

Probability of an item being defective given it was produced on machine B:

$P(\text{Defective} | B) = 2\% = 0.02$

Probability of an item being defective given it was produced on machine C:

$P(\text{Defective} | C) = 3\% = 0.03$

One item is drawn at random from the godown and is found to be defective.

To Find:

The probability that the defective item was manufactured on machine A ($P(A | \text{Defective})$).

Solution:

Let D be the event that a randomly drawn item is defective.

We need to find $P(A | D)$, which is the probability that the item was manufactured on machine A given that it is defective.

We can use Bayes' Theorem:

$P(A | D) = \frac{P(D | A) P(A)}{P(D)}$

First, we need to find the total probability of drawing a defective item, $P(D)$, using the Law of Total Probability:

$P(D) = P(D | A)P(A) + P(D | B)P(B) + P(D | C)P(C)$

Substitute the given values:

$P(D) = (0.02)(0.50) + (0.02)(0.30) + (0.03)(0.20)$

$P(D) = 0.0100 + 0.0060 + 0.0060$

$P(D) = 0.0220$

Now, substitute $P(D | A)$, $P(A)$, and $P(D)$ into Bayes' Theorem:

$P(A | D) = \frac{(0.02)(0.50)}{0.0220}$

$P(A | D) = \frac{0.0100}{0.0220}$

To remove decimals, we can multiply the numerator and denominator by 1000:

$P(A | D) = \frac{0.0100 \times 1000}{0.0220 \times 1000} = \frac{10}{22}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, 2:

$P(A | D) = \frac{\cancel{10}^{5}}{\cancel{22}_{11}}$

$P(A | D) = \frac{5}{11}$

Final Answer:

The probability that the defective item was manufactured on machine A is $\frac{5}{11}$.

Given:

The probability distribution of the discrete random variable X is defined as:

$P(X = x) = \begin{cases}k (x + 1) & \text{for } x = 1,\; 2,\; 3,\; 4 \\ 2 kx & \text{for } x = 5,\; 6,\; 7 \\ 0 & \text{otherwise} \end{cases}$

To Find:

(i) The value of k

(ii) E (X)

(iii) Standard deviation of X

Solution:

(i) To find the value of k, we use the fundamental property of a probability distribution that the sum of probabilities for all possible values of the random variable must be equal to 1. That is, $\sum\limits_{x} P(X=x) = 1$.

The possible values for X are $x = 1, 2, 3, 4, 5, 6, 7$.

We calculate the probability for each of these values using the given piecewise function:

$P(X=1) = k(1+1) = 2k$

$P(X=2) = k(2+1) = 3k$

$P(X=3) = k(3+1) = 4k$

$P(X=4) = k(4+1) = 5k$

$P(X=5) = 2k(5) = 10k$

$P(X=6) = 2k(6) = 12k$

$P(X=7) = 2k(7) = 14k$

Summing these probabilities:

$\sum\limits_{x=1}^{7} P(X=x) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7)$

$\sum\limits_{x=1}^{7} P(X=x) = 2k + 3k + 4k + 5k + 10k + 12k + 14k$

$\sum\limits_{x=1}^{7} P(X=x) = (2 + 3 + 4 + 5 + 10 + 12 + 14)k = 50k$

Setting the sum equal to 1:

From equation (i), we can solve for k:

$k = \frac{1}{50}$

(ii) To find the expected value E(X), we use the formula $E(X) = \sum\limits_{x} x P(X=x)$.

$E(X) = \sum\limits_{x=1}^{7} x P(X=x)$

$E(X) = 1 \cdot P(X=1) + 2 \cdot P(X=2) + 3 \cdot P(X=3) + 4 \cdot P(X=4) + 5 \cdot P(X=5) + 6 \cdot P(X=6) + 7 \cdot P(X=7)$

Substituting the values of $P(X=x)$ in terms of k:

$E(X) = 1(2k) + 2(3k) + 3(4k) + 4(5k) + 5(10k) + 6(12k) + 7(14k)$

$E(X) = 2k + 6k + 12k + 20k + 50k + 72k + 98k$

$E(X) = (2 + 6 + 12 + 20 + 50 + 72 + 98)k = 260k$

Now, substitute the value of $k = \frac{1}{50}$ found in part (i):

$E(X) = 260 \cdot \frac{1}{50}$

$E(X) = \frac{260}{50} = \frac{26}{5} = 5.2$

(iii) To find the standard deviation of X, we first need to calculate the variance of X using the formula $Var(X) = E(X^2) - (E(X))^2$. The standard deviation, $\sigma_X$, is the square root of the variance, $\sigma_X = \sqrt{Var(X)}$.

First, we calculate $E(X^2)$, which is the expected value of $X^2$. The formula is $E(X^2) = \sum\limits_{x} x^2 P(X=x)$.

$E(X^2) = \sum\limits_{x=1}^{7} x^2 P(X=x)$

$E(X^2) = 1^2 P(X=1) + 2^2 P(X=2) + 3^2 P(X=3) + 4^2 P(X=4) + 5^2 P(X=5) + 6^2 P(X=6) + 7^2 P(X=7)$

Substituting the values of $P(X=x)$ in terms of k:

$E(X^2) = 1(2k) + 4(3k) + 9(4k) + 16(5k) + 25(10k) + 36(12k) + 49(14k)$

$E(X^2) = 2k + 12k + 36k + 80k + 250k + 432k + 686k$

$E(X^2) = (2 + 12 + 36 + 80 + 250 + 432 + 686)k = 1498k$

Substituting the value of $k = \frac{1}{50}$:

$E(X^2) = 1498 \cdot \frac{1}{50} = \frac{1498}{50} = \frac{149.8}{5} = 29.96$

Now, we calculate the variance $Var(X)$ using the formula $Var(X) = E(X^2) - (E(X))^2$.

We have $E(X^2) = 29.96$ and from part (ii), $E(X) = 5.2$.

$Var(X) = 29.96 - (5.2)^2$

$Var(X) = 29.96 - 27.04$

$Var(X) = 2.92$

Finally, we calculate the standard deviation $\sigma_X$ by taking the square root of the variance:

$\sigma_X = \sqrt{Var(X)} = \sqrt{2.92}$

Using a calculator, the approximate value of the standard deviation is:

$\sigma_X \approx 1.7088$

Solution:

Given the probability distribution of a discrete random variable X and $E(X) = 2.94$.

(i) The value of A if E(X) = 2.94

The expected value of a discrete random variable X is given by the formula:

$E(X) = \sum X \cdot P(X)$

Using the given distribution:

$E(X) = 1 \cdot P(X=1) + 2 \cdot P(X=2) + 4 \cdot P(X=4) + 2A \cdot P(X=2A) + 3A \cdot P(X=3A) + 5A \cdot P(X=5A)$

Substitute the given probabilities and $E(X) = 2.94$:

$2.94 = 1 \left(\frac{1}{2}\right) + 2 \left(\frac{1}{5}\right) + 4 \left(\frac{3}{25}\right) + 2A \left(\frac{1}{10}\right) + 3A \left(\frac{1}{25}\right) + 5A \left(\frac{1}{25}\right)$

$2.94 = \frac{1}{2} + \frac{2}{5} + \frac{12}{25} + \frac{2A}{10} + \frac{3A}{25} + \frac{5A}{25}$

To combine the terms, find a common denominator for the fractions, which is 50:

$2.94 = \frac{25}{50} + \frac{20}{50} + \frac{24}{50} + \frac{10A}{50} + \frac{6A}{50} + \frac{10A}{50}$

$2.94 = \frac{25+20+24}{50} + \frac{10A+6A+10A}{50}$

$2.94 = \frac{69}{50} + \frac{26A}{50}$

$2.94 = 1.38 + \frac{13A}{25}$

Subtract 1.38 from both sides:

$2.94 - 1.38 = \frac{13A}{25}$

$1.56 = \frac{13A}{25}$

Multiply both sides by 25:

$13A = 1.56 \times 25$

$13A = 39$

Divide by 13:

$A = \frac{39}{13}$

Thus, the value of A is 3. With A=3, the values of the random variable X are $1, 2, 4, 2(3)=6, 3(3)=9, 5(3)=15$.

(ii) Variance of X

The variance of a discrete random variable X is given by the formula:

$Var(X) = E(X^2) - [E(X)]^2$

We are given $E(X) = 2.94$. Therefore, $[E(X)]^2 = (2.94)^2 = 8.6436$.

Now, we calculate $E(X^2) = \sum X^2 \cdot P(X)$ using the values of X with $A=3$ ($1, 2, 4, 6, 9, 15$):

$E(X^2) = (1)^2 \left(\frac{1}{2}\right) + (2)^2 \left(\frac{1}{5}\right) + (4)^2 \left(\frac{3}{25}\right) + (6)^2 \left(\frac{1}{10}\right) + (9)^2 \left(\frac{1}{25}\right) + (15)^2 \left(\frac{1}{25}\right)$

$E(X^2) = 1 \left(\frac{1}{2}\right) + 4 \left(\frac{1}{5}\right) + 16 \left(\frac{3}{25}\right) + 36 \left(\frac{1}{10}\right) + 81 \left(\frac{1}{25}\right) + 225 \left(\frac{1}{25}\right)$

$E(X^2) = \frac{1}{2} + \frac{4}{5} + \frac{48}{25} + \frac{36}{10} + \frac{81}{25} + \frac{225}{25}$

Convert fractions to a common denominator (50):

$E(X^2) = \frac{25}{50} + \frac{40}{50} + \frac{96}{50} + \frac{180}{50} + \frac{162}{50} + \frac{450}{50}$

$E(X^2) = \frac{25+40+96+180+162+450}{50}$

$E(X^2) = \frac{953}{50}$

$E(X^2) = 19.06$

Now, substitute the values of $E(X^2)$ and $[E(X)]^2$ into the variance formula:

$Var(X) = E(X^2) - [E(X)]^2$

$Var(X) = 19.06 - 8.6436$

The variance of X is 10.4164.

Final Answer:

(i) The value of A is 3.

(ii) The variance of X is 10.4164.

Solution:

Given the probability distribution of a random variable X.

First, we need to find the value of the constant k. The sum of all probabilities for a probability distribution must be equal to 1. That is, $\sum P(X=x) = 1$ for all possible values of x.

The possible values for X are 1, 2, 3, 4, 5, and 6.

$P(X=1) = k(1)^2 = k$

$P(X=2) = k(2)^2 = 4k$

$P(X=3) = k(3)^2 = 9k$

$P(X=4) = 2k(4) = 8k$

$P(X=5) = 2k(5) = 10k$

$P(X=6) = 2k(6) = 12k$

Sum of probabilities:

$P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) = 1$

$k + 4k + 9k + 8k + 10k + 12k = 1$

$(1 + 4 + 9 + 8 + 10 + 12)k = 1$

$44k = 1$

Now we can use the value of k to calculate the required quantities.

(i) E(X)

The expected value $E(X)$ is given by $E(X) = \sum x \cdot P(X=x)$.

$E(X) = 1 \cdot P(X=1) + 2 \cdot P(X=2) + 3 \cdot P(X=3) + 4 \cdot P(X=4) + 5 \cdot P(X=5) + 6 \cdot P(X=6)$

$E(X) = 1(k) + 2(4k) + 3(9k) + 4(8k) + 5(10k) + 6(12k)$

$E(X) = k + 8k + 27k + 32k + 50k + 72k$

$E(X) = (1 + 8 + 27 + 32 + 50 + 72)k$

$E(X) = 190k$

Substitute $k = \frac{1}{44}$ from (i):

$E(X) = 190 \times \frac{1}{44} = \frac{190}{44}$

Simplifying the fraction:

(ii) E(3X²)

We need to calculate $E(X^2)$ first, which is given by $E(X^2) = \sum x^2 \cdot P(X=x)$.

$E(X^2) = (1)^2 P(X=1) + (2)^2 P(X=2) + (3)^2 P(X=3) + (4)^2 P(X=4) + (5)^2 P(X=5) + (6)^2 P(X=6)$

$E(X^2) = 1(k) + 4(4k) + 9(9k) + 16(8k) + 25(10k) + 36(12k)$

$E(X^2) = k + 16k + 81k + 128k + 250k + 432k$

$E(X^2) = (1 + 16 + 81 + 128 + 250 + 432)k$

$E(X^2) = 908k$

Substitute $k = \frac{1}{44}$ from (i):

$E(X^2) = 908 \times \frac{1}{44} = \frac{908}{44}$

Simplifying the fraction:

$E(X^2) = \frac{227}{11}$

Now, we calculate $E(3X^2)$:

$E(3X^2) = 3 \cdot E(X^2)$

$E(3X^2) = 3 \times \frac{227}{11}$

(iii) P(X ≥ 4)

The probability $P(X \geq 4)$ is the sum of the probabilities for X = 4, 5, and 6.

$P(X \geq 4) = P(X=4) + P(X=5) + P(X=6)$

Using the expressions for these probabilities with k:

$P(X \geq 4) = 8k + 10k + 12k$

$P(X \geq 4) = (8 + 10 + 12)k$

$P(X \geq 4) = 30k$

Substitute $k = \frac{1}{44}$ from (i):

$P(X \geq 4) = 30 \times \frac{1}{44} = \frac{30}{44}$

Simplifying the fraction:

Final Answer:

(i) $E(X) = \frac{95}{22}$

(ii) $E(3X^2) = \frac{681}{11}$

(iii) $P(X \geq 4) = \frac{15}{22}$

Solution:

Given:

Total number of coins in the bag = $2n + 1$.

Number of double-headed coins = $n$.

Number of fair coins = $(2n+1) - n = n+1$.

Probability of getting a head when a coin is picked at random and tossed = $\frac{31}{42}$.

To Find:

The value of $n$.

Let H be the event that the toss results in a head.

Let A be the event that a double-headed coin is picked from the bag.

Let B be the event that a fair coin is picked from the bag.

The total number of coins is $2n+1$.

The number of double-headed coins is $n$.

The number of fair coins is $n+1$.

The probability of picking a double-headed coin is:

$P(A) = \frac{\text{Number of double-headed coins}}{\text{Total number of coins}} = \frac{n}{2n+1}$

The probability of picking a fair coin is:

$P(B) = \frac{\text{Number of fair coins}}{\text{Total number of coins}} = \frac{n+1}{2n+1}$

The probability of getting a head given a double-headed coin is picked is:

$P(H|A) = 1$ (Since a double-headed coin always results in a head)

The probability of getting a head given a fair coin is picked is:

$P(H|B) = \frac{1}{2}$ (Since a fair coin has a head on one side)

Using the Law of Total Probability, the probability of getting a head is:

$P(H) = P(H|A)P(A) + P(H|B)P(B)$

We are given that $P(H) = \frac{31}{42}$. Substitute the probabilities into the formula:

Now, we solve this equation for $n$:

$\frac{31}{42} = \frac{n}{2n+1} + \frac{n+1}{2(2n+1)}$

Combine the terms on the right side by finding a common denominator:

$\frac{31}{42} = \frac{2 \cdot n}{2(2n+1)} + \frac{n+1}{2(2n+1)}$

$\frac{31}{42} = \frac{2n + (n+1)}{2(2n+1)}$

$\frac{31}{42} = \frac{3n+1}{4n+2}$

Cross-multiply:

$31 \times (4n+2) = 42 \times (3n+1)$

Distribute the numbers on both sides:

$31 \times 4n + 31 \times 2 = 42 \times 3n + 42 \times 1$

$124n + 62 = 126n + 42$

Rearrange the terms to group n on one side and constants on the other:

$62 - 42 = 126n - 124n$

$20 = 2n$

Divide by 2:

The value of $n$ is 10.

Final Answer:

The value of $n$ is 10.

Solution:

Given:

A well-shuffled deck of 52 cards.

Two cards are drawn successively without replacement.

X is the random variable representing the number of aces drawn.

To Find:

(i) The mean of X, $E(X)$.

(ii) The standard deviation of X, $\sigma(X)$.

The possible values for the random variable X (number of aces drawn in two cards) are 0, 1, or 2.

Total number of cards in the deck = 52.

Number of aces in the deck = 4.

Number of non-aces in the deck = $52 - 4 = 48$.

Total number of ways to draw 2 cards from 52 without replacement is given by the number of combinations $\binom{52}{2}$.

Total number of ways = $\binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326$.

Now, we calculate the probability for each possible value of X:

P(X=0): This means drawing 0 aces and 2 non-aces. The number of ways to do this is $\binom{4}{0} \times \binom{48}{2}$.

