Chapter 13 Probability

Given:

$P(A) = \frac{7}{13}$

$P(B) = \frac{9}{13}$

$P(A \cap B) = \frac{4}{13}$

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We need to evaluate $P(A|B)$.

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The formula for the conditional probability of event A given event B is:

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

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Substitute the given values into the formula:

$P(A|B) = \frac{\frac{4}{13}}{\frac{9}{13}}$

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Simplify the expression:

$P(A|B) = \frac{4}{13} \times \frac{13}{9}$

$P(A|B) = \frac{4}{\cancel{13}} \times \frac{\cancel{13}}{9}$

$P(A|B) = \frac{4}{9}$

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Thus, the value of $P(A|B)$ is $\frac{4}{9}$.

Let S be the sample space of the gender of two children. Assuming each child can be either a Boy (B) or a Girl (G) with equal probability, the possible outcomes for two children are:

$S = \{BB, BG, GB, GG\}$

Each outcome in S is equally likely, so the probability of each outcome is $\frac{1}{4}$.

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Let A be the event that both children are boys.

$A = \{BB\}$

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

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Let B be the event that at least one child is a boy.

B includes all outcomes except GG.

$B = \{BB, BG, GB\}$

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

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We need to find the probability that both children are boys given that at least one of them is a boy. This is the conditional probability $P(A|B)$.

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The intersection of events A and B, denoted by $A \cap B$, is the event where both children are boys AND at least one child is a boy. This is simply the event that both children are boys.

$A \cap B = \{BB\}$

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

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The formula for conditional probability is:

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

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Substitute the probabilities we found:

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

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Simplify the fraction:

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

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

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

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Thus, the probability that both children are boys given that at least one of them is a boy is $\frac{1}{3}$.

Let S be the sample space of drawing one card from the ten cards numbered 1 to 10.

$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$

The total number of outcomes in S is $n(S) = 10$.

Each outcome is equally likely.

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Let A be the event that the drawn card is an even number.

$A = \{2, 4, 6, 8, 10\}$

$n(A) = 5$

$P(A) = \frac{n(A)}{n(S)} = \frac{5}{10} = \frac{1}{2}$

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Let B be the event that the number on the drawn card is more than 3.

$B = \{4, 5, 6, 7, 8, 9, 10\}$

$n(B) = 7$

$P(B) = \frac{n(B)}{n(S)} = \frac{7}{10}$

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We are asked to find the probability that the card is an even number given that the number is more than 3. This is the conditional probability $P(A|B)$.

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The intersection of events A and B, denoted by $A \cap B$, is the event that the drawn card is an even number AND the number is more than 3.

$A \cap B = \{x \in S \mid x \text{ is even and } x > 3\}$

$A \cap B = \{4, 6, 8, 10\}$

$n(A \cap B) = 4$

$P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{4}{10} = \frac{2}{5}$

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The formula for conditional probability is:

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

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Substitute the probabilities we found:

$P(A|B) = \frac{\frac{2}{5}}{\frac{7}{10}}$

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Simplify the fraction:

$P(A|B) = \frac{2}{5} \times \frac{10}{7}$

$P(A|B) = \frac{2}{\cancel{5}_1} \times \frac{\cancel{10}^2}{7}$

$P(A|B) = \frac{2 \times 2}{1 \times 7} = \frac{4}{7}$

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Alternatively, we can consider the reduced sample space when it is known that the number is more than 3. This reduced sample space is B.

The elements in B are $\{4, 5, 6, 7, 8, 9, 10\}$. $n(B) = 7$.

We are interested in the outcomes in B that are also even numbers. These are the elements in $A \cap B = \{4, 6, 8, 10\}$. The number of such outcomes is $n(A \cap B) = 4$.

The conditional probability $P(A|B)$ is the number of outcomes in $A \cap B$ divided by the number of outcomes in B (the reduced sample space).

$P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{4}{7}$

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Thus, the probability that the drawn card is an even number given that the number is more than 3 is $\frac{4}{7}$.

Let S be the sample space of choosing a student randomly from the school.

Total number of students = 1000.

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Let G be the event that the chosen student is a girl.

Number of girls = 430.

$P(G) = \frac{\text{Number of girls}}{\text{Total number of students}} = \frac{430}{1000} = 0.43$

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Let C be the event that the chosen student studies in Class XII.

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We are given that 10% of the girls study in Class XII.

This means that the number of girls in Class XII is 10% of 430.

Number of girls in Class XII $= 10\% \text{ of } 430 = \frac{10}{100} \times 430 = 0.1 \times 430 = 43$

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The event that the chosen student is a girl AND studies in Class XII is the intersection of events G and C, denoted by $G \cap C$. This is the event that the chosen student is a girl studying in Class XII.

Number of students in $G \cap C$ = Number of girls in Class XII = 43.

$P(G \cap C) = \frac{\text{Number of girls in Class XII}}{\text{Total number of students}} = \frac{43}{1000} = 0.043$

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We are asked to find the probability that a student studies in Class XII given that the chosen student is a girl. This is the conditional probability $P(C|G)$.

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The formula for conditional probability is:

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

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Substitute the probabilities we found:

$P(C|G) = \frac{0.043}{0.43} = \frac{43/1000}{430/1000} = \frac{43}{430}$

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Simplify the fraction:

$P(C|G) = \frac{\cancel{43}_1}{\cancel{430}_{10}} = \frac{1}{10} = 0.1$

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Alternatively, we are given that the chosen student is a girl. The reduced sample space is the set of girls in the school. The number of girls is 430.

We are interested in the event that the chosen student studies in Class XII, given that she is a girl. This means we are looking for the number of girls who study in Class XII within the reduced sample space of girls.

Number of girls in Class XII = 43.

The conditional probability is the number of girls in Class XII divided by the total number of girls.

$P(\text{Student studies in Class XII} | \text{Student is a girl}) = \frac{\text{Number of girls in Class XII}}{\text{Total number of girls}} = \frac{43}{430} = \frac{1}{10} = 0.1$

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Thus, the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl is $\frac{1}{10}$ or 0.1.

A die is thrown three times. The sample space S consists of all possible sequences of three outcomes, where each outcome can be any integer from 1 to 6.

The total number of outcomes in the sample space is $n(S) = 6 \times 6 \times 6 = 216$.

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Let A be the event that 4 is on the third throw.

The outcomes in event A are of the form $(x, y, 4)$, where x and y can be any number from 1 to 6.

$A = \{(x, y, 4) \mid x \in \{1, 2, 3, 4, 5, 6\}, y \in \{1, 2, 3, 4, 5, 6\}\}$

$n(A) = 6 \times 6 \times 1 = 36$

$P(A) = \frac{n(A)}{n(S)} = \frac{36}{216} = \frac{1}{6}$

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Let B be the event that 6 is on the first throw and 5 is on the second throw.

The outcomes in event B are of the form $(6, 5, z)$, where z can be any number from 1 to 6.

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

$n(B) = 1 \times 1 \times 6 = 6$

$P(B) = \frac{n(B)}{n(S)} = \frac{6}{216} = \frac{1}{36}$

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We need to find the probability of A given that B has already occurred, which is $P(A|B)$.

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The intersection of events A and B, denoted by $A \cap B$, is the event that 4 is on the third throw AND 6 is on the first and 5 on the second throw.

The outcomes in $A \cap B$ must be of the form $(6, 5, 4)$.

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

$n(A \cap B) = 1$

$P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{216}$

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The formula for conditional probability is:

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

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Substitute the probabilities we found:

$P(A|B) = \frac{\frac{1}{216}}{\frac{1}{36}}$

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Simplify the fraction:

$P(A|B) = \frac{1}{216} \times \frac{36}{1}$

$P(A|B) = \frac{\cancel{36}^1}{\cancel{216}_6} = \frac{1}{6}$

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Alternatively, given that event B (6 on the first throw and 5 on the second throw) has already occurred, the reduced sample space is the set of outcomes where the first throw is 6 and the second throw is 5. This reduced sample space is B.

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

$n(B) = 6$

We are interested in the event A (4 on the third throw) occurring within this reduced sample space B.

The outcome in B where the third throw is 4 is $(6, 5, 4)$.

The number of outcomes in B that are also in A is 1.

The conditional probability $P(A|B)$ is the number of outcomes in $A \cap B$ divided by the number of outcomes in B (the reduced sample space).

$P(A|B) = \frac{\text{Number of outcomes in } A \cap B}{\text{Number of outcomes in } B} = \frac{1}{6}$

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Thus, the probability of A given that B has already occurred is $\frac{1}{6}$.

Let S be the sample space when a die is thrown twice.

$S = \{(i, j) \mid i, j \in \{1, 2, 3, 4, 5, 6\}\}$

The total number of outcomes is $n(S) = 6 \times 6 = 36$.

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Let B be the event that the sum of the numbers appearing is observed to be 6.

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

$n(B) = 5$

$P(B) = \frac{n(B)}{n(S)} = \frac{5}{36}$

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Let A be the event that the number 4 has appeared at least once.

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

$n(A) = 11$

$P(A) = \frac{n(A)}{n(S)} = \frac{11}{36}$

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We need to find the conditional probability that the number 4 has appeared at least once, given that the sum of the numbers is 6. This is $P(A|B)$.

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The intersection of events A and B, denoted by $A \cap B$, is the event that the sum is 6 AND the number 4 has appeared at least once.

We look for outcomes in B that contain the number 4:

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

The outcomes in B that have 4 appearing at least once are (2, 4) and (4, 2).

$A \cap B = \{(2, 4), (4, 2)\}$

$n(A \cap B) = 2$

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

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The formula for conditional probability is:

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

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Substitute the probabilities we found:

$P(A|B) = \frac{\frac{1}{18}}{\frac{5}{36}}$

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Simplify the fraction:

$P(A|B) = \frac{1}{18} \times \frac{36}{5}$

$P(A|B) = \frac{1}{\cancel{18}_1} \times \frac{\cancel{36}^2}{5}$

$P(A|B) = \frac{1 \times 2}{1 \times 5} = \frac{2}{5}$

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Alternatively, given that event B (the sum of the numbers is 6) has occurred, the reduced sample space is B.

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

$n(B) = 5$

We are interested in the event A (the number 4 has appeared at least once) occurring within this reduced sample space B.

The outcomes in B that have 4 appearing at least once are $(2, 4)$ and $(4, 2)$. The number of such outcomes is 2.

The conditional probability $P(A|B)$ is the number of outcomes in $A \cap B$ divided by the number of outcomes in B (the reduced sample space).

$P(A|B) = \frac{\text{Number of outcomes in } A \cap B}{\text{Number of outcomes in } B} = \frac{2}{5}$

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Thus, the conditional probability that the number 4 has appeared at least once, given that the sum of the numbers appearing is observed to be 6, is $\frac{2}{5}$.

Let's determine the sample space S for this experiment.

If the first toss is Head (H), the coin is tossed again, resulting in outcomes HH or HT.

If the first toss is Tail (T), a die is thrown, resulting in outcomes T1, T2, T3, T4, T5, T6.

The sample space is $S = \{HH, HT, T1, T2, T3, T4, T5, T6\}$.

The outcomes are not equally likely in the simple sample space S defined this way. However, we can consider the outcomes of the first toss and second stage as compound events with probabilities. Assuming a fair coin and a fair die:

• P(H on first toss) = 1/2
• P(T on first toss) = 1/2
• If H occurs, P(H on second toss) = 1/2, P(T on second toss) = 1/2
• If T occurs, P(any number on die) = 1/6

The probabilities of the elementary outcomes in S are:

$P(HH) = P(\text{H on 1st}) \times P(\text{H on 2nd}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$P(HT) = P(\text{H on 1st}) \times P(\text{T on 2nd}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$P(T1) = P(\text{T on 1st}) \times P(\text{1 on die}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

$P(T2) = \frac{1}{12}$, $P(T3) = \frac{1}{12}$, $P(T4) = \frac{1}{12}$, $P(T5) = \frac{1}{12}$, $P(T6) = \frac{1}{12}$

The sum of probabilities is $\frac{1}{4} + \frac{1}{4} + 6 \times \frac{1}{12} = \frac{1}{2} + \frac{6}{12} = \frac{1}{2} + \frac{1}{2} = 1$. The probabilities are correctly assigned.

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Let A be the event that 'the die shows a number greater than 4'.

This event can only occur if the first toss was a Tail, followed by the die showing 5 or 6.

$A = \{T5, T6\}$

$P(A) = P(T5) + P(T6) = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$

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Let B be the event that 'there is at least one tail'.

Event B includes all outcomes except HH.

$B = \{HT, T1, T2, T3, T4, T5, T6\}$

$P(B) = P(HT) + P(T1) + P(T2) + P(T3) + P(T4) + P(T5) + P(T6)$

$P(B) = \frac{1}{4} + 6 \times \frac{1}{12} = \frac{1}{4} + \frac{6}{12} = \frac{1}{4} + \frac{1}{2} = \frac{1+2}{4} = \frac{3}{4}$

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We need to find the conditional probability of A given that B has already occurred, which is $P(A|B)$.

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The intersection of events A and B, denoted by $A \cap B$, is the event that the die shows a number greater than 4 AND there is at least one tail.

We look for outcomes that are in both sets A and B:

$A = \{T5, T6\}$

$B = \{HT, T1, T2, T3, T4, T5, T6\}$

$A \cap B = \{T5, T6\}$

$P(A \cap B) = P(T5) + P(T6) = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$

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The formula for conditional probability is:

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

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Substitute the probabilities we found:

$P(A|B) = \frac{\frac{1}{6}}{\frac{3}{4}}$

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Simplify the fraction:

$P(A|B) = \frac{1}{6} \times \frac{4}{3}$

$P(A|B) = \frac{1}{\cancel{6}_3} \times \frac{\cancel{4}^2}{3}$

$P(A|B) = \frac{1 \times 2}{3 \times 3} = \frac{2}{9}$

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Thus, the conditional probability of the event that 'the die shows a number greater than 4' given that 'there is at least one tail' is $\frac{2}{9}$.

Given:

$P(E) = 0.6$

$P(F) = 0.3$

$P(E \cap F) = 0.2$

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We need to find $P(E|F)$ and $P(F|E)$.

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The formula for the conditional probability of event E given event F is:

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

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Substitute the given values to find $P(E|F)$:

$P(E|F) = \frac{0.2}{0.3}$

$P(E|F) = \frac{2}{3}$

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The formula for the conditional probability of event F given event E is:

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

Note that $P(F \cap E) = P(E \cap F) = 0.2$.

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Substitute the given values to find $P(F|E)$:

$P(F|E) = \frac{0.2}{0.6}$

$P(F|E) = \frac{2}{6} = \frac{1}{3}$

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Thus, $P(E|F) = \frac{2}{3}$ and $P(F|E) = \frac{1}{3}$.

Given:

$P(B) = 0.5$

$P(A \cap B) = 0.32$

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We need to compute $P(A|B)$.

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The formula for the conditional probability of event A given event B is:

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

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Substitute the given values into the formula:

$P(A|B) = \frac{0.32}{0.5}$

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Simplify the expression:

$P(A|B) = \frac{0.32 \times 100}{0.5 \times 100} = \frac{32}{50}$

$P(A|B) = \frac{\cancel{32}^{16}}{\cancel{50}_{25}} = \frac{16}{25}$

Alternatively, in decimal form:

$P(A|B) = \frac{0.32}{0.5} = 0.64$

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Thus, $P(A|B) = \frac{16}{25}$ or 0.64.

Given:

$P(A) = 0.8$

$P(B) = 0.5$

$P(B|A) = 0.4$

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(i) Find P(A ∩ B)

We know the formula for conditional probability:

Rearranging the formula, we get:

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

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Substitute the given values:

$P(A \cap B) = 0.4 \times 0.8$

$P(A \cap B) = 0.32$

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(ii) Find P(A|B)

We use the formula for conditional probability:

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Substitute the value of $P(A \cap B)$ calculated in part (i) and the given value of $P(B)$:

$P(A|B) = \frac{0.32}{0.5}$

$P(A|B) = \frac{0.32 \times 100}{0.5 \times 100} = \frac{32}{50}$

$P(A|B) = \frac{16}{25}$

Alternatively, in decimal form:

$P(A|B) = 0.64$

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(iii) Find P(A ∪ B)

We use the addition rule for probability:

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Substitute the given values of $P(A)$, $P(B)$ and the calculated value of $P(A \cap B)$:

$P(A \cup B) = 0.8 + 0.5 - 0.32$

$P(A \cup B) = 1.3 - 0.32$

$P(A \cup B) = 0.98$

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Summary of results:

(i) $P(A \cap B) = 0.32$

(ii) $P(A|B) = 0.64$ (or $\frac{16}{25}$)

(iii) $P(A \cup B) = 0.98$

Given:

$2P(A) = \frac{5}{13} \implies P(A) = \frac{5}{2 \times 13} = \frac{5}{26}$

$P(B) = \frac{5}{13}$

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

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We need to evaluate $P(A \cup B)$.

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To find $P(A \cup B)$, we can use the addition rule for probability:

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

We know $P(A)$ and $P(B)$, but we need to find $P(A \cap B)$.

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We can find $P(A \cap B)$ using the formula for conditional probability $P(A|B)$:

Rearranging the formula, we get:

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

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Substitute the given values of $P(A|B)$ and $P(B)$:

$P(A \cap B) = \frac{2}{5} \times \frac{5}{13}$

$P(A \cap B) = \frac{2}{\cancel{5}_1} \times \frac{\cancel{5}_1}{13} = \frac{2}{13}$

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Now substitute the values of $P(A)$, $P(B)$, and $P(A \cap B)$ into the addition rule formula:

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

$P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$

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To simplify, find a common denominator, which is 26:

$P(A \cup B) = \frac{5}{26} + \frac{5 \times 2}{13 \times 2} - \frac{2 \times 2}{13 \times 2}$

$P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26}$

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Perform the addition and subtraction:

$P(A \cup B) = \frac{5 + 10 - 4}{26} = \frac{15 - 4}{26} = \frac{11}{26}$

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Thus, $P(A \cup B) = \frac{11}{26}$.