Number of ways to get 0 aces = $\binom{4}{0} \times \binom{48}{2} = 1 \times \frac{48 \times 47}{2 \times 1} = 1 \times 24 \times 47 = 1128$.

$P(X=0) = \frac{\text{Number of ways to get 0 aces}}{\text{Total number of ways to draw 2 cards}} = \frac{1128}{1326}$.

Simplify the fraction: $\frac{1128}{1326} = \frac{\cancel{1128}^{188}}{\cancel{1326}_{221}} = \frac{188}{221}$.

P(X=1): This means drawing 1 ace and 1 non-ace. The number of ways to do this is $\binom{4}{1} \times \binom{48}{1}$.

Number of ways to get 1 ace = $\binom{4}{1} \times \binom{48}{1} = 4 \times 48 = 192$.

$P(X=1) = \frac{\text{Number of ways to get 1 ace}}{\text{Total number of ways to draw 2 cards}} = \frac{192}{1326}$.

Simplify the fraction: $\frac{192}{1326} = \frac{\cancel{192}^{32}}{\cancel{1326}_{221}} = \frac{32}{221}$.

P(X=2): This means drawing 2 aces and 0 non-aces. The number of ways to do this is $\binom{4}{2} \times \binom{48}{0}$.

Number of ways to get 2 aces = $\binom{4}{2} \times \binom{48}{0} = \frac{4 \times 3}{2 \times 1} \times 1 = 6 \times 1 = 6$.

$P(X=2) = \frac{\text{Number of ways to get 2 aces}}{\text{Total number of ways to draw 2 cards}} = \frac{6}{1326}$.

Simplify the fraction: $\frac{6}{1326} = \frac{\cancel{6}^{1}}{\cancel{1326}_{221}} = \frac{1}{221}$.

The probability distribution of X is:

Check: $P(X=0) + P(X=1) + P(X=2) = \frac{188}{221} + \frac{32}{221} + \frac{1}{221} = \frac{188+32+1}{221} = \frac{221}{221} = 1$. The probabilities are valid.

(i) Mean of X, E(X)

The mean (expected value) of a discrete random variable X is given by $E(X) = \sum x \cdot P(X=x)$.

$E(X) = 0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2)$

$E(X) = 0 \left(\frac{188}{221}\right) + 1 \left(\frac{32}{221}\right) + 2 \left(\frac{1}{221}\right)$

$E(X) = 0 + \frac{32}{221} + \frac{2}{221}$

$E(X) = \frac{32+2}{221} = \frac{34}{221}$

Simplify the fraction:

(ii) Standard deviation of X, $\sigma(X)$

The standard deviation is the square root of the variance, $\sigma(X) = \sqrt{Var(X)}$.

The variance is given by $Var(X) = E(X^2) - [E(X)]^2$.

First, we calculate $E(X^2) = \sum x^2 \cdot P(X=x)$.

$E(X^2) = (0)^2 \cdot P(X=0) + (1)^2 \cdot P(X=1) + (2)^2 \cdot P(X=2)$

$E(X^2) = 0^2 \left(\frac{188}{221}\right) + 1^2 \left(\frac{32}{221}\right) + 2^2 \left(\frac{1}{221}\right)$

$E(X^2) = 0 \left(\frac{188}{221}\right) + 1 \left(\frac{32}{221}\right) + 4 \left(\frac{1}{221}\right)$

$E(X^2) = 0 + \frac{32}{221} + \frac{4}{221}$

$E(X^2) = \frac{32+4}{221} = \frac{36}{221}$

Now, substitute $E(X^2)$ and $E(X)$ into the variance formula:

$Var(X) = E(X^2) - [E(X)]^2$

$Var(X) = \frac{36}{221} - \left(\frac{2}{13}\right)^2$

$Var(X) = \frac{36}{221} - \frac{4}{169}$

The least common denominator for 221 (which is $13 \times 17$) and 169 (which is $13 \times 13$) is $13 \times 13 \times 17 = 169 \times 17 = 2873$.

$Var(X) = \frac{36 \times 13}{221 \times 13} - \frac{4 \times 17}{169 \times 17}$

$Var(X) = \frac{468}{2873} - \frac{68}{2873}$

$Var(X) = \frac{468 - 68}{2873} = \frac{400}{2873}$

Finally, calculate the standard deviation:

$\sigma(X) = \sqrt{Var(X)} = \sqrt{\frac{400}{2873}}$

Final Answer:

The mean of the random variable X is $E(X) = \frac{2}{13}$.

The standard deviation of the random variable X is $\sigma(X) = \frac{20}{\sqrt{2873}}$.

Solution:

Given:

A fair die is tossed twice.

A 'success' is defined as getting an even number on a single toss.

X is the random variable representing the number of successes in two tosses.

To Find:

The variance of X, $Var(X)$.

On a single toss of a fair die, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is 6.

The outcomes corresponding to a 'success' (getting an even number) are $\{2, 4, 6\}$. The number of successful outcomes is 3.

The probability of success on a single toss is:

$p = P(\text{Success}) = P(\text{Even number}) = \frac{\text{Number of even outcomes}}{\text{Total number of outcomes}} = \frac{3}{6} = \frac{1}{2}$

The probability of failure on a single toss is:

$q = P(\text{Failure}) = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$

The die is tossed twice, and each toss is an independent event. The random variable X represents the number of successes in these two tosses.

The possible values that X can take are 0, 1, or 2.

We calculate the probability for each value of X:

$P(X=0)$: Getting 0 successes in 2 tosses means getting failure on the first toss and failure on the second toss.

$P(X=0) = P(\text{Failure on 1st}) \times P(\text{Failure on 2nd}) = q \times q = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$P(X=1)$: Getting 1 success in 2 tosses means getting success on one toss and failure on the other. There are two ways this can happen: (Success, Failure) or (Failure, Success).

$P(X=1) = P(\text{Success on 1st and Failure on 2nd}) + P(\text{Failure on 1st and Success on 2nd})$

$P(X=1) = (p \times q) + (q \times p) = \left(\frac{1}{2} \times \frac{1}{2}\right) + \left(\frac{1}{2} \times \frac{1}{2}\right) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

$P(X=2)$: Getting 2 successes in 2 tosses means getting success on the first toss and success on the second toss.

$P(X=2) = P(\text{Success on 1st}) \times P(\text{Success on 2nd}) = p \times p = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

The probability distribution of X is:

Check: $\sum P(X) = \frac{1}{4} + \frac{1}{2} + \frac{1}{4} = \frac{1+2+1}{4} = \frac{4}{4} = 1$. The distribution is valid.

To find the variance $Var(X)$, we use the formula $Var(X) = E(X^2) - [E(X)]^2$.

First, calculate the expected value (mean) of X, $E(X) = \sum\limits_{x} x \cdot P(X=x)$.

$E(X) = 0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2)$

$E(X) = 0 \left(\frac{1}{4}\right) + 1 \left(\frac{1}{2}\right) + 2 \left(\frac{1}{4}\right)$

$E(X) = 0 + \frac{1}{2} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1$

Next, calculate the expected value of $X^2$, $E(X^2) = \sum\limits_{x} x^2 \cdot P(X=x)$.

$E(X^2) = (0)^2 P(X=0) + (1)^2 P(X=1) + (2)^2 P(X=2)$

$E(X^2) = 0^2 \left(\frac{1}{4}\right) + 1^2 \left(\frac{1}{2}\right) + 2^2 \left(\frac{1}{4}\right)$

$E(X^2) = 0 \left(\frac{1}{4}\right) + 1 \left(\frac{1}{2}\right) + 4 \left(\frac{1}{4}\right)$

$E(X^2) = 0 + \frac{1}{2} + \frac{4}{4} = \frac{1}{2} + 1 = \frac{1}{2} + \frac{2}{2} = \frac{3}{2}$

Now, substitute $E(X)$ and $E(X^2)$ into the variance formula:

$Var(X) = E(X^2) - [E(X)]^2$

$Var(X) = \frac{3}{2} - (1)^2$

$Var(X) = \frac{3}{2} - 1$

$Var(X) = \frac{3}{2} - \frac{2}{2}$

Alternatively, since X follows a binomial distribution $B(n, p)$ with $n=2$ and $p=\frac{1}{2}$, the variance is given by $Var(X) = npq$.

$Var(X) = 2 \times \frac{1}{2} \times \frac{1}{2} = \frac{2}{4} = \frac{1}{2}$.

Final Answer:

The variance of the number of successes is $\frac{1}{2}$.

Solution:

Given:

5 cards numbered 1, 2, 3, 4, 5.

Two cards are drawn at random without replacement.

X is the random variable representing the sum of the numbers on the two cards drawn.

To Find:

(i) The mean of X, $E(X)$.

(ii) The variance of X, $Var(X)$.

Total number of cards = 5.

Number of cards drawn = 2.

The cards are drawn without replacement, and the order of drawing does not affect the sum. So, the total number of possible pairs of cards drawn is the number of combinations of choosing 2 cards from 5.

Total number of ways to draw 2 cards = $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

The possible pairs of cards and their sums are:

• (1, 2) $\implies$ Sum = 3
• (1, 3) $\implies$ Sum = 4
• (1, 4) $\implies$ Sum = 5
• (1, 5) $\implies$ Sum = 6
• (2, 3) $\implies$ Sum = 5
• (2, 4) $\implies$ Sum = 6
• (2, 5) $\implies$ Sum = 7
• (3, 4) $\implies$ Sum = 7
• (3, 5) $\implies$ Sum = 8
• (4, 5) $\implies$ Sum = 9

The possible values for the random variable X (the sum) are {3, 4, 5, 6, 7, 8, 9}.

Now, we determine the probability for each possible value of X by counting the number of pairs that result in that sum and dividing by the total number of ways (10):

$P(X=3)$: Only one pair (1, 2). $P(X=3) = \frac{1}{10}$.

$P(X=4)$: Only one pair (1, 3). $P(X=4) = \frac{1}{10}$.

$P(X=5)$: Two pairs (1, 4) and (2, 3). $P(X=5) = \frac{2}{10} = \frac{1}{5}$.

$P(X=6)$: Two pairs (1, 5) and (2, 4). $P(X=6) = \frac{2}{10} = \frac{1}{5}$.

$P(X=7)$: Two pairs (2, 5) and (3, 4). $P(X=7) = \frac{2}{10} = \frac{1}{5}$.

$P(X=8)$: Only one pair (3, 5). $P(X=8) = \frac{1}{10}$.

$P(X=9)$: Only one pair (4, 5). $P(X=9) = \frac{1}{10}$.

The probability distribution of X is:

Check: $\sum P(X) = \frac{1}{10} + \frac{1}{10} + \frac{2}{10} + \frac{2}{10} + \frac{2}{10} + \frac{1}{10} + \frac{1}{10} = \frac{1+1+2+2+2+1+1}{10} = \frac{10}{10} = 1$. The distribution is valid.

(i) Mean of X, E(X)

The mean (expected value) of a discrete random variable X is given by $E(X) = \sum\limits_{x} x \cdot P(X=x)$.

$E(X) = 3 \left(\frac{1}{10}\right) + 4 \left(\frac{1}{10}\right) + 5 \left(\frac{2}{10}\right) + 6 \left(\frac{2}{10}\right) + 7 \left(\frac{2}{10}\right) + 8 \left(\frac{1}{10}\right) + 9 \left(\frac{1}{10}\right)$

$E(X) = \frac{3}{10} + \frac{4}{10} + \frac{10}{10} + \frac{12}{10} + \frac{14}{10} + \frac{8}{10} + \frac{9}{10}$

$E(X) = \frac{3+4+10+12+14+8+9}{10} = \frac{60}{10}$

(ii) Variance of X, Var(X)

The variance is given by $Var(X) = E(X^2) - [E(X)]^2$.

First, we calculate $E(X^2) = \sum\limits_{x} x^2 \cdot P(X=x)$.

$E(X^2) = (3)^2 \left(\frac{1}{10}\right) + (4)^2 \left(\frac{1}{10}\right) + (5)^2 \left(\frac{2}{10}\right) + (6)^2 \left(\frac{2}{10}\right) + (7)^2 \left(\frac{2}{10}\right) + (8)^2 \left(\frac{1}{10}\right) + (9)^2 \left(\frac{1}{10}\right)$

$E(X^2) = 9 \left(\frac{1}{10}\right) + 16 \left(\frac{1}{10}\right) + 25 \left(\frac{2}{10}\right) + 36 \left(\frac{2}{10}\right) + 49 \left(\frac{2}{10}\right) + 64 \left(\frac{1}{10}\right) + 81 \left(\frac{1}{10}\right)$

$E(X^2) = \frac{9}{10} + \frac{16}{10} + \frac{50}{10} + \frac{72}{10} + \frac{98}{10} + \frac{64}{10} + \frac{81}{10}$

$E(X^2) = \frac{9+16+50+72+98+64+81}{10} = \frac{390}{10}$

$E(X^2) = 39$

Now, substitute $E(X)$ and $E(X^2)$ into the variance formula:

$Var(X) = E(X^2) - [E(X)]^2$

$Var(X) = 39 - (6)^2$

$Var(X) = 39 - 36$

Final Answer:

The mean of X is $E(X) = 6$.

The variance of X is $Var(X) = 3$.

Given:

$P(A) = \frac{4}{5}$

$P(A \cap B) = \frac{7}{10}$

To Find:

The conditional probability $P(B | A)$.

Solution:

The formula for the conditional probability of event B occurring given that event A has occurred is:

$P(B | A) = \frac{P(A \cap B)}{P(A)}$

Substitute the given values into the formula:

$P(B | A) = \frac{\frac{7}{10}}{\frac{4}{5}}$

To divide by a fraction, we multiply by its reciprocal:

$P(B | A) = \frac{7}{10} \times \frac{5}{4}$

Multiply the numerators and the denominators:

$P(B | A) = \frac{7 \times 5}{10 \times 4}$

$P(B | A) = \frac{35}{40}$

Simplify the fraction by dividing the numerator and the denominator by their greatest common divisor, which is 5:

$P(B | A) = \frac{\cancel{35}^{7}}{\cancel{40}_{8}}$

$P(B | A) = \frac{7}{8}$

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C) $\frac{7}{8}$.

Given:

$P(A \cap B) = \frac{7}{10}$

$P(B) = \frac{17}{20}$

To Find:

The conditional probability $P(A | B)$.

Solution:

The formula for the conditional probability of event A occurring given that event B has occurred is:

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

Substitute the given values into the formula:

$P(A | B) = \frac{\frac{7}{10}}{\frac{17}{20}}$

To divide by a fraction, we multiply by its reciprocal:

$P(A | B) = \frac{7}{10} \times \frac{20}{17}$

Multiply the numerators and the denominators, and simplify:

$P(A | B) = \frac{7 \times \cancel{20}^{2}}{\cancel{10}_{1} \times 17}$

$P(A | B) = \frac{7 \times 2}{1 \times 17}$

$P(A | B) = \frac{14}{17}$

Comparing this result with the given options, we find that it matches option (A).

The correct answer is (A) $\frac{14}{17}$.

Given:

$P(A) = \frac{3}{10}$

$P(B) = \frac{2}{5}$

$P(A \cup B) = \frac{3}{5}$

To Find:

The sum of conditional probabilities $P(B | A) + P(A | B)$.

Solution:

First, we need to find the probability of the intersection of events A and B, $P(A \cap B)$.

We use the formula for the probability of the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Rearranging the formula to find $P(A \cap B)$:

$P(A \cap B) = P(A) + P(B) - P(A \cup B)$

Substitute the given values into the formula:

$P(A \cap B) = \frac{3}{10} + \frac{2}{5} - \frac{3}{5}$

To add and subtract these fractions, we find a common denominator, which is 10.

$\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10}$

$\frac{3}{5} = \frac{3 \times 2}{5 \times 2} = \frac{6}{10}$

So, $P(A \cap B) = \frac{3}{10} + \frac{4}{10} - \frac{6}{10}$

$P(A \cap B) = \frac{3 + 4 - 6}{10}$

$P(A \cap B) = \frac{7 - 6}{10}$

$P(A \cap B) = \frac{1}{10}$

Now, we calculate $P(B | A)$ using the formula:

$P(B | A) = \frac{P(A \cap B)}{P(A)}$

Substitute the values of $P(A \cap B)$ and $P(A)$:

$P(B | A) = \frac{\frac{1}{10}}{\frac{3}{10}}$

$P(B | A) = \frac{1}{10} \times \frac{10}{3}$

$P(B | A) = \frac{\cancel{1}^{1}}{\cancel{10}^{1}} \times \frac{\cancel{10}^{1}}{\cancel{3}^{1}}$

$P(B | A) = \frac{1}{3}$

Next, we calculate $P(A | B)$ using the formula:

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

Substitute the values of $P(A \cap B)$ and $P(B)$:

$P(A | B) = \frac{\frac{1}{10}}{\frac{2}{5}}$

$P(A | B) = \frac{1}{10} \times \frac{5}{2}$

$P(A | B) = \frac{1 \times \cancel{5}^{1}}{\cancel{10}^{2} \times 2}$

$P(A | B) = \frac{1}{2 \times 2}$

$P(A | B) = \frac{1}{4}$

Finally, we find the sum $P(B | A) + P(A | B)$:

$P(B | A) + P(A | B) = \frac{1}{3} + \frac{1}{4}$

To add these fractions, we find a common denominator, which is 12.