Given:

$P(A) = \frac{6}{11}$

$P(B) = \frac{5}{11}$

$P(A \cup B) = \frac{7}{11}$

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

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(i) Find P(A ∩ B)

We use the addition rule for probability:

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

Rearranging the formula to solve for $P(A \cap B)$:

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

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Substitute the given values:

$P(A \cap B) = \frac{6}{11} + \frac{5}{11} - \frac{7}{11}$

$P(A \cap B) = \frac{6 + 5 - 7}{11} = \frac{11 - 7}{11} = \frac{4}{11}$

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(ii) Find P(A|B)

We use the formula for conditional probability:

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Substitute the calculated value of $P(A \cap B)$ and the given value of $P(B)$:

$P(A|B) = \frac{\frac{4}{11}}{\frac{5}{11}}$

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Simplify the fraction:

$P(A|B) = \frac{4}{11} \times \frac{11}{5} = \frac{4}{\cancel{11}_1} \times \frac{\cancel{11}_1}{5} = \frac{4}{5}$

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(iii) Find P(B|A)

We use the formula for conditional probability:

Note that $P(B \cap A) = P(A \cap B) = \frac{4}{11}$.

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Substitute the calculated value of $P(A \cap B)$ and the given value of $P(A)$:

$P(B|A) = \frac{\frac{4}{11}}{\frac{6}{11}}$

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Simplify the fraction:

$P(B|A) = \frac{4}{11} \times \frac{11}{6} = \frac{4}{\cancel{11}_1} \times \frac{\cancel{11}_1}{6} = \frac{4}{6} = \frac{2}{3}$

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Summary of results:

(i) $P(A \cap B) = \frac{4}{11}$

(ii) $P(A|B) = \frac{4}{5}$

(iii) $P(B|A) = \frac{2}{3}$

When a coin is tossed three times, the sample space S is:

$S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$

The total number of outcomes is $n(S) = 2^3 = 8$. Each outcome is equally likely, with a probability of $\frac{1}{8}$.

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(i) E : head on third toss , F : heads on first two tosses

Event E: Head on the third toss.

$E = \{HHH, HTH, TTH, SlashesHTT \text{ - Mistake in thought block, it should be TTH for head on 3rd toss}\}$

$E = \{HHH, HTH, THH, TTH\}$ (Corrected from thought block, any outcome with H in the 3rd position)

$E = \{HHH, HTH, THH, TTH\}$. No, that's wrong again. The third toss is the last one.

$E = \{HHH, HTH, TTH, HTT\}$ (No, HTT has T on 3rd).

$E = \{HHH, HTH, THH, TTH\}$. Let me re-list: HHH (3rd is H), HHT (3rd is T), HTH (3rd is H), THH (3rd is H), HTT (3rd is T), THT (3rd is T), TTH (3rd is H), TTT (3rd is T).

Event E: head on third toss.

$E = \{HHH, HTH, THH, TTH\}$ No, that's still wrong. The third toss is the last one.

Let's list the outcomes again and check the third toss:

• HHH: 3rd is H
• HHT: 3rd is T
• HTH: 3rd is H
• THH: 3rd is H
• HTT: 3rd is T
• THT: 3rd is T
• TTH: 3rd is H
• TTT: 3rd is T

Event E: head on third toss.

$E = \{HHH, HTH, THH, TTH\}$. This is still wrong. TTH has H on the third toss. THH has H on the third toss. HTH has H on the third toss. HHH has H on the third toss.

$E = \{HHH, HTH, THH, TTH\}$. Okay, the first three elements are correct based on the thought block. TTH has H on the third toss. Let me check again the list of outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. The third toss is H in HHH, HTH, THH, TTH.

Let me re-read the question: E : head on third toss. Yes, this means the LAST toss is a head. My list $E = \{HHH, HTH, THH, TTH\}$ is correct. $n(E) = 4$.

Event F: heads on first two tosses.

$F = \{HHH, HHT\}$

$n(F) = 2$.

---

Find $E \cap F$:

$E = \{HHH, HTH, THH, TTH\}$

$F = \{HHH, HHT\}$

$E \cap F = \{HHH\}$

$n(E \cap F) = 1$.

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We need to find $P(E|F)$. The formula is $P(E|F) = \frac{P(E \cap F)}{P(F)}$.

$P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{1}{8}$

$P(F) = \frac{n(F)}{n(S)} = \frac{2}{8} = \frac{1}{4}$

---

$P(E|F) = \frac{1/8}{1/4} = \frac{1}{8} \times \frac{4}{1} = \frac{4}{8} = \frac{1}{2}$.

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(ii) E : at least two heads , F : at most two heads

Event E: at least two heads (2 or 3 heads).

$E = \{HHH, HHT, HTH, THH\}$

$n(E) = 4$.

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Event F: at most two heads (0, 1, or 2 heads).

F is the complement of having exactly 3 heads (HHH).

$F = S \setminus \{HHH\} = \{HHT, HTH, THH, HTT, THT, TTH, TTT\}$

$n(F) = 7$.

---

Find $E \cap F$:

$E = \{HHH, HHT, HTH, THH\}$

$F = \{HHT, HTH, THH, HTT, THT, TTH, TTT\}$

$E \cap F = \{HHT, HTH, THH\}$

$n(E \cap F) = 3$.

---

We need to find $P(E|F) = \frac{P(E \cap F)}{P(F)}$.

$P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{3}{8}$

$P(F) = \frac{n(F)}{n(S)} = \frac{7}{8}$

---

$P(E|F) = \frac{3/8}{7/8} = \frac{3}{8} \times \frac{8}{7} = \frac{3}{7}$.

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(iii) E : at most two tails , F : at least one tail

Event E: at most two tails (0, 1, or 2 tails).

E is the complement of having exactly 3 tails (TTT).

$E = S \setminus \{TTT\} = \{HHH, HHT, HTH, THH, HTT, THT, TTH\}$

$n(E) = 7$.

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Event F: at least one tail (1, 2, or 3 tails).

F is the complement of having exactly 0 tails (HHH).

$F = S \setminus \{HHH\} = \{HHT, HTH, THH, HTT, THT, TTH, TTT\}$

$n(F) = 7$.

---

Find $E \cap F$:

$E = \{HHH, HHT, HTH, THH, HTT, THT, TTH\}$

$F = \{HHT, HTH, THH, HTT, THT, TTH, TTT\}$

$E \cap F = \{HHT, HTH, THH, HTT, THT, TTH\}$

$n(E \cap F) = 6$.

---

We need to find $P(E|F) = \frac{P(E \cap F)}{P(F)}$.

$P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{6}{8} = \frac{3}{4}$

$P(F) = \frac{n(F)}{n(S)} = \frac{7}{8}$

---

$P(E|F) = \frac{3/4}{7/8} = \frac{3}{4} \times \frac{8}{7} = \frac{3}{\cancel{4}_1} \times \frac{\cancel{8}^2}{7} = \frac{3 \times 2}{1 \times 7} = \frac{6}{7}$.

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Summary of results:

(i) $P(E|F) = \frac{1}{2}$

(ii) $P(E|F) = \frac{3}{7}$

(iii) $P(E|F) = \frac{6}{7}$

When two coins are tossed once, the sample space S is:

$S = \{HH, HT, TH, TT\}$

The total number of outcomes is $n(S) = 4$. Each outcome is equally likely, with a probability of $\frac{1}{4}$.

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(i) E : tail appears on one coin, F : one coin shows head

Event E: Tail appears on one coin (exactly one tail).

$E = \{HT, TH\}$

$n(E) = 2$.

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Event F: One coin shows head (exactly one head).

$F = \{HT, TH\}$

$n(F) = 2$.

---

Find $E \cap F$:

$E \cap F = \{HT, TH\}$

$n(E \cap F) = 2$.

---

We need to find $P(E|F)$. The formula is $P(E|F) = \frac{P(E \cap F)}{P(F)}$.

$P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{2}{4} = \frac{1}{2}$

$P(F) = \frac{n(F)}{n(S)} = \frac{2}{4} = \frac{1}{2}$

---

$P(E|F) = \frac{1/2}{1/2} = 1$.

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(ii) E : no tail appears, F : no head appears

Event E: No tail appears (both are heads).

$E = \{HH\}$

$n(E) = 1$.

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Event F: No head appears (both are tails).

$F = \{TT\}$

$n(F) = 1$.

---

Find $E \cap F$:

$E \cap F = \{HH\} \cap \{TT\} = \emptyset$ (empty set)

$n(E \cap F) = 0$.

---

We need to find $P(E|F) = \frac{P(E \cap F)}{P(F)}$.

$P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{0}{4} = 0$

$P(F) = \frac{n(F)}{n(S)} = \frac{1}{4}$

---

$P(E|F) = \frac{0}{1/4} = 0$.

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Summary of results:

(i) $P(E|F) = 1$

(ii) $P(E|F) = 0$

When a die is thrown three times, the sample space S consists of all possible sequences of three outcomes, where each outcome can be any integer from 1 to 6.

$S = \{(i, j, k) \mid i, j, k \in \{1, 2, 3, 4, 5, 6\}\}$

The total number of outcomes in the sample space is $n(S) = 6 \times 6 \times 6 = 216$.

Each outcome is equally likely.

---

Let E be the event that 4 appears on the third toss.

The outcomes in event E are of the form $(i, j, 4)$, where i and j can be any number from 1 to 6.

$E = \{(i, j, 4) \mid i \in \{1, ..., 6\}, j \in \{1, ..., 6\}\}$

$n(E) = 6 \times 6 \times 1 = 36$

$P(E) = \frac{n(E)}{n(S)} = \frac{36}{216} = \frac{1}{6}$

---

Let F be the event that 6 and 5 appear respectively on first two tosses.

The outcomes in event F are of the form $(6, 5, k)$, where k can be any number from 1 to 6.

$F = \{(6, 5, k) \mid k \in \{1, ..., 6\}\}$

$n(F) = 1 \times 1 \times 6 = 6$

$P(F) = \frac{n(F)}{n(S)} = \frac{6}{216} = \frac{1}{36}$

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We need to find the probability of E given that F has already occurred, which is $P(E|F)$.

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The intersection of events E and F, denoted by $E \cap F$, is the event that 4 appears on the third toss AND 6 and 5 appear respectively on first two tosses.

The outcomes in $E \cap F$ must be of the form $(6, 5, 4)$.

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

$n(E \cap F) = 1$

$P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{1}{216}$

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The formula for conditional probability is:

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

---

Substitute the probabilities we found:

$P(E|F) = \frac{\frac{1}{216}}{\frac{1}{36}}$

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Simplify the fraction:

$P(E|F) = \frac{1}{216} \times \frac{36}{1}$

$P(E|F) = \frac{1}{\cancel{216}_6} \times \frac{\cancel{36}^1}{1}$

$P(E|F) = \frac{1}{6}$

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Alternatively, given that event F (6 on the first toss and 5 on the second toss) has already occurred, the reduced sample space is the set of outcomes where the first throw is 6 and the second throw is 5. This reduced sample space is F.

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

$n(F) = 6$

We are interested in the event E (4 appears on the third toss) occurring within this reduced sample space F.

The outcome in F where the third toss is 4 is $(6, 5, 4)$. The number of such outcomes is 1.

The conditional probability $P(E|F)$ is the number of outcomes in $E \cap F$ divided by the number of outcomes in F (the reduced sample space).

$P(E|F) = \frac{\text{Number of outcomes in } E \cap F}{\text{Number of outcomes in } F} = \frac{1}{6}$

---

Thus, the probability of E given that F has already occurred is $\frac{1}{6}$.

Let M represent Mother, F represent Father, and S represent Son. The family lines up at random, so the sample space S consists of all possible permutations of M, F, and S in a line.

The possible arrangements are:

$S = \{MFS, MSF, FMS, FSM, SMF, SFM\}$

The total number of outcomes is $n(S) = 3! = 3 \times 2 \times 1 = 6$.

Each outcome is equally likely, with a probability of $\frac{1}{6}$.

---

Let E be the event that the son is on one end.

The son can be on the left end or the right end.

Arrangements with Son on the left: S followed by any permutation of M and F. $\{SMF, SFM\}$.

Arrangements with Son on the right: Any permutation of M and F followed by S. $\{MFS, FMS\}$.

$E = \{SMF, SFM, MFS, FMS\}$

$n(E) = 4$.

$P(E) = \frac{n(E)}{n(S)} = \frac{4}{6} = \frac{2}{3}$

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Let F be the event that the father is in the middle.

The arrangements with Father in the middle are MFM or SFS. Oh, the persons are M, F, S. So, the middle position is occupied by F, and M and S can be in the end positions.

Arrangements with Father in the middle: M in the first position, F in the second, S in the third (MFS), or S in the first position, F in the second, M in the third (SFM).

$F = \{MFS, SFM\}$

$n(F) = 2$.

$P(F) = \frac{n(F)}{n(S)} = \frac{2}{6} = \frac{1}{3}$

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We need to find the probability of E given that F has already occurred, which is $P(E|F)$.

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The intersection of events E and F, denoted by $E \cap F$, is the event that the son is on one end AND the father is in the middle.

We look for outcomes that are in both sets E and F:

$E = \{SMF, SFM, MFS, FMS\}$

$F = \{MFS, SFM\}$

$E \cap F = \{MFS, SFM\}$

$n(E \cap F) = 2$.

$P(E \cap F) = \frac{n(E \cap F)}{n(S)} = \frac{2}{6} = \frac{1}{3}$

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The formula for conditional probability is:

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

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Substitute the probabilities we found:

$P(E|F) = \frac{1/3}{1/3}$

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Simplify the fraction:

$P(E|F) = 1$

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Alternatively, given that event F (father is in the middle) has occurred, the reduced sample space is F.

$F = \{MFS, SFM\}$

$n(F) = 2$

We are interested in the event E (son on one end) occurring within this reduced sample space F.

In the outcome MFS, the son (S) is on the right end. This is in E.

In the outcome SFM, the son (S) is on the left end. This is in E.

Both outcomes in F are also in E. So, $F \subseteq E$, which means $E \cap F = F$.

The number of outcomes in F that are also in E is 2.

The conditional probability $P(E|F)$ is the number of outcomes in $E \cap F$ divided by the number of outcomes in F (the reduced sample space).

$P(E|F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{2} = 1$

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Thus, the probability of E given that F has already occurred is 1.

When a black die and a red die are rolled, the sample space S consists of all ordered pairs $(b, r)$, where $b$ is the outcome on the black die and $r$ is the outcome on the red die. Each of $b$ and $r$ can be any integer from 1 to 6.

$S = \{(b, r) \mid b, r \in \{1, 2, 3, 4, 5, 6\}\}$

The total number of outcomes is $n(S) = 6 \times 6 = 36$. Each outcome is equally likely.

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(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

Let A be the event of obtaining a sum greater than 9.

$A = \{(b, r) \mid b+r > 9\}$

The outcomes where the sum is greater than 9 are:

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

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

Sum 12: (6, 6)

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

$n(A) = 6$

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Let B be the event that the black die resulted in a 5.

$B = \{(b, r) \mid b = 5\}$

The outcomes where the black die is 5 are:

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

$n(B) = 6$

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

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We need to find the conditional probability $P(A|B)$.

First, find the intersection $A \cap B$. This is the event where the sum is greater than 9 AND the black die is 5.

$A \cap B = \{(b, r) \in A \mid b=5\}$

Looking at the outcomes in A, the ones where the black die is 5 are (5, 5) and (5, 6).

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

$n(A \cap B) = 2$

$P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{36}$

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The formula for conditional probability is:

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

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Substitute the probabilities:

$P(A|B) = \frac{2/36}{6/36} = \frac{2}{6} = \frac{1}{3}$

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Alternatively, given that the black die resulted in a 5, the reduced sample space is B, with $n(B) = 6$. We are interested in the outcomes in B where the sum is greater than 9. These are (5, 5) and (5, 6), which is 2 outcomes. The conditional probability is $\frac{2}{6} = \frac{1}{3}$.

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The conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5, is $\frac{1}{3}$.

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(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

Let C be the event of obtaining the sum 8.

$C = \{(b, r) \mid b+r = 8\}$

The outcomes where the sum is 8 are:

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

$n(C) = 5$

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Let D be the event that the red die resulted in a number less than 4.

$D = \{(b, r) \mid r < 4\}$

The outcomes where the red die is 1, 2, or 3 are:

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

$n(D) = 6 \times 3 = 18$

$P(D) = \frac{n(D)}{n(S)} = \frac{18}{36} = \frac{1}{2}$

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We need to find the conditional probability $P(C|D)$.

First, find the intersection $C \cap D$. This is the event where the sum is 8 AND the red die is less than 4.

$C \cap D = \{(b, r) \in C \mid r \in \{1, 2, 3\}\}$

Looking at the outcomes in C, the ones where the red die is 1, 2, or 3 are (5, 3) and (6, 2).

$C \cap D = \{(5, 3), (6, 2)\}$

$n(C \cap D) = 2$

$P(C \cap D) = \frac{n(C \cap D)}{n(S)} = \frac{2}{36}$

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The formula for conditional probability is:

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

---

Substitute the probabilities:

$P(C|D) = \frac{2/36}{18/36} = \frac{2}{18} = \frac{1}{9}$

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Alternatively, given that the red die resulted in a number less than 4, the reduced sample space is D, with $n(D) = 18$. We are interested in the outcomes in D where the sum is 8. These are (5, 3) and (6, 2), which is 2 outcomes. The conditional probability is $\frac{2}{18} = \frac{1}{9}$.

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The conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4, is $\frac{1}{9}$.

When a fair die is rolled, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$.