$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$

$\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$

Sum $= \frac{4}{12} + \frac{3}{12}$

Sum $= \frac{4+3}{12}$

Sum $= \frac{7}{12}$

Comparing this result with the given options, we find that none of the options exactly match $\frac{7}{12}$. However, option (D) is given as $\frac{7}{2}$, which is likely a typo and should be $\frac{7}{12}$. Assuming option (D) is $\frac{7}{12}$, it is the correct answer.

Assuming the options contain a typo and option (D) is $\frac{7}{12}$, the correct answer is (D) $\frac{7}{12}$.

Given:

$P(A) = \frac{2}{5}$

$P(B) = \frac{3}{10}$

$P(A \cap B) = \frac{1}{5}$

To Find:

The value of $P(A' | B') \cdot P(B' | A')$.

Solution:

First, we need to calculate $P(A \cup B)$ using the formula:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the given values:

$P(A \cup B) = \frac{2}{5} + \frac{3}{10} - \frac{1}{5}$

Find a common denominator, which is 10:

$P(A \cup B) = \frac{2 \times 2}{5 \times 2} + \frac{3}{10} - \frac{1 \times 2}{5 \times 2}$

$P(A \cup B) = \frac{4}{10} + \frac{3}{10} - \frac{2}{10}$

$P(A \cup B) = \frac{4 + 3 - 2}{10} = \frac{5}{10} = \frac{1}{2}$

Next, we find $P(A' \cap B')$. Using De Morgan's Law, $A' \cap B' = (A \cup B)'$.

$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)$

$P(A' \cap B') = 1 - \frac{1}{2} = \frac{1}{2}$

We also need $P(A')$ and $P(B')$:

$P(A') = 1 - P(A) = 1 - \frac{2}{5} = \frac{5 - 2}{5} = \frac{3}{5}$

$P(B') = 1 - P(B) = 1 - \frac{3}{10} = \frac{10 - 3}{10} = \frac{7}{10}$

Now, calculate $P(A' | B')$ using the formula $P(X|Y) = \frac{P(X \cap Y)}{P(Y)}$:

$P(A' | B') = \frac{P(A' \cap B')}{P(B')}$

$P(A' | B') = \frac{\frac{1}{2}}{\frac{7}{10}} = \frac{1}{2} \times \frac{10}{7} = \frac{\cancel{10}^{5}}{2 \times 7} = \frac{5}{7}$

Calculate $P(B' | A')$ using the formula $P(Y|X) = \frac{P(Y \cap X)}{P(X)}$:

$P(B' | A') = \frac{P(B' \cap A')}{P(A')}$

Since $B' \cap A' = A' \cap B'$, we have:

$P(B' | A') = \frac{P(A' \cap B')}{P(A')} = \frac{\frac{1}{2}}{\frac{3}{5}} = \frac{1}{2} \times \frac{5}{3} = \frac{5}{6}$

Finally, calculate the product $P(A' | B') \cdot P(B' | A')$:

$P(A' | B') \cdot P(B' | A') = \frac{5}{7} \times \frac{5}{6}$

Multiply the numerators and denominators:

Product $= \frac{5 \times 5}{7 \times 6} = \frac{25}{42}$

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C) $\frac{25}{42}$.

Given:

$P(A) = \frac{1}{2}$

$P(B) = \frac{1}{3}$

$P(A | B) = \frac{1}{4}$

To Find:

The probability $P(A' \cap B')$.

Solution:

We are given $P(A | B) = \frac{1}{4}$. The formula for conditional probability is:

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

We can rearrange this formula to find $P(A \cap B)$:

$P(A \cap B) = P(A | B) \times P(B)$

Substitute the given values:

$P(A \cap B) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}$

Next, we need to find $P(A' \cap B')$. Using De Morgan's Law, we know that $A' \cap B' = (A \cup B)'$.

Therefore, $P(A' \cap B') = P((A \cup B)')$.

Using the complement rule, $P(E') = 1 - P(E)$, we have:

$P((A \cup B)') = 1 - P(A \cup B)$

To find $P(A \cup B)$, we use the formula for the probability of the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the values of $P(A)$, $P(B)$, and $P(A \cap B)$:

$P(A \cup B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{12}$

To add and subtract these fractions, find a common denominator, which is 12.

$P(A \cup B) = \frac{1 \times 6}{2 \times 6} + \frac{1 \times 4}{3 \times 4} - \frac{1}{12}$

$P(A \cup B) = \frac{6}{12} + \frac{4}{12} - \frac{1}{12}$

$P(A \cup B) = \frac{6 + 4 - 1}{12} = \frac{9}{12}$

Simplify the fraction:

$P(A \cup B) = \frac{3}{4}$

Now, calculate $P(A' \cap B')$:

$P(A' \cap B') = 1 - P(A \cup B)$

$P(A' \cap B') = 1 - \frac{3}{4}$

$P(A' \cap B') = \frac{4}{4} - \frac{3}{4} = \frac{4 - 3}{4} = \frac{1}{4}$

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C) $\frac{1}{4}$.

Given:

$P(A) = 0.4$

$P(B) = 0.8$

$P(B | A) = 0.6$

To Find:

The probability $P(A \cup B)$.

Solution:

We are given the conditional probability $P(B | A)$. The formula for conditional probability is:

$P(B | A) = \frac{P(A \cap B)}{P(A)}$

We can rearrange this formula to find the probability of the intersection of A and B, $P(A \cap B)$:

$P(A \cap B) = P(B | A) \times P(A)$

Substitute the given values into this formula:

$P(A \cap B) = 0.6 \times 0.4$

$P(A \cap B) = 0.24$

Now, we need to find the probability of the union of A and B, $P(A \cup B)$. We use the formula:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the values of $P(A)$, $P(B)$, and $P(A \cap B)$:

$P(A \cup B) = 0.4 + 0.8 - 0.24$

$P(A \cup B) = 1.2 - 0.24$

$P(A \cup B) = 0.96$

Comparing this result with the given options, we find that it matches option (D).

The correct answer is (D) 0.96.

Given:

A and B are two events.

$A \neq \phi$ (A is not the empty set)

$B \neq \phi$ (B is not the empty set)

To Find:

The correct statement among the given options.

Solution:

The conditional probability of event A occurring given that event B has occurred is defined for $P(B) > 0$ as:

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

Let's examine the given options:

(A) $P(A | B) = P(A) \cdot P(B)$

This statement is generally false. It is related to the multiplication rule for independent events, $P(A \cap B) = P(A) P(B)$, which implies $P(A|B) = \frac{P(A)P(B)}{P(B)} = P(A)$ for $P(B) > 0$. So, $P(A|B) = P(A)$ is the condition for independence (when $P(B)>0$), not the general definition of $P(A|B)$.

(B) $P(A | B) = \frac{P (A \cap B)}{P(B)}$

This statement is the definition of the conditional probability of A given B, provided $P(B) > 0$. The question states $B \neq \phi$, which means B is a non-empty event, but it does not strictly guarantee $P(B) > 0$. However, in the context of defining conditional probability, this formula is the standard one applied when the probability of the conditioning event is non-zero. This is the fundamental definition.

(C) $P(A | B) \cdot P(B | A) = 1$

Substitute the definitions (assuming $P(A)>0$ and $P(B)>0$):

$P(A | B) \cdot P(B | A) = \frac{P(A \cap B)}{P(B)} \cdot \frac{P(B \cap A)}{P(A)}$

Since $P(A \cap B) = P(B \cap A)$, this becomes:

$P(A | B) \cdot P(B | A) = \frac{P(A \cap B)}{P(B)} \cdot \frac{P(A \cap B)}{P(A)} = \frac{(P(A \cap B))^2}{P(A) P(B)}$

This is generally not equal to 1. For example, if $P(A)=0.5, P(B)=0.5, P(A \cap B)=0.1$, then $\frac{(0.1)^2}{(0.5)(0.5)} = \frac{0.01}{0.25} = \frac{1}{25} \neq 1$.

(D) $P(A | B) = P(A) / P(B)$

This statement is generally false. It incorrectly omits the intersection term from the numerator of the definition.

Based on the fundamental definition of conditional probability, option (B) is the correct statement, assuming the context implies $P(B) > 0$ for the definition to be applicable.

The correct answer is (B) $P(A | B) = \frac{P (A \cap B)}{P(B)}$.

Given:

$P(A) = 0.4$

$P(B) = 0.3$

$P(A \cup B) = 0.5$

To Find:

The probability $P(B' \cap A)$.

Solution:

We are given the probabilities $P(A)$, $P(B)$, and $P(A \cup B)$.

First, we find the probability of the intersection of A and B, $P(A \cap B)$, using the formula for the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Rearranging the formula to solve for $P(A \cap B)$:

$P(A \cap B) = P(A) + P(B) - P(A \cup B)$

Substitute the given values:

$P(A \cap B) = 0.4 + 0.3 - 0.5$

$P(A \cap B) = 0.7 - 0.5$

$P(A \cap B) = 0.2$

We need to find $P(B' \cap A)$. The event $B' \cap A$ is the same as $A \cap B'$, which represents the event that A occurs but B does not occur. This is the set difference $A \setminus B$.

The probability of this event is given by:

$P(A \cap B') = P(A) - P(A \cap B)$

Substitute the values of $P(A)$ and $P(A \cap B)$:

$P(B' \cap A) = 0.4 - 0.2$

$P(B' \cap A) = 0.2$

Convert the decimal to a fraction:

$0.2 = \frac{2}{10} = \frac{1}{5}$

Comparing this result with the given options, we find that it matches option (D).

The correct answer is (D) $\frac{1}{5}$.

Given:

$P(B) = \frac{3}{5}$

$P(A | B) = \frac{1}{2}$

$P(A \cup B) = \frac{4}{5}$

To Find:

The probability $P(A)$.

Solution:

We are given the conditional probability $P(A | B)$. The formula for conditional probability is:

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

We can rearrange this formula to find the probability of the intersection of A and B, $P(A \cap B)$:

$P(A \cap B) = P(A | B) \times P(B)$

Substitute the given values into this formula:

$P(A \cap B) = \frac{1}{2} \times \frac{3}{5}$

$P(A \cap B) = \frac{3}{10}$

Now, we use the formula for the probability of the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

We need to solve for $P(A)$. Rearrange the formula:

$P(A) = P(A \cup B) - P(B) + P(A \cap B)$

Substitute the values of $P(A \cup B)$, $P(B)$, and $P(A \cap B)$:

$P(A) = \frac{4}{5} - \frac{3}{5} + \frac{3}{10}$

First, subtract the fractions with the same denominator:

$P(A) = \left(\frac{4}{5} - \frac{3}{5}\right) + \frac{3}{10}$

$P(A) = \frac{4-3}{5} + \frac{3}{10}$

$P(A) = \frac{1}{5} + \frac{3}{10}$

To add these fractions, find a common denominator, which is 10:

$P(A) = \frac{1 \times 2}{5 \times 2} + \frac{3}{10}$

$P(A) = \frac{2}{10} + \frac{3}{10}$

$P(A) = \frac{2+3}{10}$

$P(A) = \frac{5}{10}$

Simplify the fraction:

$P(A) = \frac{\cancel{5}^{1}}{\cancel{10}_{2}} = \frac{1}{2}$

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C) $\frac{1}{2}$.

Given (from Exercise 64):

$P(B) = \frac{3}{5}$

$P(A | B) = \frac{1}{2}$

$P(A \cup B) = \frac{4}{5}$

From the solution to Exercise 64, we found:

$P(A) = \frac{1}{2}$

$P(A \cap B) = \frac{3}{10}$

To Find:

The conditional probability $P(B | A')$.

Solution:

The formula for the conditional probability of event B occurring given that event A' has occurred is:

$P(B | A') = \frac{P(B \cap A')}{P(A')}$

First, we need to find $P(A')$. Using the complement rule:

$P(A') = 1 - P(A)$

Substitute the value of $P(A)$ from Exercise 64's solution:

$P(A') = 1 - \frac{1}{2} = \frac{1}{2}$

Next, we need to find $P(B \cap A')$. The event $B \cap A'$ represents the elements that are in B but not in A (B minus A). The probability of this event is given by:

$P(B \cap A') = P(B) - P(A \cap B)$

Substitute the values of $P(B)$ and $P(A \cap B)$:

$P(B \cap A') = \frac{3}{5} - \frac{3}{10}$

To subtract these fractions, find a common denominator, which is 10:

$P(B \cap A') = \frac{3 \times 2}{5 \times 2} - \frac{3}{10}$

$P(B \cap A') = \frac{6}{10} - \frac{3}{10}$

$P(B \cap A') = \frac{6 - 3}{10} = \frac{3}{10}$

Now, substitute the values of $P(B \cap A')$ and $P(A')$ into the formula for $P(B | A')$:

$P(B | A') = \frac{\frac{3}{10}}{\frac{1}{2}}$

To divide by a fraction, multiply by its reciprocal:

$P(B | A') = \frac{3}{10} \times \frac{2}{1}$

$P(B | A') = \frac{3 \times \cancel{2}^{1}}{\cancel{10}_{5} \times 1}$

$P(B | A') = \frac{3}{5}$

Comparing this result with the given options, we find that it matches option (D).

The correct answer is (D) $\frac{3}{5}$.

Given (from Exercise 64):

$P(B) = \frac{3}{5}$

$P(A | B) = \frac{1}{2}$

$P(A \cup B) = \frac{4}{5}$

From the solution to Exercise 64, we found:

$P(A) = \frac{1}{2}$

$P(A \cap B) = \frac{3}{10}$

To Find:

The value of $P((A \cup B)') + P(A' \cup B)$.

Solution:

First, calculate $P((A \cup B)')$. Using the complement rule:

$P((A \cup B)') = 1 - P(A \cup B)$

Substitute the given value of $P(A \cup B)$:

$P((A \cup B)') = 1 - \frac{4}{5}$

$P((A \cup B)') = \frac{5}{5} - \frac{4}{5} = \frac{1}{5}$

Next, calculate $P(A' \cup B)$. Using the formula for the union of two events:

$P(A' \cup B) = P(A') + P(B) - P(A' \cap B)$

We need $P(A')$ and $P(A' \cap B)$.

Calculate $P(A')$ using the complement rule:

$P(A') = 1 - P(A)$

Substitute the value of $P(A)$ found in Exercise 64:

$P(A') = 1 - \frac{1}{2} = \frac{1}{2}$

Calculate $P(A' \cap B)$. The event $A' \cap B$ is the intersection of the complement of A and B, which means the event where B occurs but A does not. This can be written as $B \setminus A$. The probability is given by:

$P(A' \cap B) = P(B) - P(A \cap B)$

Substitute the values of $P(B)$ and $P(A \cap B)$:

$P(A' \cap B) = \frac{3}{5} - \frac{3}{10}$

Find a common denominator, which is 10:

$P(A' \cap B) = \frac{3 \times 2}{5 \times 2} - \frac{3}{10} = \frac{6}{10} - \frac{3}{10} = \frac{3}{10}$

Now, substitute the values of $P(A')$, $P(B)$, and $P(A' \cap B)$ into the formula for $P(A' \cup B)$:

$P(A' \cup B) = \frac{1}{2} + \frac{3}{5} - \frac{3}{10}$

Find a common denominator, which is 10:

$P(A' \cup B) = \frac{1 \times 5}{2 \times 5} + \frac{3 \times 2}{5 \times 2} - \frac{3}{10}$

$P(A' \cup B) = \frac{5}{10} + \frac{6}{10} - \frac{3}{10}$

$P(A' \cup B) = \frac{5 + 6 - 3}{10} = \frac{11 - 3}{10} = \frac{8}{10}$

Simplify the fraction:

$P(A' \cup B) = \frac{4}{5}$

Finally, calculate the sum $P((A \cup B)') + P(A' \cup B)$:

Sum $= \frac{1}{5} + \frac{4}{5}$

Sum $= \frac{1 + 4}{5} = \frac{5}{5} = 1$

Comparing this result with the given options, we find that it matches option (D).