The total number of outcomes is $n(S) = 6$. Each outcome is equally likely, with a probability of $\frac{1}{6}$.

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The given events are:

$E = \{1, 3, 5\}$, $n(E) = 3$, $P(E) = \frac{3}{6} = \frac{1}{2}$

$F = \{2, 3\}$, $n(F) = 2$, $P(F) = \frac{2}{6} = \frac{1}{3}$

$G = \{2, 3, 4, 5\}$, $n(G) = 4$, $P(G) = \frac{4}{6} = \frac{2}{3}$

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(i) Find P(E|F) and P(F|E)

First, find $E \cap F$: $E = \{1, 3, 5\}$, $F = \{2, 3\}$.

$E \cap F = \{3\}$

$n(E \cap F) = 1$, $P(E \cap F) = \frac{1}{6}$

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/6}{1/3} = \frac{1}{6} \times \frac{3}{1} = \frac{3}{6} = \frac{1}{2}$.

$P(F|E) = \frac{P(F \cap E)}{P(E)}$. Note that $P(F \cap E) = P(E \cap F) = \frac{1}{6}$.

$P(F|E) = \frac{1/6}{1/2} = \frac{1}{6} \times \frac{2}{1} = \frac{2}{6} = \frac{1}{3}$.

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(ii) Find P(E|G) and P(G|E)

First, find $E \cap G$: $E = \{1, 3, 5\}$, $G = \{2, 3, 4, 5\}$.

$E \cap G = \{3, 5\}$

$n(E \cap G) = 2$, $P(E \cap G) = \frac{2}{6} = \frac{1}{3}$

$P(E|G) = \frac{P(E \cap G)}{P(G)} = \frac{1/3}{2/3} = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2}$.

$P(G|E) = \frac{P(G \cap E)}{P(E)}$. Note that $P(G \cap E) = P(E \cap G) = \frac{1}{3}$.

$P(G|E) = \frac{1/3}{1/2} = \frac{1}{3} \times \frac{2}{1} = \frac{2}{3}$.

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(iii) Find P((E ∪ F)|G) and P((E ∩ F)|G)

First, find $E \cup F$: $E = \{1, 3, 5\}$, $F = \{2, 3\}$.

$E \cup F = \{1, 2, 3, 5\}$

$n(E \cup F) = 4$, $P(E \cup F) = \frac{4}{6} = \frac{2}{3}$

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Now, find $(E \cup F) \cap G$: $E \cup F = \{1, 2, 3, 5\}$, $G = \{2, 3, 4, 5\}$.

$(E \cup F) \cap G = \{2, 3, 5\}$

$n((E \cup F) \cap G) = 3$, $P((E \cup F) \cap G) = \frac{3}{6} = \frac{1}{2}$

$P((E \cup F)|G) = \frac{P((E \cup F) \cap G)}{P(G)} = \frac{1/2}{2/3} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$.

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Next, find $E \cap F$: $E = \{1, 3, 5\}$, $F = \{2, 3\}$.

$E \cap F = \{3\}$

$n(E \cap F) = 1$, $P(E \cap F) = \frac{1}{6}$

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Now, find $(E \cap F) \cap G$: $E \cap F = \{3\}$, $G = \{2, 3, 4, 5\}$.

$(E \cap F) \cap G = \{3\}$

$n((E \cap F) \cap G) = 1$, $P((E \cap F) \cap G) = \frac{1}{6}$

$P((E \cap F)|G) = \frac{P((E \cap F) \cap G)}{P(G)} = \frac{1/6}{2/3} = \frac{1}{6} \times \frac{3}{2} = \frac{3}{12} = \frac{1}{4}$.

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Summary of results:

(i) $P(E|F) = \frac{1}{2}$, $P(F|E) = \frac{1}{3}$

(ii) $P(E|G) = \frac{1}{2}$, $P(G|E) = \frac{2}{3}$

(iii) $P((E \cup F)|G) = \frac{3}{4}$, $P((E \cap F)|G) = \frac{1}{4}$

Let S be the sample space for the gender of two children in a family. Assuming each child is equally likely to be a boy (B) or a girl (G), the sample space is:

$S = \{BB, BG, GB, GG\}$

Here, the first letter represents the elder child and the second letter represents the younger child. The total number of outcomes is $n(S) = 4$. Each outcome is equally likely, so the probability of each outcome is $\frac{1}{4}$.

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Let A be the event that both children are girls.

$A = \{GG\}$

$n(A) = 1$, so $P(A) = \frac{n(A)}{n(S)} = \frac{1}{4}$.

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(i) Given that the youngest is a girl.

Let B be the event that the youngest child is a girl.

$B = \{BG, GG\}$

$n(B) = 2$, so $P(B) = \frac{n(B)}{n(S)} = \frac{2}{4} = \frac{1}{2}$.

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We need to find the conditional probability $P(A|B)$.

First, find the intersection $A \cap B$. This is the event that both children are girls AND the youngest is a girl.

$A \cap B = \{GG\} \cap \{BG, GG\} = \{GG\}$

$n(A \cap B) = 1$, so $P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{1}{4}$.

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Using the formula for conditional probability:

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

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Substitute the probabilities:

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

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The conditional probability that both are girls given that the youngest is a girl is $\frac{1}{2}$.

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(ii) Given that at least one is a girl.

Let C be the event that at least one child is a girl.

$C = \{BG, GB, GG\}$

$n(C) = 3$, so $P(C) = \frac{n(C)}{n(S)} = \frac{3}{4}$.

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We need to find the conditional probability $P(A|C)$.

First, find the intersection $A \cap C$. This is the event that both children are girls AND at least one is a girl.

$A \cap C = \{GG\} \cap \{BG, GB, GG\} = \{GG\}$

$n(A \cap C) = 1$, so $P(A \cap C) = \frac{n(A \cap C)}{n(S)} = \frac{1}{4}$.

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Using the formula for conditional probability:

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

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Substitute the probabilities:

$P(A|C) = \frac{1/4}{3/4} = \frac{1}{4} \times \frac{4}{3} = \frac{1}{3}$.

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The conditional probability that both are girls given that at least one is a girl is $\frac{1}{3}$.

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Summary of results:

(i) $P(\text{both are girls} | \text{youngest is a girl}) = \frac{1}{2}$

(ii) $P(\text{both are girls} | \text{at least one is a girl}) = \frac{1}{3}$

Let's first find the total number of questions in the bank.

Number of easy True/False questions = 300

Number of difficult True/False questions = 200

Number of easy multiple choice questions = 500

Number of difficult multiple choice questions = 400

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Total number of questions = $300 + 200 + 500 + 400 = 1400$.

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Let S be the sample space of selecting a question at random from the question bank.

$n(S) = 1400$.

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Let E be the event that the selected question is an easy question.

Number of easy questions = Number of easy True/False + Number of easy multiple choice

Number of easy questions = $300 + 500 = 800$.

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Let M be the event that the selected question is a multiple choice question.

Number of multiple choice questions = Number of easy multiple choice + Number of difficult multiple choice

Number of multiple choice questions = $500 + 400 = 900$.

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We need to find the probability that the question is easy given that it is a multiple choice question. This is the conditional probability $P(E|M)$.

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The formula for conditional probability is:

$P(E|M) = \frac{P(E \cap M)}{P(M)}$

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We need to find $P(E \cap M)$ and $P(M)$.

$E \cap M$ is the event that the selected question is easy AND it is a multiple choice question. This corresponds to the easy multiple choice questions.

Number of questions in $E \cap M$ = Number of easy multiple choice questions = 500.

$P(E \cap M) = \frac{\text{Number of easy multiple choice questions}}{\text{Total number of questions}} = \frac{500}{1400} = \frac{5}{14}$.

---

$P(M) = \frac{\text{Number of multiple choice questions}}{\text{Total number of questions}} = \frac{900}{1400} = \frac{9}{14}$.

---

Now substitute these probabilities into the formula for $P(E|M)$:

$P(E|M) = \frac{P(E \cap M)}{P(M)} = \frac{\frac{5}{14}}{\frac{9}{14}}$

---

Simplify the fraction:

$P(E|M) = \frac{5}{14} \times \frac{14}{9} = \frac{5}{\cancel{14}_1} \times \frac{\cancel{14}_1}{9} = \frac{5}{9}$.

---

Alternatively, we can consider the reduced sample space. Given that the question is a multiple choice question, the sample space is reduced to only the multiple choice questions.

Number of multiple choice questions = 900.

Among these multiple choice questions, we want to find the number of easy questions. These are the easy multiple choice questions.

Number of easy multiple choice questions = 500.

The conditional probability is the number of favorable outcomes in the reduced sample space divided by the size of the reduced sample space.

$P(E|M) = \frac{\text{Number of easy multiple choice questions}}{\text{Total number of multiple choice questions}} = \frac{500}{900} = \frac{5}{9}$.

---

Thus, the probability that the selected question will be an easy question given that it is a multiple choice question is $\frac{5}{9}$.

Let S be the sample space when two dice are thrown. The total number of outcomes is $6 \times 6 = 36$. Each outcome is an ordered pair $(i, j)$, where $i$ is the number on the first die and $j$ is the number on the second die, with $i, j \in \{1, 2, 3, 4, 5, 6\}$.

$n(S) = 36$.

---

Let A be the event that the sum of the numbers on the dice is 4.

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

$n(A) = 3$.

$P(A) = \frac{n(A)}{n(S)} = \frac{3}{36}$

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Let B be the event that the two numbers appearing on the dice are different.

The outcomes where the numbers are the same are the doubles: $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$. There are 6 such outcomes.

The outcomes where the numbers are different is the complement of the doubles.

$n(B) = n(S) - n(\text{doubles}) = 36 - 6 = 30$.

$P(B) = \frac{n(B)}{n(S)} = \frac{30}{36}$

---

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:

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

---

We need to find the intersection $A \cap B$. This is the event where the sum is 4 AND the two numbers are different.

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

The outcomes in A where the numbers are different are $(1, 3)$ and $(3, 1)$.

$A \cap B = \{(1, 3), (3, 1)\}$

$n(A \cap B) = 2$.

$P(A \cap B) = \frac{n(A \cap B)}{n(S)} = \frac{2}{36}$

---

Now substitute the probabilities $P(A \cap B)$ and $P(B)$ into the formula for $P(A|B)$:

$P(A|B) = \frac{\frac{2}{36}}{\frac{30}{36}}$

---

Simplify the fraction:

$P(A|B) = \frac{2}{36} \times \frac{36}{30} = \frac{2}{30} = \frac{1}{15}$.

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Alternatively, given that the two numbers appearing on the dice are different, the reduced sample space is B, which contains 30 outcomes.

We are interested in the event 'the sum of numbers on the dice is 4' within this reduced sample space B.

The outcomes in B whose sum is 4 are $(1, 3)$ and $(3, 1)$. There are 2 such outcomes.

The conditional probability $P(A|B)$ is the number of outcomes in $A \cap B$ divided by the number of outcomes in the reduced sample space B.

$P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{30} = \frac{1}{15}$.

---

Thus, the probability of the event 'the sum of numbers on the dice is 4' given that the two numbers appearing are different is $\frac{1}{15}$.

Let's define the sample space for this experiment. The experiment proceeds in stages.

Stage 1: Throw a die. The outcome can be a multiple of 3 ({3, 6}) or not a multiple of 3 ({1, 2, 4, 5}).

Stage 2:

• If the first throw is a multiple of 3 (3 or 6), throw the die again. The outcome is a pair of die results $(d_1, d_2)$.
• If the first throw is not a multiple of 3 (1, 2, 4, or 5), toss a coin. The outcome is a pair $(d_1, c)$, where $d_1$ is the first die result and $c$ is the coin result.

Assuming a fair die (probability $\frac{1}{6}$ for each face) and a fair coin (probability $\frac{1}{2}$ for Head or Tail), the probabilities of the outcomes are:

• If $d_1 \in \{3, 6\}$ (probability $\frac{2}{6} = \frac{1}{3}$), then $d_2 \in \{1, 2, 3, 4, 5, 6\}$ (probability $\frac{1}{6}$). The outcome is $(d_1, d_2)$. The probability of each such outcome is $P(d_1) \times P(d_2) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$. There are $2 \times 6 = 12$ such outcomes: $(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)$.
• If $d_1 \in \{1, 2, 4, 5\}$ (probability $\frac{4}{6} = \frac{2}{3}$), then $c \in \{H, T\}$ (probability $\frac{1}{2}$). The outcome is $(d_1, c)$. The probability of each such outcome is $P(d_1) \times P(c) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$. There are $4 \times 2 = 8$ such outcomes: $(1,H), (1,T), (2,H), (2,T), (4,H), (4,T), (5,H), (5,T)$.

The sample space S is the union of these two sets of outcomes.

$S = \{(3,1), ..., (3,6), (6,1), ..., (6,6), (1,H), (1,T), (2,H), (2,T), (4,H), (4,T), (5,H), (5,T)\}$

---

Let E be the event that ‘the coin shows a tail’.

This event occurs when the first die is not a multiple of 3 and the coin toss results in a Tail.

$E = \{(1,T), (2,T), (4,T), (5,T)\}$

$P(E) = P(1,T) + P(2,T) + P(4,T) + P(5,T) = 4 \times \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$.

---

Let F be the event that ‘at least one die shows a 3’.

This means that in the outcomes of the die throws that occurred in the experiment, at least one of them is a 3.

• If the first die is 3 ($d_1=3$), then the condition is met by the first die. All outcomes $(3, d_2)$ are in F: $(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)$.
• If the first die is 6 ($d_1=6$), the second die is thrown. The second die must be 3 for the condition to be met: $(6,3)$.
• If the first die is 1, 2, 4, or 5 ($d_1 \in \{1,2,4,5\}$), a coin is tossed. No die shows a 3 in these outcomes. These outcomes are not in F.

So, event F consists of outcomes where the first die is 3 (and the second die is any outcome), or the first die is 6 and the second die is 3.

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

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

$P(F) = 6 \times \frac{1}{36} + 1 \times \frac{1}{36} = \frac{6+1}{36} = \frac{7}{36}$.

---

We need to find the conditional probability $P(E|F) = \frac{P(E \cap F)}{P(F)}$.

---

The intersection of events E and F, denoted by $E \cap F$, is the event that the coin shows a tail AND at least one die shows a 3.

Let's examine the outcomes in E and F:

$E = \{(1,T), (2,T), (4,T), (5,T)\}$

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

Outcomes in E are pairs of (die result, coin result).

Outcomes in F are pairs of (die result, die result).

These two sets of outcomes are mutually exclusive (disjoint). There are no outcomes that belong to both E and F.

$E \cap F = \emptyset$

$P(E \cap F) = 0$.

---

Now substitute the probabilities $P(E \cap F)$ and $P(F)$ into the formula for $P(E|F)$:

$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{0}{7/36} = 0$.

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Alternatively, given that event F (at least one die shows a 3) has occurred, the reduced sample space is F. Event E (the coin shows a tail) can only occur if the first die was 1, 2, 4, or 5, leading to a coin toss. However, all outcomes in F begin with a 3 or a 6 on the first die, leading to a second die throw. Therefore, if F has occurred, it is impossible for a coin to have been tossed, and thus impossible for the coin to show a tail. The number of outcomes in F that also result in a coin showing a tail is 0. The conditional probability is therefore $\frac{0}{n(F)} = 0$ (or $\frac{P(E \cap F)}{P(F)} = \frac{0}{P(F)} = 0$).

---

Thus, the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’, is 0.

Given:

---

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

$P(B) = 0$

---

To Find:

---

$P(A|B)$

---

Solution:

---

The formula for the conditional probability of event A given event B is defined as:

This formula is valid only when $P(B) > 0$. The definition of conditional probability requires that the conditioning event must have a non-zero probability.

In this question, we are given that $P(B) = 0$.

If $P(B) = 0$, the denominator in the formula for $P(A|B)$ becomes zero.

Division by zero is undefined in mathematics.

Therefore, when $P(B) = 0$, the conditional probability $P(A|B)$ is not defined.

Let's look at the given options:

(A) 0

(B) $\frac{1}{2}$

(C) not defined

(D) 1

Our conclusion matches option (C).

---

The final answer is $\mathbf{not \ defined}$.

Given:

---

A and B are events such that $P(A|B) = P(B|A)$.

---

To Find:

---

Which of the given options (A, B, C, D) is true.

---

Solution:

---

The conditional probability of event A given event B is defined as $P(A|B) = \frac{P(A \cap B)}{P(B)}$, provided $P(B) > 0$.

The conditional probability of event B given event A is defined as $P(B|A) = \frac{P(A \cap B)}{P(A)}$, provided $P(A) > 0$.

The given condition is $P(A|B) = P(B|A)$. For these conditional probabilities to be defined using the standard formula, we must assume $P(A) > 0$ and $P(B) > 0$.

Under this assumption, we can write the given condition as:

Multiply both sides of the equation by $P(A)P(B)$:

Rearrange the terms to one side:

Factor out $P(A \cap B)$:

Equation (1) implies that either $P(A \cap B) = 0$ or $P(A) - P(B) = 0$.

So, the condition $P(A|B) = P(B|A)$ implies that $\mathbf{P(A \cap B) = 0}$ or $\mathbf{P(A) = P(B)}$.

This means that if $P(A|B) = P(B|A)$, then either the events A and B are mutually exclusive (their intersection has probability 0) or the probabilities of the events A and B are equal.

---

Now let's consider the given options:

(A) A ⊂ B but A ≠ B: This implies $P(A) \leq P(B)$. If $P(A) < P(B)$, this contradicts the possibility $P(A)=P(B)$. If $A \subset B$ and $P(A) = P(B)$, it implies $P(B \setminus A) = P(B) - P(A) = 0$, which means $P(B \setminus A)=0$, so A is effectively B in terms of probability measure if $P(B)>0$. This option is not necessarily true.

(B) A = B: If $A = B$, then $P(A) = P(B)$ and $A \cap B = A$. If $P(A) > 0$, then $P(A \cap B) = P(A) > 0$. In this case, the condition $P(A \cap B)=0$ or $P(A)=P(B)$ becomes $P(A) > 0$ or $P(A)=P(A)$, which is true. So $A=B$ satisfies the condition $P(A|B)=P(B|A)$ (if $P(A)>0$). However, the converse is not true; $P(A|B)=P(B|A)$ does not necessarily imply $A=B$ (as shown below). This option is not necessarily true.

(C) A ∩ B = φ: This implies $P(A \cap B) = 0$. This is one of the conditions derived from equation (1). If $P(A \cap B) = 0$ and $P(A)>0, P(B)>0$, then $P(A|B) = 0/P(B) = 0$ and $P(B|A) = 0/P(A) = 0$. So $P(A|B)=P(B|A)$ holds. However, $A \cap B = \phi$ is not necessarily true; $P(A)=P(B)$ is also a possibility when $A \cap B \neq \phi$ (e.g., consider two overlapping events with equal probabilities). This option is not necessarily true.

(D) P(A) = P(B): This is the other condition derived from equation (1). If $P(A) = P(B) > 0$, then $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)} = P(B|A)$. So $P(A)=P(B)$ satisfies the condition $P(A|B)=P(B|A)$. While the derivation yields ($P(A \cap B)=0$) or ($P(A)=P(B)$), option (D) $P(A)=P(B)$ is the most likely intended answer in the context of standard multiple-choice questions of this form, representing the outcome when the intersection is non-empty, or simply the most direct relationship between $P(A)$ and $P(B)$ implied by the symmetry of the conditional probability equality.

Given the options, and the fact that the condition $P(A|B)=P(B|A)$ implies that either $P(A \cap B)=0$ or $P(A)=P(B)$, option (D) is the only one that presents one of the possible consequences directly.

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The final answer is $\mathbf{P(A) = P(B)}$.

Given:

Number of black balls in the urn = 10

Number of white balls in the urn = 5

Two balls are drawn one after the other without replacement.

---

To Find:

The probability that both drawn balls are black.

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Solution:

First, we find the total number of balls in the urn.

Total number of balls = Number of black balls + Number of white balls

Total number of balls = $10 + 5 = 15$ balls.

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We want to find the probability of drawing two black balls without putting the first ball back into the urn.

Let A be the event that the first ball drawn is black.

Let B be the event that the second ball drawn is black.

---

The probability of the first ball being black is the ratio of the number of black balls to the total number of balls:

$P(A) = \frac{\text{Number of black balls}}{\text{Total number of balls}} = \frac{10}{15}$

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Since the first ball drawn is black and it is not replaced, for the second draw, the number of black balls decreases by 1, and the total number of balls also decreases by 1.

Number of black balls remaining = $10 - 1 = 9$

Total number of balls remaining = $15 - 1 = 14$

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The probability of the second ball being black, given that the first ball was black ($P(B|A)$), is the ratio of the number of remaining black balls to the total number of remaining balls:

$P(B|A) = \frac{\text{Number of remaining black balls}}{\text{Total number of remaining balls}} = \frac{9}{14}$

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The probability that both drawn balls are black is the probability of event A happening multiplied by the probability of event B happening given that A has already happened. This is given by the multiplication rule of probability:

$P(\text{Both balls are black}) = P(A \text{ and } B) = P(A) \times P(B|A)$

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Now, we simplify the product of the fractions by cancelling common factors:

We can simplify $\frac{10}{15}$ by dividing both numerator and denominator by 5: $\frac{\cancel{10}^2}{\cancel{15}_3}$.

Now we have $\frac{2}{3} \times \frac{9}{14}$. We can cancel the 2 in the numerator with the 14 in the denominator (divide both by 2), and the 3 in the denominator with the 9 in the numerator (divide both by 3).

$P(\text{Both balls are black}) = \frac{\cancel{2}^1}{\cancel{3}_1} \times \frac{\cancel{9}^3}{\cancel{14}_7}$

Multiplying the simplified fractions:

$P(\text{Both balls are black}) = \frac{1}{1} \times \frac{3}{7} = \frac{1 \times 3}{1 \times 7} = \frac{3}{7}$

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The probability that both drawn balls are black is $\frac{3}{7}$.

Given:

---

A pack of 52 well-shuffled cards.

Three cards are drawn successively, without replacement.

---

To Find:

---

The probability that the first two cards are kings and the third card drawn is an ace.

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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 A$_3$ be the event that the third card drawn is an ace.

We want to find the probability $P(\text{K}_1 \cap \text{K}_2 \cap \text{A}_3)$.

Using the multiplication rule for probability, we have:

Step 1: Calculate the probability that the first card drawn is a king, $P(\text{K}_1)$.

There are 4 kings in a standard deck of 52 cards.

Step 2: Calculate the probability that the second card drawn is a king, given that the first card drawn was a king, $P(\text{K}_2 | \text{K}_1)$.

Since the first card drawn was a king and it was not replaced, there are now $4 - 1 = 3$ kings left in the deck.

The total number of cards remaining in the deck is $52 - 1 = 51$.

Step 3: Calculate the probability that the third card drawn is an ace, given that the first two cards drawn were kings, $P(\text{A}_3 | \text{K}_1 \cap \text{K}_2)$.

The first two cards drawn were kings, and they were not replaced. The number of aces in the deck remains 4.

The total number of cards remaining in the deck is $51 - 1 = 50$.

Substitute the probabilities from (2), (3), and (4) into equation (1):

Simplify the fractions:

Multiply the numerators and the denominators:

---

The probability that the first two cards are kings and the third card drawn is an ace is $\frac{2}{5525}$.

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The final answer is $\mathbf{\frac{2}{5525}}$.

Given:

---

A die is thrown.

Event E: The number appearing is a multiple of 3.

Event F: The number appearing is even.

---

To Find:

---

Whether events E and F are independent.

---

Solution:

---

The sample space S when a die is thrown is the set of all possible outcomes:

The total number of outcomes in the sample space is $n(S) = 6$.

Event E is that the number appearing is a multiple of 3.

The number of outcomes in event E is $n(E) = 2$.

The probability of event E is:

Event F is that the number appearing is even.

The number of outcomes in event F is $n(F) = 3$.

The probability of event F is:

Now, let's find the intersection of events E and F, denoted by $E \cap F$. This is the event that the number appearing is both a multiple of 3 and even.

The number of outcomes in the intersection event is $n(E \cap F) = 1$.

The probability of the intersection event $E \cap F$ is:

For two events E and F to be independent, the following condition must be satisfied:

Let's calculate the product of $P(E)$ and $P(F)$ using values from (1) and (2):

Now, compare the value of $P(E \cap F)$ from (3) and the value of $P(E) \cdot P(F)$ from (5).

Since $P(E \cap F) = P(E) \cdot P(F)$, the events E and F are independent.

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The events E and F are independent.

Given:

---

An unbiased die is thrown twice.

Event A: An odd number on the first throw.

Event B: An odd number on the second throw.

---

To Check:

---

Whether events A and B are independent.

---

Solution:

---

When an unbiased die is thrown, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.

When the die is thrown twice, the sample space S consists of pairs of outcomes $(i, j)$, where $i$ is the outcome of the first throw and $j$ is the outcome of the second throw.

The total number of outcomes in the sample space is $n(S) = 6 \times 6 = 36$. Since the die is unbiased, each outcome $(i, j)$ is equally likely with probability $\frac{1}{36}$.

Event A is that the number on the first throw is odd.

The odd numbers in the sample space of a single throw are $\{1, 3, 5\}$.

The number of outcomes in event A is $n(A) = (\text{number of choices for } i) \times (\text{number of choices for } j) = 3 \times 6 = 18$.

The probability of event A is:

Event B is that the number on the second throw is odd.

The number of outcomes in event B is $n(B) = (\text{number of choices for } i) \times (\text{number of choices for } j) = 6 \times 3 = 18$.

The probability of event B is:

The intersection of events A and B, denoted by $A \cap B$, is the event that the number on the first throw is odd AND the number on the second throw is odd.

The number of outcomes in the intersection event is $n(A \cap B) = (\text{number of choices for } i) \times (\text{number of choices for } j) = 3 \times 3 = 9$.

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

For two events A and B to be independent, the following condition must be satisfied:

Let's calculate the product of $P(A)$ and $P(B)$ using values from (1) and (2):

Now, compare the value of $P(A \cap B)$ from (3) and the value of $P(A) \cdot P(B)$ from (5).

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

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The events A and B are independent.

Given:

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Three coins are tossed simultaneously.

Event E: 'three heads or three tails'.

Event F: 'at least two heads'.

Event G: 'at most two heads'.

---

To Find:

---

Which of the pairs (E, F), (E, G), and (F, G) are independent and which are dependent.

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Solution:

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When three coins are tossed simultaneously, the sample space S is:

The total number of outcomes in the sample space is $n(S) = 8$. Since the coins are assumed to be unbiased, each outcome is equally likely with probability $\frac{1}{8}$.

Now, let's describe the given events in terms of the sample space and find their probabilities.

Event E: 'three heads or three tails'

$n(E) = 2$

Event F: 'at least two heads' (This means two heads or three heads)

$n(F) = 4$

Event G: 'at most two heads' (This means zero heads, one head, or two heads)

$n(G) = 7$

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Now, let's find the intersections of the pairs of events and their probabilities.

Pair (E, F): Intersection $E \cap F$

$n(E \cap F) = 1$

Check for independence of (E, F): Compare $P(E \cap F)$ with $P(E) \cdot P(F)$.

From (4) and (5), $P(E \cap F) = P(E) \cdot P(F)$.

Therefore, the events E and F are independent.

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Pair (E, G): Intersection $E \cap G$

$n(E \cap G) = 1$

Check for independence of (E, G): Compare $P(E \cap G)$ with $P(E) \cdot P(G)$.

From (6) and (7), $P(E \cap G) = \frac{1}{8}$ and $P(E) \cdot P(G) = \frac{7}{32}$. Since $\frac{1}{8} \neq \frac{7}{32}$ (because $4 \times 1 = 4$ and $1 \times 7 = 7$, and $4 \neq 7$), $P(E \cap G) \neq P(E) \cdot P(G)$.

Therefore, the events E and G are dependent.

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Pair (F, G): Intersection $F \cap G$

$n(F \cap G) = 3$

Check for independence of (F, G): Compare $P(F \cap G)$ with $P(F) \cdot P(G)$.

From (8) and (9), $P(F \cap G) = \frac{3}{8}$ and $P(F) \cdot P(G) = \frac{7}{16}$. Since $\frac{3}{8} \neq \frac{7}{16}$ (because $3 \times 16 = 48$ and $8 \times 7 = 56$, and $48 \neq 56$), $P(F \cap G) \neq P(F) \cdot P(G)$.

Therefore, the events F and G are dependent.

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Summary of independence:

(E, F): Independent

(E, G): Dependent

(F, G): Dependent

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The final answer is: (E, F) are independent, (E, G) and (F, G) are dependent.

Given:

---

E and F are independent events.

---

To Prove:

---

Events E and F′ are independent.

---

Proof:

---

Since E and F are independent events, by definition, we have:

$P(E \cap F) = P(E)P(F)$

We want to prove that E and F′ are independent. This means we need to show that:

$P(E \cap F') = P(E)P(F')$

Consider the event E. The event E can be written as the union of two disjoint events: the intersection of E and F ($E \cap F$), and the intersection of E and the complement of F ($E \cap F'$). This can be represented as:

$E = (E \cap F) \cup (E \cap F')$

Since $(E \cap F)$ and $(E \cap F')$ are mutually exclusive (disjoint), the probability of their union is the sum of their probabilities:

$P(E) = P(E \cap F) + P(E \cap F')$

We can rearrange this equation to solve for $P(E \cap F')$:

$P(E \cap F') = P(E) - P(E \cap F)$

Since E and F are independent, we know that $P(E \cap F) = P(E)P(F)$. Substitute this into the equation for $P(E \cap F')$:

$P(E \cap F') = P(E) - P(E)P(F)$

Now, factor out $P(E)$ from the right side of the equation:

$P(E \cap F') = P(E) (1 - P(F))$

We also know that the probability of the complement of an event F is $P(F') = 1 - P(F)$.

Substitute this into the equation:

$P(E \cap F') = P(E) P(F')$

This result shows that the probability of the intersection of E and F' is equal to the product of the probabilities of E and F'. By the definition of independent events, this proves that E and F′ are independent.

---

Hence, if E and F are independent events, then so are E and F′.

Given:

---

A and B are two independent events.

---

To Prove:

---

The probability of occurrence of at least one of A and B is given by $1 - P(A') P(B')$.

In other words, to prove that $P(A \cup B) = 1 - P(A')P(B')$.

---

Proof:

---

The event 'at least one of A and B' is the union of events A and B, denoted by $A \cup B$.

We know the relationship between the probability of a union of events and the probability of the complement of their union:

Using De Morgan's law, the complement of the union of two events is the intersection of their complements:

Substitute this into equation (1):

We are given that A and B are independent events.

A property of independent events is that if A and B are independent, then their complements A′ and B′ are also independent events.

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

Substitute the result from equation (3) into equation (2):

This is the required result.

---

Thus, if A and B are two independent events, the probability of occurrence of at least one of A and B is given by $1 - P(A') P(B')$.

---

The statement is proven.

Given:

---

$P(A) = \frac{3}{5}$

$P(B) = \frac{1}{5}$

A and B are independent events.

---

To Find:

---

$P(A \cap B)$

---

Solution:

---

For two events A and B to be independent, the probability of their intersection is equal to the product of their individual probabilities. This is given by the formula:

We are given the probabilities of events A and B.

Substitute the given values of $P(A)$ and $P(B)$ into the formula (1):

Multiply the numerators together and the denominators together:

---

The probability of the intersection of events A and B, given that they are independent, is $\frac{3}{25}$.

---

The final answer is $\mathbf{\frac{3}{25}}$.

Given:

---

A pack of 52 playing cards.

Two cards are drawn at random and without replacement.

---

To Find:

---

The probability that both the cards drawn are black.

---

Solution:

---

A standard deck of 52 playing cards consists of 26 red cards and 26 black cards.

Total number of cards in the pack = 52.

Number of black cards in the pack = 26.

Let B$_1$ be the event that the first card drawn is black.

Let B$_2$ be the event that the second card drawn is black.

We want to find the probability of drawing two black cards in succession without replacement, which is the probability of the intersection of events B$_1$ and B$_2$, denoted as $P(B_1 \cap B_2)$.

Using the multiplication rule of probability for dependent events, we have:

Step 1: Calculate the probability that the first card drawn is black, $P(B_1)$.

There are 26 black cards out of a total of 52 cards.

Step 2: Calculate the probability that the second card drawn is black, given that the first card drawn was black, $P(B_2 | B_1)$.

Since the first card drawn was black and it was not replaced, there are now $26 - 1 = 25$ black cards left in the deck.

The total number of cards remaining in the deck is $52 - 1 = 51$.

Now, substitute the probabilities from (2) and (3) into the multiplication rule in equation (1):

Multiply the fractions:

---

Alternatively, using combinations:

Total number of ways to draw 2 cards from 52 is $\binom{52}{2}$.

Number of ways to draw 2 black cards from 26 is $\binom{26}{2}$.

We can simplify the fraction $\frac{325}{1326}$. Both are divisible by 25? No. Divisible by 13? $325 = 13 \times 25$, $1326 = 13 \times 102$.

Both methods yield the same result.

---

The probability that both drawn cards are black is $\frac{25}{102}$.

---

The final answer is $\mathbf{\frac{25}{102}}$.

Given:

---

Total number of oranges in the box = 15.

Number of good oranges = 12.

Number of bad oranges = 3.

Three oranges are drawn without replacement.

The box is approved if all three drawn oranges are good.

---

To Find:

---

The probability that the box will be approved for sale.

---

Solution:

---

Let G$_1$ be the event that the first orange drawn is good.

Let G$_2$ be the event that the second orange drawn is good.

Let G$_3$ be the event that the third orange drawn is good.

The box is approved if and only if all three drawn oranges are good. This corresponds to the event $G_1 \cap G_2 \cap G_3$.

We want to find the probability $P(G_1 \cap G_2 \cap G_3)$. Since the oranges are drawn without replacement, these are dependent events. We can use the multiplication rule for dependent events:

Step 1: Calculate the probability that the first orange drawn is good, $P(G_1)$.

There are 12 good oranges out of a total of 15 oranges.

Step 2: Calculate the probability that the second orange drawn is good, given that the first orange drawn was good, $P(G_2 | G_1)$.

Since the first orange drawn was good and it was not replaced, there are now $12 - 1 = 11$ good oranges left in the box.

The total number of oranges remaining in the box is $15 - 1 = 14$.

Step 3: Calculate the probability that the third orange drawn is good, given that the first two oranges drawn were good, $P(G_3 | G_1 \cap G_2)$.

Since the first two oranges drawn were good and they were not replaced, there are now $11 - 1 = 10$ good oranges left in the box.

The total number of oranges remaining in the box is $14 - 1 = 13$.

Now, substitute the probabilities from (2), (3), and (4) into the multiplication rule in equation (1):

Simplify the fractions:

Let's simplify the multiplication $\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}$ differently:

Mistake in previous step. Let's restart simplification carefully:

Let's recheck the simplification: $\frac{12}{15} = \frac{4}{5}$, $\frac{10}{14} = \frac{5}{7}$. So $\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13} = \frac{4}{5} \times \frac{11}{14} \times \frac{10}{13}$. Simplify 5 and 10: $\frac{4}{1} \times \frac{11}{14} \times \frac{2}{13}$. Simplify 4 and 14: $\frac{\cancel{4}^{2}}{1} \times \frac{11}{\cancel{14}_{7}} \times \frac{2}{13} = \frac{2 \times 11 \times 2}{7 \times 13} = \frac{44}{91}$.