The correct answer is (D) 1.

Given:

$P(A) = \frac{7}{13}$

$P(B) = \frac{9}{13}$

$P(A \cap B) = \frac{4}{13}$

To Find:

The conditional probability $P(A' | B)$.

Solution:

The formula for the conditional probability of event A' occurring given that event B has occurred is:

$P(A' | B) = \frac{P(A' \cap B)}{P(B)}$

We need to find $P(A' \cap B)$. The event $A' \cap B$ represents the elements that are in B but not in A. This is the set difference $B \setminus A$. The probability of this event is given by:

$P(A' \cap B) = P(B) - P(A \cap B)$

Substitute the given values of $P(B)$ and $P(A \cap B)$:

$P(A' \cap B) = \frac{9}{13} - \frac{4}{13}$

$P(A' \cap B) = \frac{9 - 4}{13} = \frac{5}{13}$

Now, substitute the values of $P(A' \cap B)$ and $P(B)$ into the formula for $P(A' | B)$:

$P(A' | B) = \frac{\frac{5}{13}}{\frac{9}{13}}$

To divide by a fraction, multiply by its reciprocal:

$P(A' | B) = \frac{5}{13} \times \frac{13}{9}$

$P(A' | B) = \frac{5 \times \cancel{13}^{1}}{\cancel{13}_{1} \times 9}$

$P(A' | B) = \frac{5}{9}$

Comparing this result with the given options, we find that it matches option (D).

The correct answer is (D) $\frac{5}{9}$.

Given:

A and B are events.

$P(A) > 0$

$P(B) \neq 1$

To Find:

The expression for $P(A' | B')$.

Solution:

The conditional probability $P(A' | B')$ is defined as:

$P(A' | B') = \frac{P(A' \cap B')}{P(B')}$

This definition is valid because we are given $P(B) \neq 1$, which implies $P(B') = 1 - P(B) \neq 0$.

Now, let's consider the numerator, $P(A' \cap B')$. Using De Morgan's Law, the intersection of the complements of A and B is equal to the complement of the union of A and B:

$A' \cap B' = (A \cup B)'$

Taking the probability of both sides:

$P(A' \cap B') = P((A \cup B)')$

Using the complement rule, $P(E') = 1 - P(E)$, the probability of the complement of the union is:

$P((A \cup B)') = 1 - P(A \cup B)$

Substitute this expression for $P(A' \cap B')$ back into the formula for $P(A' | B')$:

$P(A' | B') = \frac{1 - P(A \cup B)}{P(B')}$

This expression matches option (C).

Let's briefly look at the other options:

(A) $1 - P(A | B) = 1 - \frac{P(A \cap B)}{P(B)}$. This is not generally equal to $P(A' | B')$.

(B) $1 - P(A' | B) = 1 - \frac{P(A' \cap B)}{P(B)} = 1 - \frac{P(B) - P(A \cap B)}{P(B)} = 1 - (1 - \frac{P(A \cap B)}{P(B)}) = \frac{P(A \cap B)}{P(B)} = P(A | B)$. So, this option is equal to $P(A | B)$, not $P(A' | B')$.

(D) $\frac{P(A')}{P(B')}$. This is the definition of $P(A' | B')$ only if $A'$ and $B'$ are independent events, which is not given in the problem.

Based on the definition of conditional probability and set properties, the correct expression is $\frac{1 - P(A \cup B)}{P(B')}$.

The correct answer is (C) $\frac{1 − P (A \cup B)}{P(B’)}$.

Given:

Events A and B are independent.

$P(A) = \frac{3}{5}$

$P(B) = \frac{4}{9}$

To Find:

The probability $P(A' \cap B')$.

Solution:

Since A and B are independent events, their complements A' and B' are also independent events.

For independent events, the probability of their intersection is the product of their individual probabilities:

$P(A' \cap B') = P(A') \cdot P(B')$

First, calculate the probabilities of the complements, $P(A')$ and $P(B')$, using the complement rule $P(E') = 1 - P(E)$.

$P(A') = 1 - P(A) = 1 - \frac{3}{5}$

$P(A') = \frac{5}{5} - \frac{3}{5} = \frac{2}{5}$

$P(B') = 1 - P(B) = 1 - \frac{4}{9}$

$P(B') = \frac{9}{9} - \frac{4}{9} = \frac{5}{9}$

Now, substitute the values of $P(A')$ and $P(B')$ into the formula for $P(A' \cap B')$:

$P(A' \cap B') = \frac{2}{5} \times \frac{5}{9}$

Multiply the fractions:

$P(A' \cap B') = \frac{2 \times 5}{5 \times 9}$

Simplify by cancelling the common factor of 5:

$P(A' \cap B') = \frac{2 \times \cancel{5}^{1}}{\cancel{5}_{1} \times 9} = \frac{2}{9}$

Comparing this result with the given options, we find that it matches option (D).

The correct answer is (D) $\frac{2}{9}$.

Solution:

Let A and B be two events.

Recall the definitions of independent and mutually exclusive events.

Two events A and B are said to be independent if the occurrence of one event does not affect the probability of the occurrence of the other event. Mathematically, this is defined by the relationship:

$P(A \cap B) = P(A) \cdot P(B)$

Two events A and B are said to be mutually exclusive if they cannot occur at the same time. Mathematically, this means their intersection is empty:

$A \cap B = \phi$

The probability of their intersection is therefore:

$P(A \cap B) = 0$

Let's examine the options.

(A) they must be mutually exclusive

If A and B are independent and $P(A) > 0$ and $P(B) > 0$, then $P(A \cap B) = P(A) \cdot P(B) > 0$. For mutually exclusive events, $P(A \cap B) = 0$. Since $P(A \cap B)$ cannot be both greater than 0 and equal to 0, independent events with non-zero probabilities cannot be mutually exclusive.

The only case where independent events can be mutually exclusive is if $P(A)=0$ or $P(B)=0$. In that case, $P(A \cap B) = 0$ (mutually exclusive) and $P(A)P(B) = 0$ (independent), so the condition $P(A \cap B) = P(A)P(B)$ is satisfied. However, the statement "they must be mutually exclusive" is not true in general for all independent events (e.g., consider flipping a fair coin twice; the outcome of the first flip is independent of the outcome of the second flip, but the events 'first flip is Heads' and 'second flip is Heads' are independent but not mutually exclusive).

So, option (A) is false.

(B) the sum of their probabilities must be equal to 1

Independence ($P(A \cap B) = P(A)P(B)$) does not imply $P(A) + P(B) = 1$. Consider flipping a fair coin twice. Let A be the event the first flip is Heads ($P(A)=0.5$) and B be the event the second flip is Heads ($P(B)=0.5$). A and B are independent events, as $P(A \cap B) = P(\text{HH}) = 0.25$, and $P(A)P(B) = 0.5 \times 0.5 = 0.25$. However, $P(A) + P(B) = 0.5 + 0.5 = 1$. This example seems to support (B), but let's try another example.

Consider rolling a fair six-sided die. Let A be the event the roll is 2 or 4 ($P(A) = 2/6 = 1/3$). Let B be the event the roll is 3 or 6 ($P(B) = 2/6 = 1/3$). $A \cap B = \phi$, so $P(A \cap B) = 0$. $P(A)P(B) = (1/3)(1/3) = 1/9$. Since $0 \neq 1/9$, A and B are not independent in this case. They are mutually exclusive.

Let A be the event of rolling an even number ({2, 4, 6}) ($P(A)=3/6=1/2$) and B be the event of rolling a number greater than 4 ({5, 6}) ($P(B)=2/6=1/3$). $A \cap B = \{6\}$, $P(A \cap B) = 1/6$. $P(A)P(B) = (1/2)(1/3) = 1/6$. So A and B are independent. In this case, $P(A) + P(B) = 1/2 + 1/3 = 3/6 + 2/6 = 5/6 \neq 1$.

This counterexample shows that independent events do not necessarily have probabilities that sum to 1.

So, option (B) is false.

(C) (A) and (B) both are correct

Since both (A) and (B) are false, this option is false.

(D) None of the above is correct

Since options (A), (B), and (C) are all incorrect statements about independent events, this option is correct.

The correct answer is (D) None of the above is correct.

Given:

$P(A) = \frac{3}{8}$

$P(B) = \frac{5}{8}$

$P(A \cup B) = \frac{3}{4}$

To Find:

The value of $P(A | B) \cdot P(A' | B)$.

Solution:

First, we find the probability of the intersection of A and B, $P(A \cap B)$, using the formula for the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Rearranging the formula to solve for $P(A \cap B)$:

$P(A \cap B) = P(A) + P(B) - P(A \cup B)$

Substitute the given values:

$P(A \cap B) = \frac{3}{8} + \frac{5}{8} - \frac{3}{4}$

$P(A \cap B) = \frac{3+5}{8} - \frac{3}{4}$

$P(A \cap B) = \frac{8}{8} - \frac{3}{4}$

$P(A \cap B) = 1 - \frac{3}{4}$

$P(A \cap B) = \frac{4}{4} - \frac{3}{4} = \frac{1}{4}$

Alternatively, using a common denominator of 8:

$P(A \cap B) = \frac{3}{8} + \frac{5}{8} - \frac{3 \times 2}{4 \times 2}$

$P(A \cap B) = \frac{3}{8} + \frac{5}{8} - \frac{6}{8}$

$P(A \cap B) = \frac{3+5-6}{8} = \frac{8-6}{8} = \frac{2}{8} = \frac{1}{4}$

Now, calculate $P(A | B)$ using the conditional probability formula:

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

Substitute the values of $P(A \cap B)$ and $P(B)$:

$P(A | B) = \frac{\frac{1}{4}}{\frac{5}{8}}$

$P(A | B) = \frac{1}{4} \times \frac{8}{5}$

$P(A | B) = \frac{1 \times \cancel{8}^{2}}{\cancel{4}_{1} \times 5} = \frac{2}{5}$

Next, calculate $P(A' | B)$. The formula is:

$P(A' | B) = \frac{P(A' \cap B)}{P(B)}$

The event $A' \cap B$ represents the elements that are in B but not in A ($B \setminus A$). The probability is given by:

$P(A' \cap B) = P(B) - P(A \cap B)$

Substitute the values of $P(B)$ and $P(A \cap B)$:

$P(A' \cap B) = \frac{5}{8} - \frac{1}{4}$

Find a common denominator, which is 8:

$P(A' \cap B) = \frac{5}{8} - \frac{1 \times 2}{4 \times 2} = \frac{5}{8} - \frac{2}{8} = \frac{5 - 2}{8} = \frac{3}{8}$

Now, substitute the values of $P(A' \cap B)$ and $P(B)$ into the formula for $P(A' | B)$:

$P(A' | B) = \frac{\frac{3}{8}}{\frac{5}{8}}$

$P(A' | B) = \frac{3}{8} \times \frac{8}{5}$

$P(A' | B) = \frac{3 \times \cancel{8}^{1}}{\cancel{8}_{1} \times 5} = \frac{3}{5}$

Finally, calculate the product $P(A | B) \cdot P(A' | B)$:

Product $= \frac{2}{5} \times \frac{3}{5}$

Product $= \frac{2 \times 3}{5 \times 5} = \frac{6}{25}$

Comparing this result with the given options, we find that it matches option (D).

The correct answer is (D) $\frac{6}{25}$.

Solution:

By the definition of independent events, two events A and B are independent if the probability of their intersection is equal to the product of their individual probabilities.

That is,

$P(A \cap B) = P(A) \cdot P(B)$

Comparing this definition with the given options, we find that it matches option (C).

(A) $P(A) + P(B)$: This is related to the probability of the union for mutually exclusive events (when $P(A \cap B) = 0$), i.e., $P(A \cup B) = P(A) + P(B)$. This is not the formula for $P(A \cap B)$ for independent events unless $P(A)P(B)=0$.

(B) $P(A) - P(B)$: This is not a standard formula for $P(A \cap B)$.

(C) $P(A) \cdot P(B)$: This is the correct definition of independence.

(D) $P(A) / P(B)$: This is not a standard formula for $P(A \cap B)$. It is related to conditional probability $P(A|B) = P(A \cap B) / P(B)$, which for independent events is equal to $P(A)$.

The correct answer is (C) $P(A) \cdot P(B)$.

Given:

Events E and F are independent.

$P(E) = 0.3$

$P(E \cup F) = 0.5$

To Find:

The value of $P(E | F) - P(F | E)$.

Solution:

Since events E and F are independent, we know that:

$P(E \cap F) = P(E) \cdot P(F)$

We also know the formula for the probability of the union of two events:

$P(E \cup F) = P(E) + P(F) - P(E \cap F)$

Substitute the independence condition into the union formula:

$P(E \cup F) = P(E) + P(F) - P(E) \cdot P(F)$

Substitute the given values of $P(E)$ and $P(E \cup F)$:

$0.5 = 0.3 + P(F) - 0.3 \cdot P(F)$

Rearrange the equation to solve for $P(F)$:

$0.5 - 0.3 = P(F) (1 - 0.3)$

$0.2 = 0.7 \cdot P(F)$

$P(F) = \frac{0.2}{0.7} = \frac{2}{7}$

Since events E and F are independent, their conditional probabilities are equal to their individual probabilities (provided the conditioning event has a non-zero probability). Since $P(E)=0.3 > 0$ and $P(F)=\frac{2}{7} > 0$, we have:

$P(E | F) = P(E)$

$P(F | E) = P(F)$

Substitute the values of $P(E)$ and $P(F)$:

$P(E | F) = 0.3 = \frac{3}{10}$

$P(F | E) = \frac{2}{7}$

Now, calculate the difference $P(E | F) - P(F | E)$:

$P(E | F) - P(F | E) = \frac{3}{10} - \frac{2}{7}$

To subtract the fractions, find a common denominator, which is 70:

$\frac{3}{10} - \frac{2}{7} = \frac{3 \times 7}{10 \times 7} - \frac{2 \times 10}{7 \times 10}$

$= \frac{21}{70} - \frac{20}{70}$

$= \frac{21 - 20}{70}$

$= \frac{1}{70}$

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C) $\frac{1}{70}$.

Given:

Number of red balls = 5

Number of blue balls = 3

Total number of balls = $5 + 3 = 8$

Number of balls drawn = 3 (without replacement)

To Find:

The probability of getting exactly one red ball when drawing 3 balls.

Solution:

We are drawing 3 balls from a total of 8 balls without replacement.

The total number of ways to choose 3 balls from 8 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Total number of possible outcomes = $\binom{8}{3}$

$\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56$

So, the total number of ways to draw 3 balls is 56.

We want to find the number of ways to get exactly one red ball. This means we must draw 1 red ball and the remaining $3 - 1 = 2$ balls must be blue.

Number of ways to choose exactly 1 red ball from 5 red balls = $\binom{5}{1}$

$\binom{5}{1} = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = \frac{5 \times 4!}{1 \times 4!} = 5$

Number of ways to choose exactly 2 blue balls from 3 blue balls = $\binom{3}{2}$

$\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2!}{2! \times 1} = 3$

The number of ways to get exactly one red ball and two blue balls is the product of the number of ways to choose 1 red ball and the number of ways to choose 2 blue balls (since these choices are independent):

Number of favorable outcomes = $\binom{5}{1} \times \binom{3}{2} = 5 \times 3 = 15$

The probability of getting exactly one red ball is the ratio of the number of favorable outcomes to the total number of possible outcomes:

Probability (exactly 1 red ball) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability (exactly 1 red ball) = $\frac{15}{56}$

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C) $\frac{15}{56}$.

Given (from Question 74):

A bag contains 5 red balls and 3 blue balls.

Total number of balls = $5 + 3 = 8$.

3 balls are drawn at random without replacement.

To Find:

The probability that exactly two of the three balls were red, given that the first ball drawn was red.

Solution:

Let A be the event that exactly two of the three balls drawn are red.

Let B be the event that the first ball drawn is red.

We need to find the conditional probability $P(A | B)$.

Since we are given that the first ball drawn was red, we are conditioning on event B occurring.

After the first ball (which is red) is drawn, the contents of the bag change.

The remaining number of balls in the bag is $8 - 1 = 7$.

The remaining balls consist of $5 - 1 = 4$ red balls and 3 blue balls.

We are drawing a total of 3 balls. Since the first ball drawn was red, for exactly two of the three balls to be red, we must draw exactly $2 - 1 = 1$ red ball from the remaining 2 draws.

These remaining 2 balls are drawn from the 7 balls left in the bag (4 red and 3 blue).

We need to find the probability of drawing exactly 1 red ball and 1 blue ball in the next two draws from the remaining 4 red and 3 blue balls.

The total number of ways to choose 2 balls from the remaining 7 balls is given by the combination $\binom{7}{2}$.

Total ways to choose 2 balls = $\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \times 6}{2 \times 1} = 21$.

The number of ways to choose exactly 1 red ball from the remaining 4 red balls is $\binom{4}{1}$.

Ways to choose 1 red ball = $\binom{4}{1} = \frac{4!}{1!(4-1)!} = \frac{4!}{1!3!} = 4$.

The number of ways to choose exactly 1 blue ball from the remaining 3 blue balls is $\binom{3}{1}$.

Ways to choose 1 blue ball = $\binom{3}{1} = \frac{3!}{1!(3-1)!} = \frac{3!}{1!2!} = 3$.

The number of ways to choose exactly 1 red ball and 1 blue ball in the next two draws is the product of the number of ways to choose 1 red and the number of ways to choose 1 blue.

Number of favorable outcomes = $\binom{4}{1} \times \binom{3}{1} = 4 \times 3 = 12$.

The conditional probability $P(A | B)$ is the ratio of the number of favorable outcomes (for the remaining draws given the first was red) to the total number of outcomes (for the remaining draws given the first was red).

$P(A | B) = \frac{\text{Number of ways to get 1R and 1B in 2 draws from remaining}}{\text{Total number of ways to draw 2 balls from remaining}}$

$P(A | B) = \frac{12}{21}$

Simplify the fraction:

$P(A | B) = \frac{\cancel{12}^{4}}{\cancel{21}_{7}} = \frac{4}{7}$.

Comparing this result with the given options, we find that it matches option (B).

The correct answer is (B) $\frac{4}{7}$.

Given:

$P(\text{A hits}) = P(H_A) = 0.4$

$P(\text{B hits}) = P(H_B) = 0.3$

$P(\text{C hits}) = P(H_C) = 0.2$

To Find:

The probability of exactly two hits when A, B, and C fire in turn.

Solution:

First, find the probabilities of each person missing the target.

$P(\text{A misses}) = P(M_A) = 1 - P(H_A) = 1 - 0.4 = 0.6$

$P(\text{B misses}) = P(M_B) = 1 - P(H_B) = 1 - 0.3 = 0.7$

$P(\text{C misses}) = P(M_C) = 1 - P(H_C) = 1 - 0.2 = 0.8$

Exactly two hits can occur in three mutually exclusive scenarios:

Scenario 1: A hits, B hits, C misses ($H_A \cap H_B \cap M_C$)

Scenario 2: A hits, B misses, C hits ($H_A \cap M_B \cap H_C$)

Scenario 3: A misses, B hits, C hits ($M_A \cap H_B \cap H_C$)

Since the events of hitting or missing for each person are independent, the probability of each scenario is the product of the individual probabilities:

Probability of Scenario 1: $P(H_A \cap H_B \cap M_C) = P(H_A) \times P(H_B) \times P(M_C)$

$= 0.4 \times 0.3 \times 0.8 = 0.12 \times 0.8 = 0.096$

Probability of Scenario 2: $P(H_A \cap M_B \cap H_C) = P(H_A) \times P(M_B) \times P(H_C)$

$= 0.4 \times 0.7 \times 0.2 = 0.28 \times 0.2 = 0.056$

Probability of Scenario 3: $P(M_A \cap H_B \cap H_C) = P(M_A) \times P(H_B) \times P(H_C)$

$= 0.6 \times 0.3 \times 0.2 = 0.18 \times 0.2 = 0.036$

The probability of exactly two hits is the sum of the probabilities of these three mutually exclusive scenarios:

$P(\text{Exactly two hits}) = P(H_A \cap H_B \cap M_C) + P(H_A \cap M_B \cap H_C) + P(M_A \cap H_B \cap H_C)$

$= 0.096 + 0.056 + 0.036$

$= 0.152 + 0.036 = 0.188$

Comparing this result with the given options, we find that it matches option (B).

The correct answer is (B) 0.188.

Given:

Family with three children.

Each child is equally likely to be a boy (B) or a girl (G).

To Find:

The probability that the eldest child is a girl, given that the family has at least one girl.

Solution:

The sample space S, representing the gender of the three children in order of birth (eldest, middle, youngest), consists of $2^3 = 8$ equally likely outcomes:

$S = \{BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG\}$

Let A be the event that the eldest child is a girl.

The outcomes in event A are where the first child is a girl:

$A = \{GBB, GBG, GGB, GGG\}$

The number of outcomes in A is 4.

$P(A) = \frac{\text{Number of outcomes in A}}{\text{Total number of outcomes}} = \frac{4}{8} = \frac{1}{2}$

Let B be the event that the family has at least one girl.

The complement of event B, B', is the event that the family has no girls (all children are boys).

$B' = \{BBB\}$

$P(B') = \frac{\text{Number of outcomes in B'}}{\text{Total number of outcomes}} = \frac{1}{8}$

The probability of event B is:

$P(B) = 1 - P(B')$

$P(B) = 1 - \frac{1}{8} = \frac{8 - 1}{8} = \frac{7}{8}$

We need to find the probability of A given B, which is $P(A | B)$. The formula for conditional probability is:

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

Find the intersection of events A and B, $A \cap B$. This is the event that the eldest child is a girl AND the family has at least one girl.

Since A is the event that the eldest child is a girl, any outcome in A necessarily includes at least one girl (the eldest). Therefore, the intersection $A \cap B$ consists of all outcomes where the eldest child is a girl and there is at least one girl. This is simply event A itself.

$A \cap B = A = \{GBB, GBG, GGB, GGG\}$

$P(A \cap B) = P(A) = \frac{1}{2}$

Substitute the probabilities into the conditional probability formula:

$P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{2}}{\frac{7}{8}}$

To divide by a fraction, we multiply by its reciprocal:

$P(A | B) = \frac{1}{2} \times \frac{8}{7}$

Simplify the expression:

$P(A | B) = \frac{1 \times \cancel{8}^{4}}{\cancel{2}_{1} \times 7} = \frac{4}{7}$

Comparing this result with the given options, we find that it matches option (D).

The correct answer is (D) $\frac{4}{7}$.

Given:

A standard six-sided die is thrown.

A card is selected from a deck of 52 playing cards.

To Find:

The probability of getting an even number on the die AND a spade card.

Solution:

Let A be the event of getting an even number when a die is thrown.

The possible outcomes when rolling a die are $\{1, 2, 3, 4, 5, 6\}$. The even numbers are $\{2, 4, 6\}$.

Number of favorable outcomes for A = 3.

Total number of outcomes = 6.

The probability of event A is:

$P(A) = \frac{\text{Number of even numbers}}{\text{Total numbers}} = \frac{3}{6} = \frac{1}{2}$

Let B be the event of getting a spade card from a deck of 52 playing cards.

The total number of cards in a deck is 52.

There are 13 spades in a standard deck of cards.

Number of favorable outcomes for B = 13.

Total number of outcomes = 52.

The probability of event B is:

$P(B) = \frac{\text{Number of spades}}{\text{Total number of cards}} = \frac{13}{52} = \frac{1}{4}$

The event of throwing a die and the event of selecting a card are independent events, as the outcome of one does not affect the outcome of the other.

The probability of both events occurring is the probability of their intersection. For independent events A and B, this is given by:

$P(A \cap B) = P(A) \cdot P(B)$

Substitute the probabilities of event A and event B:

$P(\text{Even and Spade}) = P(\text{Even}) \times P(\text{Spade})$

$P(\text{Even and Spade}) = \frac{1}{2} \times \frac{1}{4}$

$P(\text{Even and Spade}) = \frac{1 \times 1}{2 \times 4} = \frac{1}{8}$

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C) $\frac{1}{8}$.

Given:

Number of orange balls = 3

Number of green balls = 3

Number of blue balls = 2

Total number of balls = $3 + 3 + 2 = 8$

Number of balls drawn = 3 (without replacement)

To Find:

The probability of drawing exactly 2 green balls and 1 blue ball.

Solution:

We are drawing 3 balls from a total of 8 balls without replacement. The order in which the balls are drawn does not matter, so we use combinations.

The total number of ways to choose 3 balls from 8 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Total number of possible outcomes = $\binom{8}{3}$

$\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56$

So, the total number of ways to draw 3 balls from 8 is 56.

We want to find the number of ways to get exactly 2 green balls and 1 blue ball. This involves choosing 2 green balls from the 3 available green balls and choosing 1 blue ball from the 2 available blue balls.

Number of ways to choose 2 green balls from 3 = $\binom{3}{2}$

$\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2!}{2! \times 1} = 3$

Number of ways to choose 1 blue ball from 2 = $\binom{2}{1}$

$\binom{2}{1} = \frac{2!}{1!(2-1)!} = \frac{2!}{1!1!} = 2$

The number of ways to get exactly 2 green balls and 1 blue ball is the product of the number of ways to choose 2 green balls and the number of ways to choose 1 blue ball (since these choices are independent events from their respective groups).

Number of favorable outcomes = (Ways to choose 2 green) $\times$ (Ways to choose 1 blue)

Number of favorable outcomes = $\binom{3}{2} \times \binom{2}{1} = 3 \times 2 = 6$

The probability of drawing exactly 2 green balls and 1 blue ball is the ratio of the number of favorable outcomes to the total number of possible outcomes:

Probability (2 green and 1 blue) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability (2 green and 1 blue) = $\frac{6}{56}$

Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2:

$\frac{6}{56} = \frac{\cancel{6}^{3}}{\cancel{56}_{28}} = \frac{3}{28}$

Comparing this result with the given options, we find that it matches option (A).

The correct answer is (A) $\frac{3}{28}$.

Given:

Total number of batteries = 8

Number of dead batteries = 3

Number of good batteries = $8 - 3 = 5$

Number of batteries selected = 2 (without replacement)

To Find:

The probability that both selected batteries are dead.

Solution:

We are selecting 2 batteries from a total of 8 batteries without replacement. The order of selection does not matter, so we can use combinations to find the total number of possible outcomes.

Total number of ways to choose 2 batteries from 8 is given by $\binom{8}{2}$.

Total outcomes $= \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8!}{2!6!} = \frac{8 \times 7 \times \cancel{6!}}{2 \times 1 \times \cancel{6!}} = \frac{8 \times 7}{2} = 4 \times 7 = 28$

We want to find the number of ways to choose exactly 2 dead batteries from the 3 available dead batteries.

Number of favorable outcomes = $\binom{3}{2}$

$\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times \cancel{2!}}{\cancel{2!} \times 1} = 3$

The probability of selecting 2 dead batteries is the ratio of the number of favorable outcomes to the total number of possible outcomes.

Probability (both dead) $= \frac{\text{Number of ways to choose 2 dead}}{\text{Total number of ways to choose 2}}$

Probability (both dead) $= \frac{3}{28}$

Comparing this result with the given options, we find that it matches option (D).

The correct answer is (D) $\frac{3}{28}$.

Given:

Number of coins tossed = 8

We want the probability of getting exactly 3 heads.

To Find:

The probability of getting exactly 3 heads in 8 tosses.

Solution:

When a single coin is tossed, there are two possible outcomes: Head (H) or Tail (T).

The probability of getting a Head on a single toss is $P(H) = \frac{1}{2}$.

The probability of getting a Tail on a single toss is $P(T) = \frac{1}{2}$.

When 8 coins are tossed, the total number of possible outcomes is $2^8$ (since each toss is independent and has 2 outcomes).

Total number of outcomes = $2^8 = 256$.

We are interested in the event of getting exactly 3 heads out of 8 tosses. This is a problem that can be solved using the binomial probability distribution.

The number of ways to choose exactly 3 positions for the heads out of 8 tosses is given by the combination formula $\binom{n}{k}$, where $n=8$ (total tosses) and $k=3$ (number of heads).

Number of ways to get exactly 3 heads = $\binom{8}{3}$

$\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!}$

$\binom{8}{3} = \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$

$\binom{8}{3} = \frac{336}{6} = 56$

For any specific sequence of 3 heads and 5 tails (e.g., HHHTT TTT), the probability is $(P(H))^3 \times (P(T))^5 = (\frac{1}{2})^3 \times (\frac{1}{2})^5 = (\frac{1}{2})^{3+5} = (\frac{1}{2})^8 = \frac{1}{256}$.

The probability of getting exactly 3 heads is the number of ways to get 3 heads multiplied by the probability of any single way of getting 3 heads:

Probability (exactly 3 heads) = (Number of ways to get 3 heads) $\times$ (Probability of a specific sequence with 3 heads)

Probability (exactly 3 heads) $= \binom{8}{3} \times (\frac{1}{2})^8$

Probability (exactly 3 heads) $= 56 \times \frac{1}{256} = \frac{56}{256}$

Simplify the fraction $\frac{56}{256}$. Both numbers are divisible by 8.

$\frac{56}{256} = \frac{56 \div 8}{256 \div 8} = \frac{7}{32}$

Alternatively, using the binomial probability formula $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$ with $n=8$, $k=3$, $p=0.5$:

$P(X=3) = \binom{8}{3} (0.5)^3 (1-0.5)^{8-3}$

$P(X=3) = \binom{8}{3} (0.5)^3 (0.5)^5$

$P(X=3) = \binom{8}{3} (0.5)^8$

As calculated above, $\binom{8}{3} = 56$ and $(0.5)^8 = \frac{1}{256}$.

$P(X=3) = 56 \times \frac{1}{256} = \frac{56}{256} = \frac{7}{32}$

Comparing this result with the given options, we find that it matches option (B).

The correct answer is (B) $\frac{7}{32}$.

Given:

Two standard six-sided dice are thrown.

The sum of the numbers on the dice was less than 6.

To Find:

The probability of getting a sum of 3, given that the sum was less than 6.

Solution:

The sample space S when throwing two dice consists of $6 \times 6 = 36$ equally likely outcomes, represented as ordered pairs (die 1 result, die 2 result):

S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}

Let A be the event that the sum of the numbers on the dice is 3.

The outcomes where the sum is 3 are:

A = \{(1,2), (2,1)\}

The number of outcomes in A is $|A| = 2$.

The probability of A is $P(A) = \frac{|A|}{|S|} = \frac{2}{36} = \frac{1}{18}$.

Let B be the event that the sum of the numbers on the dice is less than 6. The sums less than 6 are {2, 3, 4, 5}.

The outcomes corresponding to these sums are:

Sum 2: \{(1,1)\}

Sum 3: \{(1,2), (2,1)\}

Sum 4: \{(1,3), (2,2), (3,1)\}

Sum 5: \{(1,4), (2,3), (3,2), (4,1)\}

So, the outcomes in event B are:

B = \{(1,1), (1,2), (2,1), (1,3), (2,2), (3,1), (1,4), (2,3), (3,2), (4,1)\}

The number of outcomes in B is $|B| = 1 + 2 + 3 + 4 = 10$.

The probability of B is $P(B) = \frac{|B|}{|S|} = \frac{10}{36} = \frac{5}{18}$.

We need to find the conditional probability of event A occurring given that event B has occurred, $P(A | B)$. The formula is:

$P(A | B) = \frac{P(A \cap B)}{P(B)}$

Find the intersection of events A and B, $A \cap B$. This is the event where the sum is 3 AND the sum is less than 6. All outcomes with a sum of 3 also have a sum less than 6. Thus, $A \cap B = A$.

$A \cap B = \{(1,2), (2,1)\}$

The number of outcomes in $A \cap B$ is $|A \cap B| = 2$.

The probability of $A \cap B$ is $P(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{2}{36} = \frac{1}{18}$.

Substitute the probabilities $P(A \cap B)$ and $P(B)$ into the conditional probability formula:

$P(A | B) = \frac{\frac{1}{18}}{\frac{5}{18}}$

To divide by a fraction, multiply by its reciprocal:

$P(A | B) = \frac{1}{18} \times \frac{18}{5}$

$P(A | B) = \frac{1 \times \cancel{18}^{1}}{\cancel{18}_{1} \times 5} = \frac{1}{5}$

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C) $\frac{1}{5}$.

Solution:

A binomial distribution arises from a sequence of independent trials, each with exactly two possible outcomes.