Apologies for the repeated error in simplification. The correct simplification is:

Let's simplify again. $\frac{12}{15}=\frac{4}{5}$. $\frac{10}{14}=\frac{5}{7}$. No, $\frac{10}{14}$ is from the third draw. $\frac{12}{15} \times \frac{11}{14} \times \frac{10}{13}$.

$\frac{12}{15} = \frac{4}{5}$.

$\frac{10}{14} = \frac{5}{7}$.

So, $\frac{4}{5} \times \frac{11}{14} \times \frac{10}{13}$.

Multiply numerators: $4 \times 11 \times 10 = 440$.

Multiply denominators: $5 \times 14 \times 13 = 70 \times 13 = 910$.

$P(\text{approved}) = \frac{440}{910} = \frac{44}{91}$.

---

Alternative approach using combinations:

The total number of ways to choose 3 oranges from 15 is given by $\binom{15}{3}$.

The number of ways to choose 3 good oranges from the 12 good oranges is given by $\binom{12}{3}$.

The probability of drawing 3 good oranges is the ratio of the number of ways to draw 3 good oranges to the total number of ways to draw 3 oranges:

Simplify the fraction $\frac{220}{455}$. Both numerator and denominator are divisible by 5.

Both methods yield the same result. The probability that the box will be approved for sale is $\frac{44}{91}$.

---

The final answer is $\mathbf{\frac{44}{91}}$.

Given:

---

A fair coin and an unbiased die are tossed.

Event A: 'head appears on the coin'.

Event B: '3 on the die'.

---

To Check:

---

Whether events A and B are independent events or not.

---

Solution:

---

The outcomes when a fair coin is tossed are Head (H) and Tail (T).

The outcomes when an unbiased die is thrown are $\{1, 2, 3, 4, 5, 6\}$.

When a fair coin and an unbiased die are tossed simultaneously, the sample space S consists of pairs of outcomes (coin outcome, die outcome):

$S = \{(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)\}$

The total number of outcomes in the sample space is $n(S) = 2 \times 6 = 12$. Since the coin is fair and the die is unbiased, each outcome $(c, d)$ is equally likely with probability $\frac{1}{12}$.

Event A: 'head appears on the coin'.

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

The number of outcomes in event A is $n(A) = 6$.

The probability of event A is:

Event B: '3 on the die'.

$B = \{(H, 3), (T, 3)\}$

The number of outcomes in event B is $n(B) = 2$.

The probability of event B is:

The intersection of events A and B, denoted by $A \cap B$, is the event that a head appears on the coin AND a 3 appears on the die.

$A \cap B = \{(H, 3)\}$

The number of outcomes in the intersection event is $n(A \cap B) = 1$.

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

For two events A and B to be independent, the following condition must be satisfied:

Let's calculate the product of $P(A)$ and $P(B)$ using values from (1) and (2):

Now, compare the value of $P(A \cap B)$ from (3) and the value of $P(A) \cdot P(B)$ from (5).

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

---

The events A and B are independent.

Given:

---

A die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed.

Event A: 'the number is even'.

Event B: 'the number is red'.

---

To Check:

---

Whether events A and B are independent.

---

Solution:

---

The sample space S when the die is tossed is the set of all possible outcomes:

The total number of outcomes in the sample space is $n(S) = 6$. Since the die is unbiased, each outcome is equally likely with probability $\frac{1}{6}$.

Event A: 'the number is even'.

The number of outcomes in event A is $n(A) = 3$.

The probability of event A is:

Event B: 'the number is red'.

The numbers marked in red are 1, 2, 3.

The number of outcomes in event B is $n(B) = 3$.

The probability of event B is:

Now, let's find the intersection of events A and B, denoted by $A \cap B$. This is the event that the number appearing is both even and red.

From the die markings, the even numbers are {2, 4, 6} and the red numbers are {1, 2, 3}. The common number is 2.

The number of outcomes in the intersection event is $n(A \cap B) = 1$.

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

For two events A and B to be independent, the following condition must be satisfied:

Let's calculate the product of $P(A)$ and $P(B)$ using values from (1) and (2):

Now, compare the value of $P(A \cap B)$ from (3) and the value of $P(A) \cdot P(B)$ from (5).

Since $\frac{1}{6} \neq \frac{1}{4}$, the condition for independence $P(A \cap B) = P(A) \cdot P(B)$ is not satisfied.

Therefore, the events A and B are dependent.

---

The events A and B are dependent.

Given:

---

$P(E) = \frac{3}{5}$

$P(F) = \frac{3}{10}$

$P(E \cap F) = \frac{1}{5}$

---

To Check:

---

Whether events E and F are independent.

---

Solution:

---

Two events E and F are independent if and only if the probability of their intersection is equal to the product of their individual probabilities. That is:

We are given the values of $P(E)$, $P(F)$, and $P(E \cap F)$.

Let's calculate the product of $P(E)$ and $P(F)$:

Multiply the numerators together and the denominators together:

We are given that $P(E \cap F) = \frac{1}{5}$.

To check for independence, we compare the value of $P(E \cap F)$ from (3) with the value of $P(E) \cdot P(F)$ from (2).

We need to compare $\frac{1}{5}$ and $\frac{9}{50}$.

To compare these fractions, we can express $\frac{1}{5}$ with a denominator of 50:

So, $P(E \cap F) = \frac{10}{50}$ and $P(E) \cdot P(F) = \frac{9}{50}$.

Clearly, $\frac{10}{50} \neq \frac{9}{50}$.

Since $P(E \cap F) \neq P(E) \cdot P(F)$, the condition for independence is not satisfied.

Therefore, the events E and F are dependent.

---

The events E and F are dependent.

Given:

---

Events A and B.

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

$P(A \cup B) = \frac{3}{5}$

$P(B) = p$

---

To Find:

---

The value of p if A and B are:

(i) mutually exclusive

(ii) independent

---

Solution:

---

(i) A and B are mutually exclusive events.

If A and B are mutually exclusive events, their intersection is the empty set, i.e., $A \cap B = \phi$.

The probability of their intersection is $P(A \cap B) = 0$.

The formula for the probability of the union of two mutually exclusive events is:

Substitute the given values into equation (1):

To find p, subtract $\frac{1}{2}$ from both sides of the equation:

Find a common denominator, which is 10:

So, if A and B are mutually exclusive, $p = \frac{1}{10}$.

---

(ii) A and B are independent events.

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

Given $P(A) = \frac{1}{2}$ and $P(B) = p$, we have $P(A \cap B) = \frac{1}{2} \cdot p = \frac{p}{2}$.

The general formula for the probability of the union of two events is:

Substitute the given values and the expression for $P(A \cap B)$ into equation (2):

Combine the terms involving p:

Subtract $\frac{1}{2}$ from both sides of the equation:

Calculate the left side (which we already did in part (i)):

Multiply both sides by 2 to solve for p:

So, if A and B are independent, $p = \frac{1}{5}$.

---

Final values of p:

(i) If A and B are mutually exclusive, $p = \frac{1}{10}$.

(ii) If A and B are independent, $p = \frac{1}{5}$.

---

The final answer is $\mathbf{p = \frac{1}{10}}$ for mutually exclusive and $\mathbf{p = \frac{1}{5}}$ for independent.

Given:

---

A and B are independent events.

$P(A) = 0.3$

$P(B) = 0.4$

---

To Find:

---

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

(ii) $P(A \cup B)$

(iii) $P (A|B)$

(iv) $P (B|A)$

---

Solution:

---

(i) Find $P(A \cap B)$

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

Substitute the given values $P(A) = 0.3$ and $P(B) = 0.4$ into equation (1):

---

(ii) Find $P(A \cup B)$

The general formula for the probability of the union of two events is:

Substitute the given values $P(A) = 0.3$, $P(B) = 0.4$, and the calculated value $P(A \cap B) = 0.12$ into equation (2):

---

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

The definition of conditional probability is:

Substitute the calculated value $P(A \cap B) = 0.12$ and the given value $P(B) = 0.4$ into equation (3):

Alternatively, since A and B are independent events, the occurrence of event B does not affect the probability of event A occurring. Thus, by the property of independent events:

Given $P(A) = 0.3$, so $P(A|B) = 0.3$.

---

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

The definition of conditional probability is:

Substitute the calculated value $P(A \cap B) = 0.12$ and the given value $P(A) = 0.3$ into equation (4):

Alternatively, since A and B are independent events, the occurrence of event A does not affect the probability of event B occurring. Thus, by the property of independent events:

Given $P(B) = 0.4$, so $P(B|A) = 0.4$.

---

Final answers:

(i) $P(A \cap B) = 0.12$

(ii) $P(A \cup B) = 0.58$

(iii) $P (A|B) = 0.3$

(iv) $P (B|A) = 0.4$

---

The final answer is $\mathbf{(i) \ 0.12, (ii) \ 0.58, (iii) \ 0.3, (iv) \ 0.4}$.

Given:

---

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

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

$P(A \cap B) = \frac{1}{8}$

---

To Find:

---

$P(\text{not A and not B})$, which is $P(A' \cap B')$

---

Solution:

---

We want to find the probability that neither event A nor event B occurs. This corresponds to the probability of the intersection of the complements of A and B, denoted as $P(A' \cap B')$.

Using De Morgan's Law, the intersection of the complements of two events is equivalent to the complement of their union:

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

So, the probability we need to find is $P((A \cup B)')$.

The probability of the complement of an event is 1 minus the probability of the event:

First, we need to find the probability of the union of events A and B, $P(A \cup B)$. The general formula for the probability of the union of two events is:

Substitute the given probabilities into equation (2):

To add and subtract these fractions, find a common denominator, which is 8.

Now, substitute the calculated value of $P(A \cup B)$ into equation (1):

---

Note: We can check if the events A and B are independent based on the given probabilities.

Since the given $P(A \cap B) = \frac{1}{8}$, and $P(A \cap B) = P(A) \cdot P(B)$, the events A and B are indeed independent.

If A and B are independent, then their complements A′ and B′ are also independent. In this case, the probability of their intersection is the product of their probabilities:

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

We can find $P(A')$ and $P(B')$:

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

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

Then, $P(A' \cap B') = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$.

This confirms the result obtained using De Morgan's Law and the union formula.

---

The final answer is $\mathbf{\frac{3}{8}}$.

Given:

---

Events A and B.

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

$P(B) = \frac{7}{12}$

$P(\text{not A or not B}) = P(A' \cup B') = \frac{1}{4}$

---

To State:

---

Whether A and B are independent events.

---

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, i.e., $P(A \cap B) = P(A) \cdot P(B)$.

We are given $P(A)$, $P(B)$, and $P(A' \cup B')$. We need to find $P(A \cap B)$ and compare it with $P(A) \cdot P(B)$.

Using De Morgan's Law, we know that the complement of the intersection of two events is the union of their complements:

Therefore, the probability of the complement of the intersection is equal to the probability of the union of the complements:

We know that $P((A \cap B)') = 1 - P(A \cap B)$. So,

Substitute the given value $P(A' \cup B') = \frac{1}{4}$ into equation (1):

Now, solve for $P(A \cap B)$:

Next, let's calculate the product of $P(A)$ and $P(B)$ using the given values:

Now, we compare the value of $P(A \cap B)$ from equation (2) with the value of $P(A) \cdot P(B)$ from equation (3).

We have $P(A \cap B) = \frac{3}{4}$ and $P(A) \cdot P(B) = \frac{7}{24}$.

To compare $\frac{3}{4}$ and $\frac{7}{24}$, we can express $\frac{3}{4}$ with a common denominator of 24:

So, $P(A \cap B) = \frac{18}{24}$ and $P(A) \cdot P(B) = \frac{7}{24}$.

Since $\frac{18}{24} \neq \frac{7}{24}$, it follows that $P(A \cap B) \neq P(A) \cdot P(B)$.

According to the definition of independent events, if $P(A \cap B) \neq P(A) \cdot P(B)$, then the events A and B are not independent; they are dependent.

---

The events A and B are dependent.

Given:

---

A and B are independent events.

$P(A) = 0.3$

$P(B) = 0.6$

---

To Find:

---

(i) $P(A \text{ and } B)$

(ii) $P(A \text{ and not } B)$

(iii) $P(A \text{ or } B)$

(iv) $P(\text{neither } A \text{ nor } B)$

---

Solution:

---

Since A and B are independent events, we know that:

Also, if A and B are independent, then the following pairs of events are also independent:

$\quad$ - A and B'

$\quad$ - A' and B

$\quad$ - A' and B'

We will use these properties to find the required probabilities.

---

(i) Find $P(A \text{ and } B)$

The event 'A and B' is the intersection $A \cap B$.

Using the property of independent events:

Substitute the given values $P(A) = 0.3$ and $P(B) = 0.6$:

---

(ii) Find $P(A \text{ and not } B)$

The event 'A and not B' is the intersection $A \cap B'$.

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

Therefore, the probability of their intersection is the product of their probabilities:

First, find the probability of the complement of B, $P(B')$:

Now, calculate $P(A \cap B')$:

---

(iii) Find $P(A \text{ or } B)$

The event 'A or B' is the union $A \cup B$.

Using the formula for the probability of the union of two events:

We have $P(A) = 0.3$, $P(B) = 0.6$, and we found $P(A \cap B) = 0.18$ in part (i).

Substitute these values:

Alternatively, for independent events, $P(A \cup B) = 1 - P(A' \cap B')$. We will calculate $P(A' \cap B')$ in the next part.

---

(iv) Find $P(\text{neither } A \text{ nor } B)$

The event 'neither A nor B' means that event A does not occur AND event B does not occur. This is the intersection of the complements of A and B, denoted as $A' \cap B'$.

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

The probability of the complement of the union is $1 - P(A \cup B)$.

Using the result from part (iii), $P(A \cup B) = 0.72$:

Alternatively, since A and B are independent, A' and B' are also independent. Therefore, the probability of their intersection is the product of their probabilities:

First, find the probability of the complement of A, $P(A')$:

We already found $P(B') = 0.4$ in part (ii).

Now, calculate $P(A' \cap B')$:

Both methods yield the same result.

---

Final probabilities:

(i) $P(A \text{ and } B) = 0.18$

(ii) $P(A \text{ and not } B) = 0.12$

(iii) $P(A \text{ or } B) = 0.72$

(iv) $P(\text{neither } A \text{ nor } B) = 0.28$

---

The final answer is $\mathbf{(i) \ 0.18, (ii) \ 0.12, (iii) \ 0.72, (iv) \ 0.28}$.

Given:

---

A die is tossed thrice.

---

To Find:

---

The probability of getting an odd number at least once.

---

Solution:

---

When a single die is tossed, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is 6.

The odd numbers are $\{1, 3, 5\}$. The probability of getting an odd number in a single toss is:

The even numbers are $\{2, 4, 6\}$. The probability of getting an even number in a single toss is:

Let A be the event of getting an odd number at least once in three tosses.

The complement of event A, denoted by A′, is the event of getting an odd number zero times in three tosses, which means getting an even number on all three tosses.

Let E$_1$ be the event of getting an even number on the first toss.

Let E$_2$ be the event of getting an even number on the second toss.

Let E$_3$ be the event of getting an even number on the third toss.

The event A′ is the intersection of E$_1$, E$_2$, and E$_3$, i.e., $A' = E_1 \cap E_2 \cap E_3$.

Since each toss of a die is an independent event, the probability of the intersection of E$_1$, E$_2$, and E$_3$ is the product of their individual probabilities:

The probability of getting an even number on any single toss is $P(\text{Even}) = \frac{1}{2}$.

So, $P(E_1) = \frac{1}{2}$, $P(E_2) = \frac{1}{2}$, and $P(E_3) = \frac{1}{2}$.

Substitute these probabilities into equation (1):

The probability of getting an odd number at least once (event A) is $1$ minus the probability of getting no odd numbers (event A′):

Substitute the value of $P(A')$ into equation (2):

---

The probability of getting an odd number at least once when a die is tossed thrice is $\frac{7}{8}$.

---

The final answer is $\mathbf{\frac{7}{8}}$.

Given:

---

A box contains 10 black and 8 red balls.

Total number of balls = $10 + 8 = 18$.

Two balls are drawn at random with replacement.

---

To Find:

---

(i) Probability that both balls are red.

(ii) Probability that the first ball is black and the second is red.

(iii) Probability that one of them is black and other is red.

---

Solution:

---

Let B be the event of drawing a black ball in a single draw.

Let R be the event of drawing a red ball in a single draw.

The probability of drawing a black ball in a single draw is:

The probability of drawing a red ball in a single draw is:

Since the balls are drawn with replacement, the outcome of the first draw does not affect the outcome of the second draw. The events of drawing each ball are independent.

Let $R_1$ be the event that the first ball is red, $R_2$ be the event that the second ball is red, $B_1$ be the event that the first ball is black, and $B_2$ be the event that the second ball is black.

$P(R_1) = P(R_2) = P(R) = \frac{4}{9}$

$P(B_1) = P(B_2) = P(B) = \frac{5}{9}$

---

(i) Probability that both balls are red:

This is the probability of the event $R_1 \cap R_2$. Since $R_1$ and $R_2$ are independent events:

Substitute the probabilities:

---

(ii) Probability that the first ball is black and the second is red:

This is the probability of the event $B_1 \cap R_2$. Since $B_1$ and $R_2$ are independent events:

Substitute the probabilities:

---

(iii) Probability that one of them is black and other is red:

This event can occur in two mutually exclusive ways:

1. The first ball is black and the second ball is red ($B_1 \cap R_2$).

2. The first ball is red and the second ball is black ($R_1 \cap B_2$).

The probability of this event is the sum of the probabilities of these two mutually exclusive events:

Since the individual draws are independent, we have:

Summing these probabilities:

---

Final probabilities:

(i) Probability that both balls are red = $\frac{16}{81}$.

(ii) Probability that the first ball is black and second is red = $\frac{20}{81}$.

(iii) Probability that one of them is black and other is red = $\frac{40}{81}$.

---

The final answer is $\mathbf{(i) \ \frac{16}{81}, (ii) \ \frac{20}{81}, (iii) \ \frac{40}{81}}$.

Given:

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Probability of A solving the problem, $P(A) = \frac{1}{2}$.

Probability of B solving the problem, $P(B) = \frac{1}{3}$.

A and B try to solve the problem independently.

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To Find:

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(i) The probability that the problem is solved.

(ii) The probability that exactly one of them solves the problem.

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Solution:

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Let A be the event that A solves the problem, and B be the event that B solves the problem.

We are given that A and B are independent events.

The probabilities of their complements (not solving the problem) are:

Since A and B are independent, A' and B' are also independent. Similarly, A and B' are independent, and A' and B are independent.

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(i) The probability that the problem is solved.

The problem is solved if at least one of A or B solves it. This is the event $A \cup B$.

The probability of the union of two events is given by:

Since A and B are independent, $P(A \cap B) = P(A) \cdot P(B)$.

Substitute the probabilities into the union formula:

Find a common denominator (6):

Alternatively, the event 'problem is solved' is the complement of the event 'neither A nor B solves the problem' ($A' \cap B'$).

The probability that the problem is solved is $\frac{2}{3}$.

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(ii) Probability that exactly one of them solves the problem.

Exactly one of them solves the problem means either (A solves AND B does not solve) or (A does not solve AND B solves).

This corresponds to the event $(A \cap B') \cup (A' \cap B)$.

Since the events $(A \cap B')$ and $(A' \cap B)$ are mutually exclusive, the probability of their union is the sum of their probabilities:

Since A and B are independent, A and B' are independent, and A' and B are independent.

Summing these probabilities:

Find a common denominator (6):

The probability that exactly one of them solves the problem is $\frac{1}{2}$.

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Final Answers:

(i) The probability that the problem is solved is $\frac{2}{3}$.

(ii) The probability that exactly one of them solves the problem is $\frac{1}{2}$.

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The final answer is $\mathbf{(i) \ \frac{2}{3}, (ii) \ \frac{1}{2}}$.

Given:

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A well-shuffled deck of 52 cards.

One card is drawn at random.

Different pairs of events E and F.

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To Find:

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In which cases are the events E and F independent.

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Solution:

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Two events E and F are independent if and only if $P(E \cap F) = P(E) \cdot P(F)$. We will check this condition for each pair of events.

Total number of cards in the deck = 52.

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(i) E : ‘the card drawn is a spade’, F : ‘the card drawn is an ace’.

Event E: The card drawn is a spade.

Number of spades in a deck = 13.

Event F: The card drawn is an ace.

Number of aces in a deck = 4.

Event $E \cap F$: The card drawn is a spade AND an ace.

There is only one card that is both a spade and an ace: the Ace of Spades.

$n(E \cap F) = 1$.

Check the independence condition: $P(E) \cdot P(F)$.

Comparing (3) and (4), we see that $P(E \cap F) = P(E) \cdot P(F) = \frac{1}{52}$.

Thus, for case (i), events E and F are independent.

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(ii) E : ‘the card drawn is black’, F : ‘the card drawn is a king’.

Event E: The card drawn is black.

Number of black cards in a deck (Clubs and Spades) = 26.

Event F: The card drawn is a king.

Number of kings in a deck = 4.

Event $E \cap F$: The card drawn is black AND a king.

The black kings are the King of Clubs and the King of Spades. There are 2 black kings.

$n(E \cap F) = 2$.

Check the independence condition: $P(E) \cdot P(F)$.

Comparing (7) and (8), we see that $P(E \cap F) = P(E) \cdot P(F) = \frac{1}{26}$.

Thus, for case (ii), events E and F are independent.

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(iii) E : ‘the card drawn is a king or queen’, F : ‘the card drawn is a queen or jack’.

Event E: The card drawn is a king or a queen.

Number of kings = 4.

Number of queens = 4.

The events 'drawing a king' and 'drawing a queen' are mutually exclusive.

$n(E) = \text{Number of kings} + \text{Number of queens} = 4 + 4 = 8$.

Event F: The card drawn is a queen or a jack.

Number of queens = 4.

Number of jacks = 4.

The events 'drawing a queen' and 'drawing a jack' are mutually exclusive.

$n(F) = \text{Number of queens} + \text{Number of jacks} = 4 + 4 = 8$.

Event $E \cap F$: The card drawn is (a king or a queen) AND (a queen or a jack).

The cards that are both a king/queen AND a queen/jack are the queens.

$n(E \cap F) = \text{Number of queens} = 4$.

Check the independence condition: $P(E) \cdot P(F)$.

Comparing (11) and (12), we have $P(E \cap F) = \frac{1}{13}$ and $P(E) \cdot P(F) = \frac{4}{169}$.

To compare them, we can write $\frac{1}{13}$ with a denominator of 169: $\frac{1}{13} = \frac{1 \times 13}{13 \times 13} = \frac{13}{169}$.

Since $\frac{13}{169} \neq \frac{4}{169}$, it follows that $P(E \cap F) \neq P(E) \cdot P(F)$.

Thus, for case (iii), events E and F are dependent.

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Summary:

(i) Independent

(ii) Independent

(iii) Dependent

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The events E and F are independent in cases (i) and (ii).

Given:

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Let H be the event that the selected student reads Hindi newspaper.

Let E be the event that the selected student reads English newspaper.

We are given the following probabilities:

$P(H) = 60\% = \frac{60}{100} = 0.60$

$P(E) = 40\% = \frac{40}{100} = 0.40$

$P(\text{H and E}) = P(H \cap E) = 20\% = \frac{20}{100} = 0.20$

---

To Find:

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(a) Probability that she reads neither Hindi nor English newspapers ($P(H' \cap E')$).

(b) Probability that she reads English newspaper, given she reads Hindi newspaper ($P(E|H)$).

(c) Probability that she reads Hindi newspaper, given she reads English newspaper ($P(H|E)$).

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Solution:

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(a) Find the probability that she reads neither Hindi nor English newspapers ($P(H' \cap E')$).

The event 'neither Hindi nor English' is the complement of the event 'reads Hindi or English'. Using De Morgan's Law, $(H \cup E)' = H' \cap E'$.

So, we want to find $P(H' \cap E') = P((H \cup E)')$.

The probability of the complement of an event is $1$ minus the probability of the event:

First, we find the probability of the union of H and E using the formula:

Substitute the given probabilities:

Now, substitute this value back into equation (1):

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(b) If she reads Hindi newspaper, find the probability that she reads English newspaper ($P(E|H)$).

This is a conditional probability. The formula for the probability of event E given event H is:

Substitute the given probabilities $P(E \cap H) = 0.20$ and $P(H) = 0.60$:

Simplify the fraction:

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(c) If she reads English newspaper, find the probability that she reads Hindi newspaper ($P(H|E)$).

This is also a conditional probability. The formula for the probability of event H given event E is:

Substitute the given probabilities $P(H \cap E) = 0.20$ and $P(E) = 0.40$:

Simplify the fraction:

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Final Answers:

(a) Probability of reading neither Hindi nor English newspapers = $0.20$ or $\frac{1}{5}$.

(b) Probability of reading English newspaper, given she reads Hindi newspaper = $\frac{1}{3}$.

(c) Probability of reading Hindi newspaper, given she reads English newspaper = $\frac{1}{2}$.

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The final answer is $\mathbf{(a) \ 0.20, (b) \ \frac{1}{3}, (c) \ \frac{1}{2}}$.

Given:

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A pair of unbiased dice is rolled.

---

To Find:

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The probability of obtaining an even prime number on each die.

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Solution:

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When a single unbiased die is rolled, the possible outcomes are the numbers from 1 to 6:

The total number of outcomes for a single die roll is 6.

An event involves getting an 'even prime number'. Let's identify the numbers that are both even and prime from the possible outcomes.

Prime numbers are natural numbers greater than 1 that have no positive divisors other than 1 and themselves. The prime numbers in the set $\{1, 2, 3, 4, 5, 6\}$ are $\{2, 3, 5\}$.

Even numbers in the set $\{1, 2, 3, 4, 5, 6\}$ are $\{2, 4, 6\}$.

The number that is both even and prime is the number that is in both sets, which is $\{2\}$.

So, the event 'obtaining an even prime number' on a single die roll corresponds to getting the number 2.

The probability of getting an even prime number (which is 2) on a single die roll is:

Now, a pair of dice is rolled. This means we are rolling two dice independently.

Let A be the event of getting an even prime number on the first die.

Let B be the event of getting an even prime number on the second die.

We want to find the probability of obtaining an even prime number on each die, which means getting an even prime number on the first die AND getting an even prime number on the second die. This is the intersection of events A and B, i.e., $A \cap B$.

Since the outcomes of the two dice rolls are independent events, the probability of their intersection is the product of their individual probabilities:

From equation (1), we know that the probability of getting an even prime number on a single die is $\frac{1}{6}$. So, $P(A) = \frac{1}{6}$ and $P(B) = \frac{1}{6}$.

Substitute these probabilities:

Now, compare this result with the given options:

(A) 0

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

(C) $\frac{1}{12}$

(D) $\frac{1}{36}$

Our calculated probability matches option (D).

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The final answer is $\mathbf{\frac{1}{36}}$.

Given:

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Two events A and B.

The question asks for the condition under which A and B will be independent.

---

To Find:

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The condition from the given options that implies A and B are independent.

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Solution:

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By definition, two events A and B are independent if and only if:

Let's examine each given option:

(A) A and B are mutually exclusive:

If A and B are mutually exclusive, then $A \cap B = \phi$, which means $P(A \cap B) = 0$.

For mutually exclusive events to be independent, we must have $P(A) \cdot P(B) = P(A \cap B) = 0$. This implies that either $P(A) = 0$ or $P(B) = 0$ (or both).

If $P(A) > 0$ and $P(B) > 0$, mutually exclusive events are not independent.

Therefore, mutual exclusivity is not a general condition for independence.

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(B) P(A′B′) = [1 – P(A)] [1 – P(B)]:

The notation $P(A'B')$ represents $P(A' \cap B')$.

The condition given is $P(A' \cap B') = (1 - P(A))(1 - P(B))$.

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

So, the condition can be written as $P(A' \cap B') = P(A')P(B')$.

This is the definition of independence for the events A' and B'.

A property of independence states that events A and B are independent if and only if their complements A' and B' are independent.

Let's prove this equivalence. If $P(A' \cap B') = P(A')P(B')$, using De Morgan's Law ($A' \cap B' = (A \cup B)'$), we have:

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

Rearranging the terms:

Comparing this with the general formula for the union of two events $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we must have:

Subtracting $P(A) + P(B)$ from both sides gives:

This is the definition of independence for events A and B.

Thus, the condition in option (B) is equivalent to the independence of A and B.

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(C) P(A) = P(B):

This condition states that the probabilities of events A and B are equal. This does not imply independence. For example, consider rolling a fair die. Let A = {1, 2, 3} and B = {2, 4, 6}. Then $P(A) = 3/6 = 1/2$ and $P(B) = 3/6 = 1/2$, so $P(A) = P(B)$. However, $A \cap B = \{2\}$, so $P(A \cap B) = 1/6$. The product $P(A)P(B) = (1/2)(1/2) = 1/4$. Since $1/6 \neq 1/4$, A and B are dependent.

Therefore, $P(A) = P(B)$ is not a condition for independence.

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(D) P(A) + P(B) = 1:

This condition means the sum of the probabilities of A and B is 1. This does not imply independence. For example, consider rolling a fair die. Let A = {1, 2} and B = {3, 4, 5, 6}. Then $P(A) = 2/6 = 1/3$ and $P(B) = 4/6 = 2/3$. $P(A) + P(B) = 1/3 + 2/3 = 1$. However, $A \cap B = \phi$, so $P(A \cap B) = 0$. The product $P(A)P(B) = (1/3)(2/3) = 2/9$. Since $0 \neq 2/9$, A and B are dependent.

Therefore, $P(A) + P(B) = 1$ is not a condition for independence.

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Based on the analysis, the only condition from the given options that is equivalent to the independence of A and B is option (B).

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The final answer is $\mathbf{P(A′B′) = [1 – P(A)] [1 – P(B)]}$.

Given:

Let S be the event that there is a strike.

Let C be the event that the construction job will be completed on time.

We are given the following probabilities:

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To Find:

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The probability that the construction job will be completed on time, $P(C)$.

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Solution:

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The event of having a strike (S) and the event of having no strike (S') are complementary events. This means that either one happens or the other happens, and they cannot happen at the same time. The sum of their probabilities is 1.

The probability of no strike is:

Substitute the given value of $P(S)$:

$P(S') = 0.35$

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The events S (strike) and S' (no strike) form a partition of the sample space. This means that exactly one of these events must occur.

To find the probability of the construction job being completed on time, $P(C)$, we can use the Law of Total Probability. This law helps us find the overall probability of an event by considering different cases that could lead to that event.

The Law of Total Probability states that for an event C, if the sample space is divided into mutually exclusive and exhaustive events $B_1, B_2, ..., B_n$ (like S and S'), then:

In our case, the events S and S' divide the possibilities. So, the probability of completion on time is the probability of completion given a strike times the probability of a strike, plus the probability of completion given no strike times the probability of no strike.

Substitute the given probabilities $P(S) = 0.65$, $P(C|S) = 0.32$, $P(C|S') = 0.80$, and the calculated value $P(S') = 0.35$ into the formula:

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Now, we perform the multiplication and addition:

$0.32 \times 0.65 = 0.2080$

$0.80 \times 0.35 = 0.2800$

Substitute these values back into equation (ii):

Adding the two values:

$0.2080 + 0.2800 = 0.4880$

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The probability that the construction job will be completed on time is 0.488.

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The final answer is $\mathbf{0.488}$.

Given:

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Bag I: 3 red balls, 4 black balls. Total balls in Bag I = $3 + 4 = 7$.

Bag II: 5 red balls, 6 black balls. Total balls in Bag II = $5 + 6 = 11$.

One ball is drawn at random from one of the bags.

The drawn ball is found to be red.

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To Find:

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The probability that the red ball was drawn from Bag II.

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Solution:

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Let B$_1$ be the event that the ball is drawn from Bag I.

Let B$_2$ be the event that the ball is drawn from Bag II.

Since a bag is chosen at random, the probability of choosing either bag is equal:

Let R be the event that the drawn ball is red.

We are given that the drawn ball is red, and we want to find the probability that it was drawn from Bag II. This is the conditional probability $P(B_2 | R)$.

We can use Bayes' Theorem to find this probability. Bayes' Theorem states:

We need to find $P(R | B_2)$, $P(B_2)$, and $P(R)$.

$P(B_2)$ is given as $\frac{1}{2}$.

$P(R | B_2)$ is the probability of drawing a red ball given that the ball is drawn from Bag II.

In Bag II, there are 5 red balls and a total of 11 balls.

To find $P(R)$, the probability of drawing a red ball, we can use the Law of Total Probability. The events B$_1$ and B$_2$ form a partition of the sample space (the ball is drawn from either Bag I or Bag II, and these are mutually exclusive).

We need to find $P(R | B_1)$, which is the probability of drawing a red ball given that the ball is drawn from Bag I.

In Bag I, there are 3 red balls and a total of 7 balls.

Now, substitute the probabilities $P(R | B_1) = \frac{3}{7}$, $P(B_1) = \frac{1}{2}$, $P(R | B_2) = \frac{5}{11}$, and $P(B_2) = \frac{1}{2}$ into the Law of Total Probability formula:

To add these fractions, find a common denominator. The least common multiple of 14 and 22 is 154 (since $14 = 2 \times 7$ and $22 = 2 \times 11$, LCM is $2 \times 7 \times 11 = 154$).

Simplify the fraction by dividing numerator and denominator by their greatest common divisor, which is 2.

Now, we have all the components for Bayes' Theorem: $P(R | B_2) = \frac{5}{11}$, $P(B_2) = \frac{1}{2}$, and $P(R) = \frac{34}{77}$.

Substitute these values into the formula for $P(B_2 | R)$:

To divide by a fraction, multiply by its reciprocal:

Simplify the multiplication by cancelling common factors. 22 and 77 are both divisible by 11.

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The probability that the red ball was drawn from Bag II, given that it is red, is $\frac{35}{68}$.

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The final answer is $\mathbf{\frac{35}{68}}$.

Given:

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Three identical boxes:

Box I: Contains 2 gold coins (GG).

Box II: Contains 2 silver coins (SS).

Box III: Contains 1 gold coin and 1 silver coin (GS).

A person chooses a box at random and takes out a coin.

The drawn coin is gold.

---

To Find:

---

The probability that the other coin in the box is also of gold, given that the first coin drawn is gold.

---

Solution:

---

Let B$_1$ be the event that Box I is chosen.

Let B$_2$ be the event that Box II is chosen.

Let B$_3$ be the event that Box III is chosen.

Since the boxes are identical and chosen at random, the probability of choosing each box is equal:

Let G be the event that the drawn coin is of gold.

We are given that the drawn coin is gold, and we want to find the probability that the other coin in the box is also gold. If the other coin in the box is gold, it means the chosen box was Box I (which contains GG). So, we want to find the conditional probability $P(B_1 | G)$.

We can use Bayes' Theorem to find this probability. Bayes' Theorem states:

We need to find $P(G | B_1)$, $P(B_1)$, and $P(G)$.

$P(B_1)$ is given as $\frac{1}{3}$.

$P(G | B_1)$ is the probability of drawing a gold coin given that Box I is chosen.

In Box I, both coins are gold. So, drawing a gold coin from Box I is certain.

To find $P(G)$, the probability of drawing a gold coin, we can use the Law of Total Probability. The events B$_1$, B$_2$, and B$_3$ form a partition of the sample space (one of the boxes must be chosen, and they are mutually exclusive).