The standard requirements (conditions) for a probability experiment to follow a binomial distribution are:

1. There is a fixed number of trials (denoted by $n$).

2. Each trial has only two possible, mutually exclusive outcomes. These outcomes are typically labeled "success" and "failure".

3. The trials are independent. The outcome of one trial does not affect the outcome of any other trial.

4. The probability of success ($p$) is the same for each trial. Consequently, the probability of failure ($1-p$) is also the same for each trial.

Let's examine the given options:

(A) There are 2 outcomes for each trial

This is a requirement (condition 2).

(B) There is a fixed number of trials

This is a requirement (condition 1).

(C) The outcomes must be dependent on each other

This statement contradicts the requirement that the trials are independent (condition 3).

(D) The probability of success must be the same for all the trials

This is a requirement (condition 4).

Therefore, the statement that is NOT a requirement of a binomial distribution is that the outcomes must be dependent on each other.

The correct answer is (C) The outcomes must be dependent on each other.

Given:

A well-shuffled deck of 52 playing cards.

Number of cards drawn = 2.

Drawing is done with replacement.

To Find:

The probability that both cards drawn are queens.

Solution:

A standard deck of 52 playing cards contains 4 queens (one in each suit: spades, clubs, hearts, diamonds).

Let A be the event that the first card drawn is a queen.

The probability of drawing a queen on the first draw is:

$P(A) = \frac{\text{Number of queens}}{\text{Total number of cards}} = \frac{4}{52}$

$P(A) = \frac{\cancel{4}^{1}}{\cancel{52}_{13}} = \frac{1}{13}$

Since the first card is replaced before the second card is drawn, the deck remains unchanged for the second draw. The outcome of the first draw does not affect the outcome of the second draw. Thus, the two drawing events are independent.

Let B be the event that the second card drawn is a queen.

The probability of drawing a queen on the second draw is the same as the first draw:

$P(B) = \frac{\text{Number of queens}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13}$

For independent events, the probability that both events A and B occur (i.e., the first card is a queen AND the second card is a queen) is the product of their individual probabilities:

$P(A \cap B) = P(A) \cdot P(B)$

Substitute the probabilities calculated above:

$P(\text{both queens}) = \frac{1}{13} \times \frac{1}{13}$

Comparing this result with the given options, we find that it matches option (A).

The correct answer is (A) $\frac{1}{13} \times \frac{1}{13}$.

Given:

Number of true-false questions = 10

For each question, there are two possible outcomes: guessing correctly or guessing incorrectly.

The trials (guessing each question) are independent.

The probability of guessing correctly on a single true-false question is $p = \frac{1}{2}$.

The probability of guessing incorrectly is $1-p = 1 - \frac{1}{2} = \frac{1}{2}$.

This is a binomial distribution problem with $n=10$ trials and probability of success $p = \frac{1}{2}$.

To Find:

The probability of guessing correctly at least 8 out of 10 answers.

Solution:

Let X be the number of correctly guessed answers in 10 trials. X follows a binomial distribution with parameters $n=10$ and $p=\frac{1}{2}$.

The probability mass function of a binomial distribution is given by:

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$

For this problem, $n=10$, $p=\frac{1}{2}$, and $1-p=\frac{1}{2}$. So,

$P(X=k) = \binom{10}{k} (\frac{1}{2})^k (\frac{1}{2})^{10-k} = \binom{10}{k} (\frac{1}{2})^{10}$

$P(X=k) = \binom{10}{k} \frac{1}{1024}$

We need to find the probability of getting at least 8 correct answers, which means $P(X \ge 8)$. This is the sum of the probabilities of getting exactly 8, 9, or 10 correct answers.

$P(X \ge 8) = P(X=8) + P(X=9) + P(X=10)$

Calculate each term:

For $k=8$:

$P(X=8) = \binom{10}{8} \frac{1}{1024}$

$\binom{10}{8} = \binom{10}{10-8} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45$

$P(X=8) = 45 \times \frac{1}{1024} = \frac{45}{1024}$

For $k=9$:

$P(X=9) = \binom{10}{9} \frac{1}{1024}$

$\binom{10}{9} = \binom{10}{10-9} = \binom{10}{1} = 10$

$P(X=9) = 10 \times \frac{1}{1024} = \frac{10}{1024}$

For $k=10$:

$P(X=10) = \binom{10}{10} \frac{1}{1024}$

$\binom{10}{10} = 1$

$P(X=10) = 1 \times \frac{1}{1024} = \frac{1}{1024}$

Sum the probabilities:

$P(X \ge 8) = \frac{45}{1024} + \frac{10}{1024} + \frac{1}{1024}$

$P(X \ge 8) = \frac{45 + 10 + 1}{1024}$

$P(X \ge 8) = \frac{56}{1024}$

Simplify the fraction $\frac{56}{1024}$. Both numerator and denominator are divisible by 8.

$\frac{\cancel{56}^{7}}{\cancel{1024}_{128}}$

$P(X \ge 8) = \frac{7}{128}$

Comparing this result with the given options, we find that it matches option (B).

The correct answer is (B) $\frac{7}{128}$.

Given:

Probability that a person is not a swimmer = 0.3

Total number of persons = 5

Number of swimmers required = 4

To Find:

The probability that out of 5 persons, exactly 4 are swimmers.

Solution:

Let S be the event that a person is a swimmer.

Let NS be the event that a person is not a swimmer.

We are given the probability that a person is not a swimmer:

P(NS) = 0.3

The probability that a person is a swimmer is therefore:

P(S) = 1 - P(NS)

P(S) = 1 - 0.3

P(S) = 0.7

This problem can be modeled using a binomial distribution, as we have a fixed number of independent trials (checking each person), and for each trial, there are only two outcomes (swimmer or not a swimmer) with constant probabilities.

Let n be the total number of trials (persons), so $n = 5$.

Let k be the number of successes (swimmers) we are interested in, so $k = 4$.

Let p be the probability of success (a person being a swimmer) in a single trial, so $p = P(S) = 0.7$.

Let q be the probability of failure (a person not being a swimmer) in a single trial, so $q = P(NS) = 0.3$.

The probability of getting exactly k successes in n trials in a binomial distribution is given by the formula:

Substitute the values $n = 5$, $k = 4$, $p = 0.7$, and $q = 0.3$ into the formula:

P(exactly 4 swimmers) = $\binom{5}{4} (0.7)^4 (0.3)^{5-4}$

P(exactly 4 swimmers) = $\binom{5}{4} (0.7)^4 (0.3)^1$

The term $\binom{5}{4}$ is the binomial coefficient, which can also be written as ⁵C₄.

So, P(exactly 4 swimmers) = ⁵C₄ (0.7)⁴ (0.3)

Now, we compare this result with the given options:

(A) ⁵C₄ (0.7)⁴ (0.3)

(B) ⁵C₁ (0.7) (0.3)⁴

(C) ⁵C₄ (0.7) (0.3)⁴

(D) (0.7)⁴ (0.3)

Our calculated probability matches option (A).

Final Answer: The correct option is (A) ⁵C₄ (0.7)⁴ (0.3).

Given:

The probability distribution of a discrete random variable X is given by the table:

$P(X=2) = \frac{5}{k}$

$P(X=3) = \frac{7}{k}$

$P(X=4) = \frac{9}{k}$

$P(X=5) = \frac{11}{k}$

To Find:

The value of the constant k.

Solution:

For any valid probability distribution of a discrete random variable, the sum of the probabilities of all possible values must be equal to 1.

That is, $\sum P(X=x_i) = 1$.

Summing the given probabilities:

$P(X=2) + P(X=3) + P(X=4) + P(X=5) = 1$

$\frac{5}{k} + \frac{7}{k} + \frac{9}{k} + \frac{11}{k} = 1$

Since all fractions have the same denominator k, we can add the numerators:

$\frac{5 + 7 + 9 + 11}{k} = 1$

Calculate the sum of the numerators:

$5 + 7 + 9 + 11 = 12 + 9 + 11 = 21 + 11 = 32$

Substitute the sum back into the equation:

$\frac{32}{k} = 1$

To solve for k, multiply both sides of the equation by k:

$32 = 1 \times k$

$k = 32$

Comparing this result with the given options, we find that it matches option (C).

The correct answer is (C) 32.

Given:

The probability distribution of a random variable X is given by the table:

To Find:

The expected value E(X).

Solution:

The expected value E(X) of a discrete random variable X is calculated as the sum of the products of each possible value of X and its corresponding probability P(X).

The formula for the expected value is:

$E(X) = \sum x_i P(X=x_i)$

Using the values from the given table, we calculate the product of each value of X and its probability:

• For X = -4, $x_1 P(X=x_1) = (-4) \times 0.1 = -0.4$
• For X = -3, $x_2 P(X=x_2) = (-3) \times 0.2 = -0.6$
• For X = -2, $x_3 P(X=x_3) = (-2) \times 0.3 = -0.6$
• For X = -1, $x_4 P(X=x_4) = (-1) \times 0.2 = -0.2$
• For X = 0, $x_5 P(X=x_5) = (0) \times 0.2 = 0$

Now, we sum these products to find E(X):

$E(X) = (-0.4) + (-0.6) + (-0.6) + (-0.2) + (0)$

$E(X) = -0.4 - 0.6 - 0.6 - 0.2 + 0$

$E(X) = -1.0 - 0.6 - 0.2$

$E(X) = -1.6 - 0.2$

$E(X) = -1.8$

Comparing the calculated value with the given options:

(A) 0

(B) –1

(C) –2

(D) –1.8

The calculated expected value is -1.8, which corresponds to option (D).

Final Answer: The correct option is (D) –1.8.

Given:

The probability distribution of a random variable X is given by the table:

To Find:

The expected value of X squared, E(X$^2$).

Solution:

The expected value of a function of a discrete random variable, $g(X)$, is given by the formula:

$E(g(X)) = \sum\limits_{i} g(x_i) P(X=x_i)$

In this case, $g(X) = X^2$. So, we need to calculate $E(X^2)$ using the formula:

$E(X^2) = \sum\limits_{i} x_i^2 P(X=x_i)$

We need to calculate $x_i^2$ for each value of $x_i$ and multiply it by the corresponding probability $P(X=x_i)$.

For $x_1 = 1$: $x_1^2 P(X=x_1) = (1)^2 \times \frac{1}{10} = 1 \times \frac{1}{10} = \frac{1}{10}$

For $x_2 = 2$: $x_2^2 P(X=x_2) = (2)^2 \times \frac{1}{5} = 4 \times \frac{1}{5} = \frac{4}{5}$

For $x_3 = 3$: $x_3^2 P(X=x_3) = (3)^2 \times \frac{3}{10} = 9 \times \frac{3}{10} = \frac{27}{10}$

For $x_4 = 4$: $x_4^2 P(X=x_4) = (4)^2 \times \frac{2}{5} = 16 \times \frac{2}{5} = \frac{32}{5}$

Now, sum these values to find E(X$^2$):

$E(X^2) = \frac{1}{10} + \frac{4}{5} + \frac{27}{10} + \frac{32}{5}$

To add these fractions, we find a common denominator, which is 10.

Convert the fractions with denominator 5 to have a denominator of 10:

$\frac{4}{5} = \frac{4 \times 2}{5 \times 2} = \frac{8}{10}$

$\frac{32}{5} = \frac{32 \times 2}{5 \times 2} = \frac{64}{10}$

Now, substitute these back into the sum for E(X$^2$):

$E(X^2) = \frac{1}{10} + \frac{8}{10} + \frac{27}{10} + \frac{64}{10}$

$E(X^2) = \frac{1 + 8 + 27 + 64}{10}$

$E(X^2) = \frac{100}{10}$

$E(X^2) = 10$

Comparing the calculated value with the given options:

(A) 3

(B) 5

(C) 7

(D) 10

The calculated expected value E(X$^2$) is 10, which corresponds to option (D).

Final Answer: The correct option is (D) 10.

Given:

A random variable X follows a binomial distribution with parameters n and p, where $0 < p < 1$.

The ratio $\frac{P(X=r)}{P(X=n-r)}$ is independent of n and r.

To Find:

The value of p.

Solution:

For a binomial distribution, the probability mass function (PMF) is given by:

$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$

Using this formula, we can write the expressions for P(X=r) and P(X=n-r):

$P(X=r) = \binom{n}{r} p^r (1-p)^{n-r}$

$P(X=n-r) = \binom{n}{n-r} p^{n-r} (1-p)^{n-(n-r)} = \binom{n}{n-r} p^{n-r} (1-p)^r$

Now, let's form the ratio $\frac{P(X=r)}{P(X=n-r)}$:

$\frac{P(X=r)}{P(X=n-r)} = \frac{\binom{n}{r} p^r (1-p)^{n-r}}{\binom{n}{n-r} p^{n-r} (1-p)^r}$

We know that the binomial coefficient $\binom{n}{k}$ has the property $\binom{n}{k} = \binom{n}{n-k}$. Therefore, $\binom{n}{r} = \binom{n}{n-r}$.

Substituting this into the ratio, the binomial coefficients cancel out:

$\frac{P(X=r)}{P(X=n-r)} = \frac{\cancel{\binom{n}{r}} p^r (1-p)^{n-r}}{\cancel{\binom{n}{r}} p^{n-r} (1-p)^r}$

$\frac{P(X=r)}{P(X=n-r)} = \frac{p^r (1-p)^{n-r}}{p^{n-r} (1-p)^r}$

Using the properties of exponents $\frac{a^x}{a^y} = a^{x-y}$, we can simplify the terms with p and (1-p):

$\frac{p^r}{p^{n-r}} = p^{r - (n-r)} = p^{r - n + r} = p^{2r - n}$

$\frac{(1-p)^{n-r}}{(1-p)^r} = (1-p)^{(n-r) - r} = (1-p)^{n - 2r}$

So the ratio becomes:

$\frac{P(X=r)}{P(X=n-r)} = p^{2r - n} (1-p)^{n - 2r}$

We can rewrite the second term as $(1-p)^{-(2r-n)}$.

$\frac{P(X=r)}{P(X=n-r)} = p^{2r - n} (1-p)^{-(2r - n)}$

$\frac{P(X=r)}{P(X=n-r)} = \frac{p^{2r - n}}{(1-p)^{2r - n}}$

$\frac{P(X=r)}{P(X=n-r)} = \left(\frac{p}{1-p}\right)^{2r - n}$

We are given that this ratio is independent of n and r.

The expression is $\left(\frac{p}{1-p}\right)^{2r - n}$. For this expression to be constant and not depend on n or r, the base $\left(\frac{p}{1-p}\right)$ must be equal to 1 (since the exponent $2r-n$ varies with n and r).

Set the base equal to 1:

$\frac{p}{1-p} = 1$

Multiply both sides by $(1-p)$:

$p = 1 \times (1-p)$

$p = 1 - p$

Add p to both sides:

$p + p = 1$

$2p = 1$

Divide by 2:

$p = \frac{1}{2}$

The value $p = \frac{1}{2}$ satisfies the condition $0 < p < 1$.

Comparing the result with the given options:

(A) $\frac{1}{2}$

(B) $\frac{1}{3}$

(C) $\frac{1}{5}$

(D) $\frac{1}{7}$

The calculated value of p is $\frac{1}{2}$, which matches option (A).

Final Answer: The correct option is (A) $\frac{1}{2}$.

Given:

Let P be the event that a student fails in Physics.

Let M be the event that a student fails in Mathematics.

Probability that a student fails in Physics, P(P) = 30% = 0.30

Probability that a student fails in Mathematics, P(M) = 25% = 0.25

Probability that a student fails in both Physics and Mathematics, P(P $\cap$ M) = 10% = 0.10

To Find:

The probability that a student fails in physics if she has failed in mathematics. This is the conditional probability P(P | M).

Solution:

We need to find the conditional probability P(P | M), which is the probability of failing in Physics given that the student has failed in Mathematics.

The formula for conditional probability P(A | B) is given by $\frac{P(A \cap B)}{P(B)}$.

In this case, A is the event of failing in Physics (P) and B is the event of failing in Mathematics (M).

Substitute the given values into the formula:

$P(P \cap M) = 0.10$

$P(M) = 0.25$

$P(P | M) = \frac{0.10}{0.25}$

To simplify this fraction, we can write the decimals as fractions:

$0.10 = \frac{10}{100}$

$0.25 = \frac{25}{100}$

So, the probability is:

$P(P | M) = \frac{\frac{10}{100}}{\frac{25}{100}}$

$P(P | M) = \frac{10}{100} \times \frac{100}{25}$

$P(P | M) = \frac{10 \times \cancel{100}}{\cancel{100} \times 25}$

$P(P | M) = \frac{10}{25}$

Now, simplify the fraction $\frac{10}{25}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 5.