$P(G | B_2)$ is the probability of drawing a gold coin given that Box II is chosen. Box II contains two silver coins.

$P(G | B_3)$ is the probability of drawing a gold coin given that Box III is chosen. Box III contains one gold and one silver coin.

Now, substitute the probabilities $P(G | B_1) = 1$, $P(B_1) = \frac{1}{3}$, $P(G | B_2) = 0$, $P(B_2) = \frac{1}{3}$, $P(G | B_3) = \frac{1}{2}$, and $P(B_3) = \frac{1}{3}$ into the Law of Total Probability formula for $P(G)$:

Find a common denominator (6):

Now, we have all the components for Bayes' Theorem: $P(G | B_1) = 1$, $P(B_1) = \frac{1}{3}$, and $P(G) = \frac{1}{2}$.

Substitute these values into the formula for $P(B_1 | G)$:

To divide by a fraction, multiply by its reciprocal:

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The question asks for the probability that the other coin in the box is also of gold, given that the first coin drawn is gold. This happens if and only if the chosen box is Box I (GG), from which we drew a gold coin. The probability that the chosen box is Box I, given that a gold coin was drawn, is exactly what we calculated $P(B_1 | G)$.

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The probability that the other coin in the box is also of gold, given the drawn coin is gold, is $\frac{2}{3}$.

---

The final answer is $\mathbf{\frac{2}{3}}$.

Given:

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Let H be the event that a person has HIV.

Let H' be the event that a person does not have HIV.

Let P be the event that the HIV test reports HIV+ive.

Let N be the event that the HIV test reports HIV-ive.

We are given the following probabilities:

Probability of having HIV: $P(H) = 0.1\% = \frac{0.1}{100} = 0.001$

Probability of not having HIV: $P(H') = 1 - P(H) = 1 - 0.001 = 0.999$

Probability of testing HIV+ive given the person has HIV (True Positive Rate): $P(P | H) = 90\% = \frac{90}{100} = 0.90$

Probability of testing HIV-ive given the person has HIV (False Negative Rate): $P(N | H) = 10\% = \frac{10}{100} = 0.10$

Probability of testing HIV-ive given the person does not have HIV (True Negative Rate): $P(N | H') = 99\% = \frac{99}{100} = 0.99$

Probability of testing HIV+ive given the person does not have HIV (False Positive Rate): $P(P | H') = 1\% = \frac{1}{100} = 0.01$

---

To Find:

---

The probability that the person actually has HIV, given that the pathologist reports him/her as HIV+ive. This is the conditional probability $P(H | P)$.

---

Solution:

---

We are looking for the probability $P(H | P)$. We can use Bayes' Theorem for this:

To use this formula, we need to find the value of $P(P)$, the overall probability that the test reports HIV+ive. The event P (testing HIV+ive) can occur in two mutually exclusive ways: either the person has HIV and tests positive (event $P \cap H$), or the person does not have HIV and tests positive (event $P \cap H'$).

The events H and H' form a partition of the sample space (a person either has HIV or does not). We can use the Law of Total Probability to find $P(P)$:

Substitute the given probabilities into equation (2):

Calculate the products:

Add these values to find $P(P)$:

Now, substitute the given values $P(P | H) = 0.90$, $P(H) = 0.001$, and the calculated value $P(P) = 0.01089$ into Bayes' Theorem formula (1):

To simplify this fraction, multiply the numerator and the denominator by 1,000,000 (or 100,000 to shift the decimal by 5 places):

Now, simplify the fraction $\frac{90}{1089}$. Both numerator and denominator are divisible by 9.

The probability that the person actually has HIV, given a positive test result, is $\frac{10}{121}$.

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The final answer is $\mathbf{\frac{10}{121}}$.

Given:

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Let $M_A$ be the event that a bolt is manufactured by machine A.

Let $M_B$ be the event that a bolt is manufactured by machine B.

Let $M_C$ be the event that a bolt is manufactured by machine C.

We are given the proportion of bolts manufactured by each machine:

The sum of these probabilities is $0.25 + 0.35 + 0.40 = 1.00$. The events $M_A$, $M_B$, $M_C$ form a partition of the sample space, meaning every bolt comes from exactly one of these machines.

Let D be the event that a bolt is defective.

We are given the percentage of defective bolts produced by each machine:

A bolt is drawn at random and is found to be defective (event D has occurred).

---

To Find:

---

The probability that this defective bolt was manufactured by machine B. This is the conditional probability $P(M_B | D)$.

---

Solution:

---

We need to find $P(M_B | D)$. We can use Bayes' Theorem, which relates conditional probabilities.

To use this formula, we first need to find $P(D)$, the overall probability of drawing a defective bolt. We can find this using the Law of Total Probability. Since the events $M_A$, $M_B$, and $M_C$ cover all possibilities for where a bolt comes from, the total probability of a bolt being defective is the sum of the probabilities of it being defective and from A, or defective and from B, or defective and from C.

Using the multiplication rule for probability ($P(A \cap B) = P(A|B)P(B)$), we get:

Substitute the given probabilities into this formula:

Calculate the products:

$0.05 \times 0.25 = 0.0125$

$0.04 \times 0.35 = 0.0140$

$0.02 \times 0.40 = 0.0080$

Add these values to find the total probability of a defective bolt $P(D)$:

$P(D) = 0.0345$

---

Now, substitute the given value $P(D | M_B) = 0.04$, $P(M_B) = 0.35$, and the calculated value $P(D) = 0.0345$ into the Bayes' Theorem formula (1):

We already calculated $(0.04) \times (0.35) = 0.0140$.

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

Now, simplify the fraction $\frac{140}{345}$ by dividing the numerator and the denominator by their greatest common divisor. Both numbers are divisible by 5 (since they end in 0 and 5).

$140 \div 5 = 28$

$345 \div 5 = 69$

Check if 28 and 69 have common factors. The factors of 28 are 1, 2, 4, 7, 14, 28. The factors of 69 are 1, 3, 23, 69. The only common factor is 1, so the fraction is in its simplest form.

---

The probability that the defective bolt was manufactured by machine B is $\frac{28}{69}$.

---

The final answer is $\mathbf{\frac{28}{69}}$.

Given:

---

Let T be the event that the doctor comes by train.

Let B be the event that the doctor comes by bus.

Let S be the event that the doctor comes by scooter.

Let O be the event that the doctor comes by other means of transport.

We are given the probabilities of each mode of transport:

$P(T) = \frac{3}{10}$

$P(B) = \frac{1}{5} = \frac{2}{10}$

$P(S) = \frac{1}{10}$

$P(O) = \frac{2}{5} = \frac{4}{10}$

Check if these probabilities sum to 1: $\frac{3}{10} + \frac{2}{10} + \frac{1}{10} + \frac{4}{10} = \frac{3+2+1+4}{10} = \frac{10}{10} = 1$. The events T, B, S, O form a partition of the sample space.

Let L be the event that the doctor is late.

We are given the probabilities of being late, conditional on the mode of transport:

Probability of being late given coming by train: $P(L | T) = \frac{1}{4}$

Probability of being late given coming by bus: $P(L | B) = \frac{1}{3}$

Probability of being late given coming by scooter: $P(L | S) = \frac{1}{12}$

Probability of being late given coming by other means: $P(L | O) = 0$ (since he will not be late)

The doctor arrives and is late (event L has occurred).

---

To Find:

---

The probability that the doctor comes by train, given that he is late. This is the conditional probability $P(T | L)$.

---

Solution:

---

We are looking for the probability $P(T | L)$. We can use Bayes' Theorem for this:

To use this formula, we need to find the value of $P(L)$, the overall probability that the doctor is late. We can use the Law of Total Probability. Since the events T, B, S, and O form a partition of the sample space, the Law of Total Probability for event L is:

Substitute the given probabilities into equation (2):

To add these fractions, find a common denominator. The denominators are 40, 15, and 120. The least common multiple of 40, 15, and 120 is 120 (since $40 \times 3 = 120$ and $15 \times 8 = 120$).

Simplify the fraction $\frac{18}{120}$. Both numerator and denominator are divisible by 6.

Now, substitute the given value $P(L | T) = \frac{1}{4}$, $P(T) = \frac{3}{10}$, and the calculated value $P(L) = \frac{3}{20}$ into Bayes' Theorem formula (1):

To divide by a fraction, multiply by its reciprocal:

Simplify the multiplication by cancelling common factors. 3 in the numerator and denominator cancel out. 40 and 20 are both divisible by 20.

The probability that the doctor comes by train, given that he is late, is $\frac{1}{2}$.

---

The final answer is $\mathbf{\frac{1}{2}}$.

Given:

---

Probability that the man speaks the truth = $\frac{3}{4}$.

Probability that the man tells a lie = $1 - \frac{3}{4} = \frac{1}{4}$.

The man throws a die and reports that the outcome is a six.

---

To Find:

---

The probability that the outcome of the die throw is actually a six, given that the man reports it is a six.

---

Solution:

---

Let S be the event that the die actually shows a six.

Let S' be the event that the die does not show a six.

When an unbiased die is thrown, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.

The probability of getting a six is $P(S) = \frac{1}{6}$.

The probability of not getting a six is $P(S') = 1 - P(S) = 1 - \frac{1}{6} = \frac{5}{6}$.

Let R be the event that the man reports that the outcome is a six.

We are given that the man reports that the outcome is a six, and we want to find the probability that it is actually a six. This is the conditional probability $P(S | R)$.

We can use Bayes' Theorem for this:

We need to find $P(R | S)$, $P(S)$, and $P(R)$.

$P(S)$ is calculated as $\frac{1}{6}$.

$P(R | S)$ is the probability that the man reports a six, given that the die actually shows a six. If the die actually shows a six, the man reports it is a six if and only if he speaks the truth. The probability that he speaks the truth is $\frac{3}{4}$.

To find $P(R)$, the overall probability that the man reports that the outcome is a six, we can use the Law of Total Probability. The event R (reporting a six) can occur in two mutually exclusive ways:

1. The die is actually a six, and he reports it as a six (event $R \cap S$).

2. The die is not a six, and he reports it as a six (event $R \cap S'$). (This means he lies).

The events S and S' form a partition of the sample space. The Law of Total Probability for event R is:

We need to find $P(R | S')$. This is the probability that the man reports a six, given that the die does not show a six. If the die does not show a six (i.e., the outcome is 1, 2, 3, 4, or 5), the man reports it is a six if and only if he tells a lie. The probability that he tells a lie is $\frac{1}{4}$.

Substitute the probabilities $P(R | S) = \frac{3}{4}$, $P(S) = \frac{1}{6}$, $P(R | S') = \frac{1}{4}$, and $P(S') = \frac{5}{6}$ into equation (3):

Simplify the fraction:

Now, we have all the components for Bayes' Theorem: $P(R | S) = \frac{3}{4}$, $P(S) = \frac{1}{6}$, and $P(R) = \frac{1}{3}$.

Substitute these values into the formula for $P(S | R)$ (equation 1):

To divide by a fraction, multiply by its reciprocal:

The probability that the outcome is actually a six, given that the man reports it is a six, is $\frac{3}{8}$.

---

The final answer is $\mathbf{\frac{3}{8}}$.

Given:

---

Initial contents of the urn: 5 red balls, 5 black balls. Total = 10 balls.

A ball is drawn at random, its colour is noted, and it is returned to the urn (replacement).

2 additional balls of the colour drawn are added to the urn.

Then, a second ball is drawn at random.

---

To Find:

---

The probability that the second ball drawn is red.

---

Solution:

---

Let R$_1$ be the event that the first ball drawn is red.

Let B$_1$ be the event that the first ball drawn is black.

Let R$_2$ be the event that the second ball drawn is red.

Let B$_2$ be the event that the second ball drawn is black.

The event that the second ball is red ($R_2$) can occur in two mutually exclusive ways:

1. The first ball was red, and the second ball drawn is red ($R_1 \cap R_2$).

2. The first ball was black, and the second ball drawn is red ($B_1 \cap R_2$).

Using the Law of Total Probability, the probability of the second ball being red is:

First, let's find the probabilities of the first draw:

Initial number of red balls = 5

Initial number of black balls = 5

Total initial balls = 10

Now, let's consider the contents of the urn after the first draw and adding two balls:

Case 1: The first ball drawn was red ($R_1$).

The red ball is returned, and 2 additional red balls are added. The urn now contains:

Number of red balls = $5 + 2 = 7$

Number of black balls = 5

Total balls = $7 + 5 = 12$

The probability of drawing a red ball on the second draw, given the first was red ($P(R_2 | R_1)$), is:

Case 2: The first ball drawn was black ($B_1$).

The black ball is returned, and 2 additional black balls are added. The urn now contains:

Number of red balls = 5

Number of black balls = $5 + 2 = 7$

Total balls = $5 + 7 = 12$

The probability of drawing a red ball on the second draw, given the first was black ($P(R_2 | B_1)$), is:

Now, substitute the probabilities from (2), (3), (4), and (5) into the Law of Total Probability formula (1):

The probability that the second ball is red is $\frac{1}{2}$.

---

The final answer is $\mathbf{\frac{1}{2}}$.

Given:

---

Bag 1: Contains 4 red and 4 black balls. Total balls in Bag 1 = $4 + 4 = 8$.

Bag 2: Contains 2 red and 6 black balls. Total balls in Bag 2 = $2 + 6 = 8$.

One of the two bags is selected at random.

A ball is drawn from the selected bag, and it is found to be red.

---

To Find:

---

The probability that the red ball was drawn from the first bag.

---

Solution:

---

Let $B_1$ be the event that Bag 1 is selected.

Let $B_2$ be the event that Bag 2 is selected.

Since one of the two bags is selected at random, the probability of selecting each bag is equal:

Let R be the event that the drawn ball is red.

We are given that the drawn ball is red (event R has occurred), and we want to find the probability that it was drawn from Bag 1. This is the conditional probability $P(B_1 | R)$.

We can use Bayes' Theorem to find this probability. Bayes' Theorem states:

We need to find $P(R | B_1)$, $P(B_1)$, and $P(R)$.

$P(B_1)$ is given as $\frac{1}{2}$.

$P(R | B_1)$ is the probability of drawing a red ball given that Bag 1 is selected.

In Bag 1, there are 4 red balls and a total of 8 balls.

To find $P(R)$, the overall probability of drawing a red ball, we can use the Law of Total Probability. The events $B_1$ and $B_2$ form a partition of the sample space (the ball must be drawn from either Bag 1 or Bag 2, and these are mutually exclusive).

We need to find $P(R | B_2)$, which is the probability of drawing a red ball given that Bag 2 is selected.

In Bag 2, there are 2 red balls and a total of 8 balls.

Now, substitute the probabilities $P(R | B_1) = \frac{1}{2}$, $P(B_1) = \frac{1}{2}$, $P(R | B_2) = \frac{1}{4}$, and $P(B_2) = \frac{1}{2}$ into the Law of Total Probability formula for $P(R)$:

To add these fractions, find a common denominator, which is 8.

Now, we have all the components for Bayes' Theorem: $P(R | B_1) = \frac{1}{2}$, $P(B_1) = \frac{1}{2}$, and $P(R) = \frac{3}{8}$.

Substitute these values into the formula for $P(B_1 | R)$:

To divide by a fraction, multiply by its reciprocal:

Simplify the multiplication by cancelling common factors. 4 in the denominator and 8 in the numerator are both divisible by 4.

The probability that the red ball was drawn from the first bag, given that it is red, is $\frac{2}{3}$.

---

The final answer is $\mathbf{\frac{2}{3}}$.

Given:

---

Let H be the event that a student resides in hostel (hostler).

Let D be the event that a student is a day scholar (not residing in hostel).

Let A be the event that a student attains an A grade.

We are given the following probabilities:

Proportion of hostlers: $P(H) = 60\% = \frac{60}{100} = 0.60$

Proportion of day scholars: $P(D) = 40\% = \frac{40}{100} = 0.40$

Probability of attaining A grade given the student is a hostler: $P(A | H) = 30\% = \frac{30}{100} = 0.30$

Probability of attaining A grade given the student is a day scholar: $P(A | D) = 20\% = \frac{20}{100} = 0.20$

Note that H and D are mutually exclusive and exhaustive events, so they form a partition of the sample space ($P(H) + P(D) = 0.60 + 0.40 = 1$).

A student is chosen at random and has an A grade (event A has occurred).

---

To Find:

---

The probability that the student is a hostler, given that he has an A grade. This is the conditional probability $P(H | A)$.

---

Solution:

---

We are looking for the probability $P(H | A)$. We can use Bayes' Theorem for this:

To use this formula, we need to find the value of $P(A)$, the overall probability that a student attains an A grade. We can use the Law of Total Probability. Since the events H and D form a partition of the sample space, the Law of Total Probability for event A is:

Substitute the given probabilities into equation (2):

Calculate the products:

Add these values to find $P(A)$:

Now, substitute the given values $P(A | H) = 0.30$, $P(H) = 0.60$, and the calculated value $P(A) = 0.26$ into Bayes' Theorem formula (1):

From previous calculation, $(0.30)(0.60) = 0.18$.

To simplify this fraction, multiply the numerator and the denominator by 100:

Simplify the fraction $\frac{18}{26}$ by dividing the numerator and denominator by their greatest common divisor, which is 2.

The probability that the student is a hostler, given that he has an A grade, is $\frac{9}{13}$.

---

The final answer is $\mathbf{\frac{9}{13}}$.

Given:

---

Let K be the event that the student knows the answer.

Let G be the event that the student guesses the answer.

Let C be the event that the student answers the question correctly.

We are given the following probabilities:

Probability that the student knows the answer: $P(K) = \frac{3}{4}$

Probability that the student guesses the answer: $P(G) = \frac{1}{4}$

Note that K and G are mutually exclusive and exhaustive events, so they form a partition of the sample space ($P(K) + P(G) = \frac{3}{4} + \frac{1}{4} = 1$).

Probability of answering correctly given the student guesses: $P(C | G) = \frac{1}{4}$ (This implies it's a 4-option multiple-choice question, although not explicitly stated, this value suggests it).

---

To Find:

---

The probability that the student knows the answer, given that he answered it correctly. This is the conditional probability $P(K | C)$.

---

Solution:

---

We are looking for the probability $P(K | C)$. We can use Bayes' Theorem for this:

We need to find $P(C | K)$, $P(K)$, and $P(C)$.

$P(K)$ is given as $\frac{3}{4}$.

$P(C | K)$ is the probability of answering correctly given that the student knows the answer. If a student knows the answer, the probability of answering it correctly is 1 (assuming no errors in writing, etc.).

To find $P(C)$, the overall probability that the student answers correctly, we can use the Law of Total Probability. Since the events K and G form a partition of the sample space, the Law of Total Probability for event C is:

Substitute the given and derived probabilities into equation (3):

To add these fractions, find a common denominator, which is 16.

Now, we have all the components for Bayes' Theorem: $P(C | K) = 1$, $P(K) = \frac{3}{4}$, and $P(C) = \frac{13}{16}$.

Substitute these values into the formula for $P(K | C)$ (equation 1):

To divide by a fraction, multiply by its reciprocal:

Simplify the multiplication by cancelling common factors. 4 in the denominator and 16 in the numerator are both divisible by 4.

The probability that the student knows the answer, given that he answered it correctly, is $\frac{12}{13}$.

---

The final answer is $\mathbf{\frac{12}{13}}$.

Given:

---

Let D be the event that a randomly selected person actually has the disease.

Let H be the event that a randomly selected person is healthy (does not have the disease). Note that H is the complement of D, so $H = D'$.

Let P be the event that the blood test result is positive.

Let N be the event that the blood test result is negative.

We are given the following probabilities:

Probability that a person has the disease: $P(D) = 0.1\% = \frac{0.1}{100} = 0.001$

Probability that a person is healthy: $P(H) = P(D') = 1 - P(D) = 1 - 0.001 = 0.999$

Probability of a positive test given the person has the disease (True Positive Rate): $P(P | D) = 99\% = \frac{99}{100} = 0.99$

Probability of a positive test given the person is healthy (False Positive Rate): $P(P | H) = 0.5\% = \frac{0.5}{100} = 0.005$

---

To Find:

---

The probability that a person actually has the disease, given that his test result is positive. This is the conditional probability $P(D | P)$.

---

Solution:

---

We are looking for the probability $P(D | P)$. We can use Bayes' Theorem for this. Bayes' Theorem relates the conditional probability we want to find to known probabilities:

To use this formula, we need the values of $P(P | D)$, $P(D)$, and $P(P)$. We already have $P(P | D) = 0.99$ and $P(D) = 0.001$ from the given information.

We need to find $P(P)$, the overall probability of a positive test result. A positive test result can occur in two mutually exclusive ways:

1. The person has the disease and the test is positive ($P \cap D$).

2. The person does not have the disease (is healthy) and the test is positive ($P \cap H$).

Since the events D and H form a partition of the sample space (every person either has the disease or is healthy), we can use the Law of Total Probability to find $P(P)$:

Substitute the given probabilities into the Law of Total Probability formula:

Calculate the products:

$\begin{array}{cc}& & 0 & . & 0 & 0 & 5 \\ \times & & 0 & . & 9 & 9 & 9 \\ \hline && & & & 4 & 5 \\ && & & 4 & 5 & \times \\ && & 4 & 5 & \times & \times \\ \hline & 0 & . & 0 & 0 & 4 & 9 & 9 & 5 \\ \hline \end{array}$

Add these results to find $P(P)$:

$\begin{array}{ccccccc} & 0 & . & 0 & 0 & 0 & 9 & 9 & 0 \\ + & 0 & . & 0 & 0 & 4 & 9 & 9 & 5 \\ \hline & 0 & . & 0 & 0 & 5 & 9 & 8 & 5 \\ \hline \end{array}$

Now, substitute the values into Bayes' Theorem formula for $P(D | P)$:

To simplify this fraction, multiply the numerator and the denominator by 1,000,000 to remove the decimal points:

Simplify the fraction $\frac{990}{5985}$. Both numerator and denominator are divisible by 5:

Check for divisibility by 9 (sum of digits: $1+9+8=18$, $1+1+9+7=18$):

Check for common factors for 22 ($2 \times 11$) and 133 ($7 \times 19$). There are no common factors other than 1.

The probability that a person actually has the disease, given a positive test result, is $\frac{22}{133}$.

---

The final answer is $\mathbf{\frac{22}{133}}$.

Given:

---

Three coins:

• Coin 1: A two-headed coin (Head on both faces).
• Coin 2: A biased coin that comes up Heads 75% of the time.
• Coin 3: An unbiased coin that comes up Heads 50% of the time.

One of the three coins is chosen at random and tossed.

The outcome of the toss is heads.

---

To Find:

---

The probability that the coin chosen was the two-headed coin, given that the toss resulted in heads.

---

Solution:

---

Let $C_1$ be the event that the two-headed coin is chosen.

Let $C_2$ be the event that the biased coin is chosen.

Let $C_3$ be the event that the unbiased coin is chosen.

Since one of the three coins is chosen at random, the probability of choosing each coin is equal:

These events $C_1$, $C_2$, and $C_3$ form a partition of the sample space (one and only one of the coins must be chosen).

Let H be the event that the toss shows heads.

We are given that the toss resulted in heads (event H has occurred), and we want to find the probability that the chosen coin was the two-headed coin ($C_1$). This is the conditional probability $P(C_1 | H)$.

We can use Bayes' Theorem to find this probability. Bayes' Theorem states:

We need to find the values of $P(H | C_1)$, $P(C_1)$, and $P(H)$.

$P(C_1)$ is given as $\frac{1}{3}$.

$P(H | C_1)$ is the probability of getting a head given that the two-headed coin is chosen. Since the two-headed coin has heads on both faces, the probability of getting a head is 1 (certain event).

$P(H | C_2)$ is the probability of getting a head given that the biased coin is chosen. This is given as 75%.

$P(H | C_3)$ is the probability of getting a head given that the unbiased coin is chosen. For a fair coin, the probability of heads is 50%.

To find $P(H)$, the overall probability of getting a head, we use the Law of Total Probability. Since the events $C_1$, $C_2$, and $C_3$ form a partition, the probability of H is the sum of the probabilities of H occurring with each coin, weighted by the probability of choosing that coin:

Substitute the probabilities we know into this formula:

To add these fractions, find a common denominator, which is the least common multiple of 3, 12, and 6. The LCM is 12.

Simplify the fraction $\frac{9}{12}$ by dividing the numerator and denominator by their greatest common divisor, 3:

Now, we have all the components for Bayes' Theorem: $P(H | C_1) = 1$, $P(C_1) = \frac{1}{3}$, and $P(H) = \frac{3}{4}$.

Substitute these values into the formula for $P(C_1 | H)$:

To divide by a fraction, we multiply by its reciprocal:

The probability that the chosen coin was the two-headed coin, given that the toss resulted in heads, is $\frac{4}{9}$.

---

The final answer is $\mathbf{\frac{4}{9}}$.

Given:

---

Number of insured scooter drivers = 2000

Number of insured car drivers = 4000

Number of insured truck drivers = 6000

Total number of insured persons = $2000 + 4000 + 6000 = 12000$.

Let S be the event that the insured person is a scooter driver.

Let C be the event that the insured person is a car driver.

Let T be the event that the insured person is a truck driver.

The probabilities of selecting each type of driver at random are:

These events S, C, and T form a partition of the sample space.

Let A be the event that the insured person meets with an accident.

We are given the conditional probabilities of having an accident for each type of driver:

An insured person meets with an accident (event A has occurred).

---

To Find:

---

The probability that the person is a scooter driver, given that he met with an accident. This is the conditional probability $P(S | A)$.

---

Solution:

---

We are looking for the probability $P(S | A)$. We can use Bayes' Theorem for this:

To use this formula, we need the values of $P(A | S)$, $P(S)$, and $P(A)$. We already have $P(A | S) = 0.01$ and $P(S) = \frac{1}{6}$ from the given information.

We need to find $P(A)$, the overall probability that an insured person meets with an accident. We can use the Law of Total Probability. Since the events S, C, and T form a partition of the sample space, the Law of Total Probability for event A is:

Substitute the given and calculated probabilities into equation (2):

Convert decimals to fractions:

Find a common denominator for 600, 300, and 200. The least common multiple is 600.

Simplify the fraction $\frac{52}{600}$ by dividing the numerator and denominator by their greatest common divisor, which is 4.

Now, substitute the value $P(A | S) = 0.01 = \frac{1}{100}$, $P(S) = \frac{1}{6}$, and the calculated value $P(A) = \frac{13}{150}$ into Bayes' Theorem formula (1):

To divide by a fraction, multiply by its reciprocal:

Simplify the multiplication by cancelling common factors. 150 in the numerator and 600 in the denominator are both divisible by 150 ($\frac{150}{150} = 1$, $\frac{600}{150} = 4$).

The probability that the insured person is a scooter driver, given that they met with an accident, is $\frac{1}{52}$.

---

The final answer is $\mathbf{\frac{1}{52}}$.

Given:

---

Let $M_A$ be the event that an item is produced by machine A.

Let $M_B$ be the event that an item is produced by machine B.

We are given the proportion of items produced by each machine:

$P(M_A) = 60\% = \frac{60}{100} = 0.60$

$P(M_B) = 40\% = \frac{40}{100} = 0.40$

Check if these probabilities sum to 1: $0.60 + 0.40 = 1.00$. The events $M_A$ and $M_B$ form a partition of the sample space.

Let D be the event that an item is defective.

We are given the percentage of defective items from each machine's output:

Probability of a defective item given it's from machine A: $P(D | M_A) = 2\% = \frac{2}{100} = 0.02$

Probability of a defective item given it's from machine B: $P(D | M_B) = 1\% = \frac{1}{100} = 0.01$

An item is chosen at random from the stockpile and is found to be defective (event D has occurred).

---

To Find:

---

The probability that the defective item was produced by machine B. This is the conditional probability $P(M_B | D)$.

---

Solution:

---

We are looking for the probability $P(M_B | D)$. We can use Bayes' Theorem for this:

To use this formula, we need the values of $P(D | M_B)$, $P(M_B)$, and $P(D)$. We already have $P(D | M_B) = 0.01$ and $P(M_B) = 0.40$ from the given information.

We need to find $P(D)$, the overall probability of choosing a defective item from the stockpile. We can use the Law of Total Probability. Since the events $M_A$ and $M_B$ form a partition of the sample space, the Law of Total Probability for event D is:

Substitute the given probabilities into equation (2):

Calculate the products:

Add these values to find $P(D)$:

Now, substitute the given value $P(D | M_B) = 0.01$, $P(M_B) = 0.40$, and the calculated value $P(D) = 0.0160$ into Bayes' Theorem formula (1):

From previous calculation, $(0.01)(0.40) = 0.0040$.

To simplify this fraction, multiply the numerator and the denominator by 10000 to remove the decimal points:

Simplify the fraction $\frac{40}{160}$ by dividing the numerator and denominator by their greatest common divisor, which is 40.

The probability that the defective item was produced by machine B, given that it is defective, is $\frac{1}{4}$.

---

The final answer is $\mathbf{\frac{1}{4}}$.

Given:

---

Let $G_1$ be the event that the first group wins.

Let $G_2$ be the event that the second group wins.

We are given the probabilities that each group will win:

$P(G_1) = 0.6$

$P(G_2) = 0.4$

Check if these probabilities sum to 1: $0.6 + 0.4 = 1.0$. The events $G_1$ and $G_2$ form a partition of the sample space.

Let N be the event that a new product is introduced.

We are given the conditional probabilities of introducing a new product based on which group wins:

Probability of introducing a new product if the first group wins: $P(N | G_1) = 0.7$

Probability of introducing a new product if the second group wins: $P(N | G_2) = 0.3$

Assume that a new product is introduced (event N has occurred). The question asks for the probability that it was introduced by the second group.

---

To Find:

---

The probability that the new product introduced was by the second group, given that a new product was introduced. This is the conditional probability $P(G_2 | N)$.

---

Solution:

---

We are looking for the probability $P(G_2 | N)$. We can use Bayes' Theorem for this:

To use this formula, we need the values of $P(N | G_2)$, $P(G_2)$, and $P(N)$. We already have $P(N | G_2) = 0.3$ and $P(G_2) = 0.4$ from the given information.

We need to find $P(N)$, the overall probability that a new product is introduced. We can use the Law of Total Probability. Since the events $G_1$ and $G_2$ form a partition of the sample space, the Law of Total Probability for event N is:

Substitute the given probabilities into equation (2):

Calculate the products:

Add these values to find $P(N)$:

Now, substitute the given value $P(N | G_2) = 0.3$, $P(G_2) = 0.4$, and the calculated value $P(N) = 0.54$ into Bayes' Theorem formula (1):

From previous calculation, $(0.3)(0.4) = 0.12$.

To simplify this fraction, multiply the numerator and the denominator by 100 to remove the decimal points:

Simplify the fraction $\frac{12}{54}$ by dividing the numerator and denominator by their greatest common divisor, which is 6.

The probability that the new product introduced was by the second group, given that a new product was introduced, is $\frac{2}{9}$.

---

The final answer is $\mathbf{\frac{2}{9}}$.

Given:

---

A girl throws a die.

If the outcome is 5 or 6, she tosses a coin three times.

If the outcome is 1, 2, 3, or 4, she tosses a coin once.

She obtained exactly one head.

---

To Find:

---

The probability that she threw 1, 2, 3, or 4 with the die, given that she obtained exactly one head.

---

Solution:

---

Let $D_1$ be the event that the die shows 5 or 6.

Let $D_2$ be the event that the die shows 1, 2, 3, or 4.

Let H be the event that she obtains exactly one head in the subsequent coin toss(es).

When a fair die is thrown, the possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.

Note that $D_1$ and $D_2$ form a partition of the sample space of the die roll ($P(D_1) + P(D_2) = \frac{1}{3} + \frac{2}{3} = 1$).

We need to find the conditional probability of obtaining exactly one head given each of the die outcomes.

If the die shows 5 or 6 (event $D_1$), she tosses a coin three times.

The sample space for three coin tosses is $\{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$. The total number of outcomes is 8.

The outcomes with exactly one head are $\{HTT, THT, TTH\}$. The number of favorable outcomes is 3.

If the die shows 1, 2, 3, or 4 (event $D_2$), she tosses a coin once.

The sample space for one coin toss is $\{H, T\}$. The total number of outcomes is 2.

The outcome with exactly one head is $\{H\}$. The number of favorable outcomes is 1.

We are given that she obtained exactly one head (event H occurred), and we want to find the probability that she threw 1, 2, 3, or 4 (event $D_2$). This is the conditional probability $P(D_2 | H)$.

We can use Bayes' Theorem to find this probability:

To use this formula, we need the values of $P(H | D_2)$, $P(D_2)$, and $P(H)$. We have $P(H | D_2) = \frac{1}{2}$ from (4) and $P(D_2) = \frac{2}{3}$ from (2).

We need to find $P(H)$, the overall probability of obtaining exactly one head. We can use the Law of Total Probability. Since the events $D_1$ and $D_2$ form a partition of the sample space, the Law of Total Probability for event H is:

Substitute the probabilities from (1), (2), (3), and (4) into the Law of Total Probability formula for $P(H)$:

Simplify the first fraction and find a common denominator (24):

Now, substitute the values $P(H | D_2) = \frac{1}{2}$, $P(D_2) = \frac{2}{3}$, and the calculated value $P(H) = \frac{11}{24}$ into Bayes' Theorem formula for $P(D_2 | H)$:

Calculate the numerator:

Substitute the numerator back into the Bayes' formula:

To divide by a fraction, multiply by its reciprocal:

Simplify the multiplication by cancelling common factors. 3 in the denominator and 24 in the numerator are both divisible by 3 ($\frac{3}{3}=1$, $\frac{24}{3}=8$).

The probability that she threw 1, 2, 3, or 4 with the die, given that she obtained exactly one head, is $\frac{8}{11}$.

---

The final answer is $\mathbf{\frac{8}{11}}$.

Given:

---

Let $M_A$ be the event that an item is produced by machine A.

Let $M_B$ be the event that an item is produced by machine B.

Let $M_C$ be the event that an item is produced by machine C.

We are given the proportion of time each machine is on the job, which represents the probability that a randomly selected item was produced by that machine:

$P(M_A) = 50\% = \frac{50}{100} = 0.50$

$P(M_B) = 30\% = \frac{30}{100} = 0.30$

$P(M_C) = 20\% = \frac{20}{100} = 0.20$

The sum of these probabilities is $P(M_A) + P(M_B) + P(M_C) = 0.50 + 0.30 + 0.20 = 1.00$. The events $M_A$, $M_B$, and $M_C$ form a partition of the sample space (every item is produced by exactly one of these machines).

Let D be the event that an item is defective.

We are given the conditional probabilities of producing a defective item for each machine:

Probability of a defective item given it's from machine A: $P(D | M_A) = 1\% = \frac{1}{100} = 0.01$

Probability of a defective item given it's from machine B: $P(D | M_B) = 5\% = \frac{5}{100} = 0.05$

Probability of a defective item given it's from machine C: $P(D | M_C) = 7\% = \frac{7}{100} = 0.07$

A defective item is found (event D has occurred).

---

To Find:

---

The probability that the defective item was produced by machine A. This is the conditional probability $P(M_A | D)$.

---

Solution:

---

We are looking for the probability $P(M_A | D)$. We can use Bayes' Theorem for this. Bayes' Theorem states:

To use this formula, we need the values of $P(D | M_A)$, $P