$P(P | M) = \frac{\cancel{10}^2}{\cancel{25}_5}$

$P(P | M) = \frac{2}{5}$

Comparing the calculated probability with the given options:

(A) $\frac{1}{10}$

(B) $\frac{2}{5}$

(C) $\frac{9}{20}$

(D) $\frac{1}{3}$

The calculated probability $\frac{2}{5}$ matches option (B).

Final Answer: The correct option is (B) $\frac{2}{5}$.

Given:

Probability that student A solves the problem correctly, P(A$_C$) = $\frac{1}{3}$.

Probability that student B solves the problem correctly, P(B$_C$) = $\frac{1}{4}$.

Probability of them making a common error, which leads to the same wrong answer, is related to $\frac{1}{20}$.

To Find:

The probability that their answer is correct, given that they obtain the same answer.

Solution:

Let A$_C$ be the event that student A solves the problem correctly.

Let B$_C$ be the event that student B solves the problem correctly.

Let A$_W$ be the event that student A solves the problem incorrectly. P(A$_W$) = $1 - P(A_C) = 1 - \frac{1}{3} = \frac{2}{3}$.

Let B$_W$ be the event that student B solves the problem incorrectly. P(B$_W$) = $1 - P(B_C) = 1 - \frac{1}{4} = \frac{3}{4}$.

We assume that the events of solving correctly are independent for A and B.

The probability that both A and B solve the problem correctly is:

P(A$_C \cap$ B$_C$) = $\frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$.

This is the probability that they get the same answer and that answer is correct.

The probability that both A and B solve the problem incorrectly is (assuming independence of solving incorrectly):

P(A$_W \cap$ B$_W$) = $\frac{2}{3} \times \frac{3}{4} = \frac{6}{12} = \frac{1}{2}$.

Let S be the event that they obtain the same answer. The same answer can be obtained in two mutually exclusive ways:

1. Both solve the problem correctly (event A$_C \cap$ B$_C$).

2. Both solve the problem incorrectly, but make the same error, leading to the same wrong answer. Let this event be E$_{\text{same\_wrong}}$.

The phrasing "the probability of their making a common error is, $\frac{1}{20}$" is often interpreted in such problems as the conditional probability that they make the same type of error given that they are both wrong. Let's assume this interpretation is intended to match the options.

So, we interpret the given information as P(E$_{\text{same\_error\_type}}$ | A$_W \cap$ B$_W$) = $\frac{1}{20}$, where E$_{\text{same\_error\_type}}$ is the event they make the same type of error.

The event E$_{\text{same\_wrong}}$ (both wrong and same wrong answer) is the event (A$_W \cap$ B$_W$) AND E$_{\text{same\_error\_type}}$.

P(E$_{\text{same\_wrong}}$) = $\frac{1}{20} \times \frac{1}{2} = \frac{1}{40}$.

The event S (same answer) is the union of the event (A$_C \cap$ B$_C$) and the event E$_{\text{same\_wrong}}$. These two events are mutually exclusive.

P(S) = $\frac{1}{12} + \frac{1}{40}$.

To add these fractions, find a common denominator. The least common multiple of 12 and 40 is 120.

P(S) = $\frac{1 \times 10}{12 \times 10} + \frac{1 \times 3}{40 \times 3} = \frac{10}{120} + \frac{3}{120} = \frac{10 + 3}{120} = \frac{13}{120}$.

We need to find the probability that their answer is correct, given that they obtained the same answer. This is P(A$_C \cap$ B$_C$ | S).

Using the formula for conditional probability:

The event (A$_C \cap$ B$_C$), which means both are correct, implies that they obtained the same answer (the correct one). Therefore, the event (A$_C \cap$ B$_C$) is a subset of the event S. This means the intersection (A$_C \cap$ B$_C$) $\cap$ S is simply (A$_C \cap$ B$_C$).

So, the formula becomes:

Substitute the calculated values:

P(A$_C \cap$ B$_C$ | S) = $\frac{1/12}{13/120}$

P(A$_C \cap$ B$_C$ | S) = $\frac{1}{12} \times \frac{120}{13}$

P(A$_C \cap$ B$_C$ | S) = $\frac{120}{12 \times 13}$

Cancel out the common factor of 12:

P(A$_C \cap$ B$_C$ | S) = $\frac{\cancel{120}^{10}}{\cancel{12}_1 \times 13}$

P(A$_C \cap$ B$_C$ | S) = $\frac{10}{13}$

Comparing the result with the given options:

(A) $\frac{1}{12}$

(B) $\frac{1}{40}$

(C) $\frac{13}{120}$

(D) $\frac{10}{13}$

The calculated probability is $\frac{10}{13}$, which matches option (D).

Note: This solution relies on the interpretation that the given probability of a common error ($1/20$) is the probability of making the same type of error *given* that both students are wrong.

Final Answer: The correct option is (D) $\frac{10}{13}$.

Given:

Total number of pens in the box = 100.

Number of defective pens = 10.

Number of non-defective pens = $100 - 10 = 90$.

Sample size = 5 pens.

Drawing is done one by one with replacement.

To Find:

The probability that out of the sample of 5 pens, at most one is defective.

Solution:

Let D be the event of drawing a defective pen.

Let ND be the event of drawing a non-defective pen.

The probability of drawing a defective pen is P(D) = $\frac{\text{Number of defective pens}}{\text{Total number of pens}} = \frac{10}{100} = \frac{1}{10}$.

The probability of drawing a non-defective pen is P(ND) = $\frac{\text{Number of non-defective pens}}{\text{Total number of pens}} = \frac{90}{100} = \frac{9}{10}$.

Since the pens are drawn with replacement, the probability of drawing a defective or non-defective pen remains constant for each draw. This scenario fits a binomial distribution.

Let n be the number of trials (pens drawn), so $n = 5$.

Let X be the random variable representing the number of defective pens in the sample of 5 draws.

Let p be the probability of success (drawing a defective pen) in a single trial, so $p = P(D) = \frac{1}{10}$.

Let q be the probability of failure (drawing a non-defective pen) in a single trial, so $q = P(ND) = \frac{9}{10}$.

The probability mass function for a binomial distribution is given by:

$P(X=k) = \binom{n}{k} p^k q^{n-k}$

In this case, $n=5$, $p=\frac{1}{10}$, and $q=\frac{9}{10}$.

$P(X=k) = \binom{5}{k} \left(\frac{1}{10}\right)^k \left(\frac{9}{10}\right)^{5-k}$

We need to find the probability that at most one pen is defective. This means the number of defective pens is either 0 or 1. So, we need to calculate $P(X \leq 1) = P(X=0) + P(X=1)$.

Calculate P(X=0): Probability of drawing exactly 0 defective pens.

Here, $k=0$.

$P(X=0) = \binom{5}{0} \left(\frac{1}{10}\right)^0 \left(\frac{9}{10}\right)^{5-0}$

$P(X=0) = \left(\frac{9}{10}\right)^5$

Calculate P(X=1): Probability of drawing exactly 1 defective pen.

Here, $k=1$.

$P(X=1) = \binom{5}{1} \left(\frac{1}{10}\right)^1 \left(\frac{9}{10}\right)^{5-1}$

$P(X=1) = \frac{5}{10} \times \left(\frac{9}{10}\right)^4$

Simplify $\frac{5}{10}$ to $\frac{1}{2}$:

$P(X=1) = \frac{1}{2} \times \left(\frac{9}{10}\right)^4$

Now, add P(X=0) and P(X=1) to find $P(X \leq 1)$.

$P(X \leq 1) = P(X=0) + P(X=1) = \left(\frac{9}{10}\right)^5 + \frac{1}{2} \left(\frac{9}{10}\right)^4$

Comparing the result with the given options:

(A) $\left( \frac{9}{10} \right)^5$

(B) $\frac{1}{2} \left( \frac{9}{10} \right)^4$

(C) $\frac{1}{2} \left( \frac{9}{10} \right)^4$

(D) $\left( \frac{9}{10} \right)^5 + \frac{1}{2} \left( \frac{9}{10} \right)^4$

Our calculated probability matches option (D).

Final Answer: The correct option is (D) $\left( \frac{9}{10} \right)^5 + \frac{1}{2} \left( \frac{9}{10} \right)^4$.

Given:

Events A and B with probabilities P(A) > 0 and P(B) > 0.

To Determine:

Whether events A and B can be both mutually exclusive and independent under the given conditions.

Solution:

Let's consider the definitions of mutually exclusive events and independent events.

Mutually Exclusive Events: Two events A and B are mutually exclusive if they cannot occur simultaneously. This means their intersection is an empty set ($A \cap B = \emptyset$).

If A and B are mutually exclusive, the probability of their intersection is:

Independent Events: Two events A and B are independent if the occurrence of one event does not affect the probability of the occurrence of the other event.

If A and B are independent, the probability of their intersection is given by the product of their individual probabilities:

Now, let's consider the condition for A and B to be both mutually exclusive and independent.

If A and B are mutually exclusive, then P(A $\cap$ B) = 0.

If A and B are also independent, then P(A $\cap$ B) must equal P(A) $\times$ P(B).

For both conditions to hold simultaneously, we must have:

However, the problem statement gives us that P(A) > 0 and P(B) > 0.

If P(A) > 0 and P(B) > 0, then their product must be strictly positive:

We have arrived at a contradiction: P(A) $\times$ P(B) must be 0 for the events to be both mutually exclusive and independent, but the given conditions state that P(A) $\times$ P(B) is greater than 0.

Therefore, if P(A) > 0 and P(B) > 0, the events A and B cannot be both mutually exclusive and independent.

• If they are mutually exclusive, P(A $\cap$ B) = 0. Since P(A)P(B) > 0, P(A $\cap$ B) $\neq$ P(A)P(B), so they are not independent.
• If they are independent, P(A $\cap$ B) = P(A)P(B). Since P(A)P(B) > 0, P(A $\cap$ B) > 0, so their intersection is not empty ($A \cap B \neq \emptyset$), meaning they are not mutually exclusive.

The only way for P(A) $\times$ P(B) to be 0 is if at least one of P(A) or P(B) is 0. But the question specifies P(A) > 0 and P(B) > 0.

Conclusion: The statement that A and B can be both mutually exclusive and independent when P(A) > 0 and P(B) > 0 is false.

Answer: False

Given:

Events A and B are independent.

To Determine:

Whether A′ and B′ (the complements of A and B) are also independent.

Solution:

Two events A and B are independent if and only if the probability of their intersection is equal to the product of their individual probabilities:

We are given that A and B are independent, so this condition holds.

We need to check if A′ and B′ are independent, which means we need to check if P(A′ $\cap$ B′) = P(A′) P(B′).

By De Morgan's Law, the intersection of the complements A′ $\cap$ B′ is equal to the complement of the union of A and B:

$A' \cap B' = (A \cup B)'$

Using the property of complements, the probability of $(A \cup B)'$ is $1 - P(A \cup B)$:

$P(A' \cap B') = P((A \cup B)')$

$P(A' \cap B') = 1 - P(A \cup B)$

The probability of the union of two events is given by the formula:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute this into the expression for $P(A' \cap B')$:

$P(A' \cap B') = 1 - [P(A) + P(B) - P(A \cap B)]$

$P(A' \cap B') = 1 - P(A) - P(B) + P(A \cap B)$

Since A and B are independent, we can substitute P(A $\cap$ B) = P(A) P(B):

$P(A' \cap B') = 1 - P(A) - P(B) + P(A) P(B)$

Now, let's factor the expression on the right side:

$1 - P(A) - P(B) + P(A) P(B) = (1 - P(A)) - P(B) (1 - P(A))$

$= (1 - P(A)) (1 - P(B))$

We know that $1 - P(A) = P(A')$ and $1 - P(B) = P(B')$.

So, the expression becomes:

$P(A' \cap B') = P(A') P(B')$

This result shows that the probability of the intersection of A′ and B′ is equal to the product of their individual probabilities.

By the definition of independent events, A′ and B′ are independent.

Conclusion: The statement is true.

Answer: True

Given:

A statement claiming that if two events A and B are mutually exclusive, then they are also independent.

To Determine:

Whether the given statement is True or False.

Solution:

Let's recall the definitions of mutually exclusive and independent events.

Mutually Exclusive Events: Two events A and B are mutually exclusive if their intersection is an empty set, meaning they cannot occur at the same time.

Mathematically, this is expressed as:

This implies that the probability of their intersection is zero:

Independent Events: Two events A and B are independent if the occurrence of one event does not influence the probability of the occurrence of the other event.

Mathematically, this is expressed as:

The statement says that if A and B are mutually exclusive (i.e., P(A $\cap$ B) = 0), then they must also be independent (i.e., P(A $\cap$ B) = P(A) P(B)).

So, if A and B are mutually exclusive, the condition for independence becomes:

This equation holds if and only if at least one of the probabilities, P(A) or P(B), is zero.

Consider the case where P(A) > 0 and P(B) > 0. If A and B are mutually exclusive, then P(A $\cap$ B) = 0. However, since P(A) > 0 and P(B) > 0, their product P(A) P(B) > 0.

In this case, $P(A \cap B) = 0$ and $P(A)P(B) > 0$. Since $0 \neq P(A)P(B)$, the events A and B are not independent.

This shows that if two mutually exclusive events both have non-zero probabilities, they cannot be independent.

For example, consider flipping a coin. Let A be the event of getting a Head, and B be the event of getting a Tail. These events are mutually exclusive (you cannot get both a Head and a Tail on a single flip). $P(A) = 1/2$ and $P(B) = 1/2$. Both are greater than 0.

$P(A \cap B) = P(\emptyset) = 0$.

$P(A) P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Since $0 \neq \frac{1}{4}$, A and B are not independent. The outcome of getting a Head *does* affect the probability of getting a Tail (it makes it 0 for that flip, whereas if A did not occur, the probability of B would be 1/2).

The statement is only true in the special case where at least one of the events A or B has a probability of 0 (i.e., it is an impossible event or has probability 0). If P(A) = 0, then $P(A \cap B) = 0$ (since $A \cap B \subseteq A$), and $P(A)P(B) = 0 \cdot P(B) = 0$. In this specific case, $P(A \cap B) = P(A)P(B)$, so they are independent. However, the statement claims it is true for *all* mutually exclusive events.

Conclusion: The statement is false because mutually exclusive events are not independent if both events have positive probabilities.

Answer: False

Given:

A statement claiming that two independent events are always mutually exclusive.

To Determine:

Whether the given statement is True or False.

Solution:

Let's define independent events and mutually exclusive events.

Independent Events: Two events A and B are independent if the probability of their intersection is equal to the product of their individual probabilities.

Mutually Exclusive Events: Two events A and B are mutually exclusive if they cannot occur at the same time, meaning their intersection is an empty set.

This implies that the probability of their intersection is zero:

The statement says that if P(A $\cap$ B) = P(A) P(B), then it must always be true that P(A $\cap$ B) = 0.

Substituting the condition for independence into the condition for mutually exclusive events, we would need:

This equation P(A)P(B) = 0 is true if and only if P(A) = 0 or P(B) = 0 (or both).

However, independent events do not require P(A) or P(B) to be zero. For example, consider flipping a fair coin twice.

Let A be the event of getting a Head on the first flip. P(A) = $\frac{1}{2}$.

Let B be the event of getting a Head on the second flip. P(B) = $\frac{1}{2}$.

These two events are independent, as the outcome of the first flip does not affect the outcome of the second flip.

The event A $\cap$ B is getting a Head on the first flip AND a Head on the second flip (HH). The probability of this is P(A $\cap$ B) = $\frac{1}{4}$.

Let's check the independence condition:

$P(A) P(B) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Since $P(A \cap B) = \frac{1}{4}$ and $P(A) P(B) = \frac{1}{4}$, we have $P(A \cap B) = P(A) P(B)$. Thus, events A and B are independent.

Now, let's check if these events are mutually exclusive. Mutually exclusive means $P(A \cap B) = 0$.

In this example, $P(A \cap B) = \frac{1}{4}$, which is not equal to 0.

Therefore, the events of getting a Head on the first flip and getting a Head on the second flip are independent, but they are not mutually exclusive.

This counterexample shows that two independent events are not always mutually exclusive. They are mutually exclusive only in the trivial case where at least one of the events has probability 0.

Conclusion: The statement is false.

Answer: False

Given:

A statement relating the probability of the intersection of two events to their individual probabilities, given that the events are independent.

To Determine:

Whether the statement "If A and B are two independent events then P(A and B) = P(A) . P(B)" is True or False.

Solution:

In probability theory, two events A and B are defined as independent if the occurrence of one event does not affect the probability of the occurrence of the other event.

Mathematically, this definition of independence is formally stated as:

Two events A and B are independent if and only if the probability of their joint occurrence (the probability of A and B both happening), denoted as P(A $\cap$ B) or P(A and B), is equal to the product of their individual probabilities P(A) and P(B).

The statement given in the question, "If A and B are two independent events then P(A and B) = P(A) . P(B)", is precisely this definition of independence.

Therefore, the statement is correct by the definition of independent events.

Conclusion: The statement accurately reflects the definition of independent events.

Answer: True

Given:

A statement claiming that "Another name for the mean of a probability distribution is expected value."

To Determine:

Whether the given statement is True or False.

Solution:

In probability and statistics, the mean of a probability distribution is a measure of the central tendency of the distribution. It represents the average value that a random variable is expected to take over many trials.

For a discrete probability distribution with possible values $x_1, x_2, x_3, \ldots$ and corresponding probabilities $P(X=x_1), P(X=x_2), P(X=x_3), \ldots$, the mean (often denoted by $\mu$) is calculated as:

$\mu = \sum\limits_{i} x_i P(X=x_i)$

The expected value (often denoted by E(X) or $\mu$) of a random variable X is defined as the sum of the products of each possible value of the random variable and its probability.

For a discrete random variable X, the expected value is calculated as:

$E(X) = \sum\limits_{i} x_i P(X=x_i)$

Comparing the formulas for the mean ($\mu$) and the expected value (E(X)), we see that they are identical.

The term "expected value" is indeed synonymous with the "mean" of a random variable's probability distribution. It represents the long-term average value of the random variable.

Conclusion: The statement accurately describes the relationship between the mean and the expected value of a probability distribution.

Answer: True

Given:

Events A and B′, the complement of B, are independent.

To Determine:

Whether the statement P(A' $\cup$ B) = 1 – P(A) P(B') is True or False.

Solution:

We are given that events A and B′ are independent. By the definition of independent events, this means:

We need to evaluate the probability of the union of the complement of A (A′) and event B, i.e., P(A' $\cup$ B).

We can use the property that the probability of the complement of an event E is P(E') = 1 - P(E).

Consider the event $(A' \cup B)$. The complement of this event is $(A' \cup B)'$.

Using De Morgan's Law, $(A' \cup B)' = (A')' \cap B' = A \cap B'$.

So, the probability of the complement of $(A' \cup B)$ is P($(A' \cup B)'$) = P($A \cap B'$).

Now, using the complement rule:

P(A' $\cup$ B) = 1 - P($(A' \cup B)'$)

P(A' $\cup$ B) = 1 - P($A \cap B'$)

Since we are given that A and B′ are independent, we know that $P(A \cap B') = P(A) P(B')$.

Substitute this into the expression for P(A' $\cup$ B):

P(A' $\cup$ B) = 1 - P(A) P(B')

This result matches the statement given in the question.

Conclusion: The statement is true because if A and B′ are independent, the probability of their intersection is the product of their individual probabilities, and using De Morgan's law and the complement rule leads directly to the given expression for P(A' $\cup$ B).

Answer: True

Given:

Events A and B are independent.

To Determine:

Whether the statement P(exactly one of A, B occurs) = P(A) P(B’) + P(A’) is True or False.

Solution:

The event "exactly one of A, B occurs" means that either event A occurs and event B does not occur, OR event B occurs and event A does not occur.

In terms of set notation, this event can be written as the union of two mutually exclusive events:

(A $\cap$ B′) $\cup$ (A′ $\cap$ B)

where A′ is the complement of A and B′ is the complement of B.

The probability of the union of two mutually exclusive events is the sum of their probabilities:

We are given that A and B are independent events. A property of independent events is that if A and B are independent, then the following pairs of events are also independent:

• A and B′ are independent.
• A′ and B are independent.
• A′ and B′ are independent.

Using the definition of independence for the pairs (A, B′) and (A′, B):

Substitute these probabilities back into the expression for P(exactly one of A, B occurs):

P(exactly one of A, B occurs) = P(A) P(B′) + P(A′) P(B)

Now, let's compare this derived expression with the statement given in the question:

Given Statement: P(exactly one of A, B occurs) = P(A) P(B’) + P(A’)

Derived Correct Expression: P(exactly one of A, B occurs) = P(A) P(B’) + P(A’) P(B)

For the given statement to be true for all independent events A and B, the following must hold:

P(A) P(B′) + P(A′) P(B) = P(A) P(B′) + P(A′)

Subtracting P(A) P(B′) from both sides gives:

P(A′) P(B) = P(A′)

This equation P(A′) P(B) = P(A′) implies either P(A′) = 0 or P(B) = 1 (assuming P(A') $\neq$ 0).

• If P(A′) = 0, then P(A) = 1. If P(A)=1, then A is the certain event (sample space). If A is the certain event and A and B are independent, then B must also be the certain event (or an impossible event) to satisfy independence for all subsets. If P(A)=1, then A' is the impossible event, P(A')=0. In this case, the statement becomes P(exactly one of A, B occurs) = P(A)P(B') + 0. The event "exactly one of A, B occurs" means $(A \cap B') \cup (A' \cap B) = (\Omega \cap B') \cup (\emptyset \cap B) = B' \cup \emptyset = B'$. So P(exactly one of A, B occurs) = P(B'). The statement becomes P(B') = P(A)P(B') = 1 * P(B') = P(B'). So if P(A)=1, the statement holds.
• If P(A′) $\neq$ 0, then we must have P(B) = 1 for the equation P(A′) P(B) = P(A′) to hold. If P(B)=1, then B is the certain event, P(B')=0. The event "exactly one of A, B occurs" means $(A \cap B') \cup (A' \cap B) = (A \cap \emptyset) \cup (A' \cap \Omega) = \emptyset \cup A' = A'$. So P(exactly one of A, B occurs) = P(A'). The statement becomes P(A') = P(A)P(B') + P(A') = P(A)*0 + P(A') = P(A'). So if P(B)=1, the statement holds.

However, the statement is given as a general truth for *any* two independent events A and B, not just those where P(A)=1 or P(B)=1.

Consider independent events where P(A) > 0, P(A) < 1, P(B) > 0, and P(B) < 1 (e.g., two coin flips). Let A be getting heads on the first flip, P(A) = 0.5. Let B be getting heads on the second flip, P(B) = 0.5. A and B are independent.

P(A') = 0.5, P(B') = 0.5.

Correct probability of exactly one occurring = P(A)P(B') + P(A')P(B) = (0.5)(0.5) + (0.5)(0.5) = 0.25 + 0.25 = 0.5.

Statement's claimed probability = P(A)P(B') + P(A') = (0.5)(0.5) + 0.5 = 0.25 + 0.5 = 0.75.

Since 0.5 $\neq$ 0.75, the statement is false for these independent events.

Conclusion: The statement is not true for all independent events; it is only true in specific cases (when P(A)=1 or P(B)=1). Therefore, the general statement is false.

Answer: False

Given:

Events A and B such that P(A) > 0 and P(A) + P(B) > 1.

The statement is P(B | A) $\geq$ 1 - $\frac{P(B’)}{P(A)}$.

To Determine:

Whether the given statement is True or False.

Solution:

The definition of conditional probability P(B | A) is:

We need to analyze the term P(A $\cap$ B).

The probability of the union of two events is given by:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Rearranging this formula to solve for P(A $\cap$ B):

$P(A \cap B) = P(A) + P(B) - P(A \cup B)$

We know that the probability of any event is at most 1. Thus, $P(A \cup B) \leq 1$.

Using this inequality, we can find a lower bound for $P(A \cap B)$:

$P(A \cap B) = P(A) + P(B) - P(A \cup B) \geq P(A) + P(B) - 1$

This inequality is true because subtracting a larger value (or equal) from $P(A) + P(B)$ results in a smaller (or equal) value for the intersection probability.

Now, substitute this lower bound for P(A $\cap$ B) into the conditional probability formula P(B | A) = $\frac{P(A \cap B)}{P(A)}$:

$P(B | A) \geq \frac{P(A) + P(B) - 1}{P(A)}$

We can split the fraction on the right side:

$P(B | A) \geq \frac{P(A)}{P(A)} + \frac{P(B) - 1}{P(A)}$

$P(B | A) \geq 1 + \frac{P(B) - 1}{P(A)}$

$P(B | A) \geq 1 - \frac{1 - P(B)}{P(A)}$

We know that $1 - P(B) = P(B')$, the probability of the complement of B.

Substitute P(B') for $1 - P(B)$:

This is the inequality given in the statement.

The condition P(A) + P(B) > 1 is important here. If $P(A) + P(B) \leq 1$, it would still be possible for the events to have a non-empty intersection (or even be mutually exclusive), but the lower bound $P(A) + P(B) - 1$ would be less than or equal to 0, which doesn't give a useful lower bound for $P(A \cap B)$ (since $P(A \cap B) \geq 0$). The condition $P(A) + P(B) > 1$ guarantees that the intersection probability $P(A \cap B)$ must be strictly positive, as $P(A \cap B) = P(A) + P(B) - P(A \cup B) \geq P(A) + P(B) - 1 > 1 - 1 = 0$.

Conclusion: The derivation shows that the inequality P(B | A) $\geq$ 1 - $\frac{P(B’)}{P(A)}$ is a consequence of the formula for the probability of the union of events and the fact that the probability of the union is at most 1, given that P(A) > 0. The condition P(A) + P(B) > 1 ensures that this inequality provides a meaningful lower bound for P(B|A).

Answer: True

Given:

Events A, B, and C are independent events.

P(A) = P(B) = P(C) = p.

To Determine:

Whether the statement P(At least two of A, B, C occur) = $3p^2 - 2p^3$ is True or False.

Solution:

The event "At least two of A, B, C occur" means that either exactly two of the events occur, or exactly three of the events occur.

The possible ways for exactly two events to occur are:

• A and B occur, but C does not ($A \cap B \cap C'$)
• A and C occur, but B does not ($A \cap B' \cap C$)
• B and C occur, but A does not ($A' \cap B \cap C$)

The way for exactly three events to occur is:

• A, B, and C occur ($A \cap B \cap C$)

These four specific outcomes are mutually exclusive (they cannot happen at the same time).

The event "At least two of A, B, C occur" is the union of these four mutually exclusive events:

$(A \cap B \cap C') \cup (A \cap B' \cap C) \cup (A' \cap B \cap C) \cup (A \cap B \cap C)$

Since A, B, and C are independent, their complements A′, B′, and C′ are also independent of each other and of A, B, and C.

The probability of the intersection of independent events is the product of their individual probabilities.

P(A) = p, P(B) = p, P(C) = p.

P(A') = 1 - P(A) = 1 - p.

P(B') = 1 - P(B) = 1 - p.

P(C') = 1 - P(C) = 1 - p.

Now, calculate the probabilities of the four mutually exclusive events:

P($A \cap B \cap C'$) = P(A)P(B)P(C') = $p \times p \times (1-p) = p^2(1-p)$

P($A \cap B' \cap C$) = P(A)P(B')P(C) = $p \times (1-p) \times p = p^2(1-p)$

P($A' \cap B \cap C$) = P(A')P(B)P(C) = $(1-p) \times p \times p = p^2(1-p)$

P($A \cap B \cap C$) = P(A)P(B)P(C) = $p \times p \times p = p^3$

The probability of "At least two of A, B, C occur" is the sum of these probabilities:

P(At least two of A, B, C occur) = P($A \cap B \cap C'$) + P($A \cap B' \cap C$) + P($A' \cap B \cap C$) + P($A \cap B \cap C$)

P(At least two of A, B, C occur) = $p^2(1-p) + p^2(1-p) + p^2(1-p) + p^3$

P(At least two of A, B, C occur) = $3p^2(1-p) + p^3$

Expand the expression:

P(At least two of A, B, C occur) = $3p^2 - 3p^3 + p^3$

P(At least two of A, B, C occur) = $3p^2 - 2p^3$

This calculated probability matches the expression given in the statement.

Conclusion: The derived probability for "At least two of A, B, C occur" is $3p^2 - 2p^3$, which is exactly the expression given in the statement.

Answer: True

Given:

P(A | B) = p

P(A) = p

P(B) = $\frac{1}{3}$

P(A $\cup$ B) = $\frac{5}{9}$

To Find:

The value of p.

Solution:

We can use the formula for conditional probability, which states:

Substitute the given values into this formula:

$p = \frac{P(A \cap B)}{\frac{1}{3}}$

Multiplying both sides by $\frac{1}{3}$, we get an expression for $P(A \cap B)$:

Next, we use the Addition Rule of Probability, which relates the probability of the union of two events to their individual probabilities and the probability of their intersection:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substitute the given values and the expression for $P(A \cap B)$ into this formula:

$\frac{5}{9} = p + \frac{1}{3} - \frac{p}{3}$

Now, we solve this equation for p. First, combine the terms involving p:

$p - \frac{p}{3} = \frac{3p}{3} - \frac{p}{3} = \frac{3p - p}{3} = \frac{2p}{3}$

The equation becomes:

$\frac{5}{9} = \frac{2p}{3} + \frac{1}{3}$

Subtract $\frac{1}{3}$ from both sides of the equation:

$\frac{5}{9} - \frac{1}{3} = \frac{2p}{3}$

Find a common denominator for the subtraction on the left side. The common denominator for 9 and 3 is 9. $\frac{1}{3} = \frac{1 \times 3}{3 \times 3} = \frac{3}{9}$.

$\frac{5}{9} - \frac{3}{9} = \frac{2p}{3}$

$\frac{5 - 3}{9} = \frac{2p}{3}$

$\frac{2}{9} = \frac{2p}{3}$

To isolate p, multiply both sides of the equation by $\frac{3}{2}$:

$\frac{2}{9} \times \frac{3}{2} = \frac{2p}{3} \times \frac{3}{2}$

On the left side, cancel out common factors:

$\frac{\cancel{2}}{\cancel{9}_3} \times \frac{\cancel{3}}{\cancel{2}} = p$

$\frac{1}{3} = p$

So, the value of p is $\frac{1}{3}$.

Answer: $\frac{1}{3}$

Given:

P(A' $\cup$ B') = $\frac{2}{3}$

P(A $\cup$ B) = $\frac{5}{9}$

To Find:

The value of P(A') + P(B').

Solution:

We are given the probability of the union of the complements of events A and B, P(A' $\cup$ B').

Using De Morgan's Law, the union of the complements is equal to the complement of the intersection:

$A' \cup B' = (A \cap B)'$

So, $P(A' \cup B') = P((A \cap B)')$.

Using the complement rule, $P(E') = 1 - P(E)$, where E is any event (in this case, $E = A \cap B$).

$P((A \cap B)') = 1 - P(A \cap B)$

We are given $P(A' \cup B') = \frac{2}{3}$. Equating the two expressions:

Solving for $P(A \cap B)$:

Now, we use the Addition Rule of Probability, which states:

We are given $P(A \cup B) = \frac{5}{9}$ and we found $P(A \cap B) = \frac{1}{3}$. Substitute these values into the Addition Rule:

To find the sum $P(A) + P(B)$, add $\frac{1}{3}$ to both sides of the equation:

To add the fractions, find a common denominator. The least common multiple of 9 and 3 is 9. Convert $\frac{1}{3}$ to an equivalent fraction with denominator 9: $\frac{1}{3} = \frac{1 \times 3}{3 \times 3} = \frac{3}{9}$.

We need to find the value of $P(A') + P(B')$.

Using the complement rule for P(A') and P(B'):

$P(A') = 1 - P(A)$

$P(B') = 1 - P(B)$

Adding these two expressions:

$P(A') + P(B') = (1 - P(A)) + (1 - P(B))$

$P(A') + P(B') = 2 - P(A) - P(B)$

$P(A') + P(B') = 2 - (P(A) + P(B))$

Substitute the value of $P(A) + P(B)$ we found ($ \frac{8}{9}$) into this expression:

To perform the subtraction, write 2 as a fraction with denominator 9: $2 = \frac{2 \times 9}{9} = \frac{18}{9}$.

Final Answer: The value of P(A') + P(B') is $\frac{10}{9}$.

Given:

A random variable X follows a binomial distribution with parameters n=5 and p.

The relationship between probabilities is P(X = 2) = 9 $\times$ P(X = 3).

To Find:

The value of the parameter p.

Solution: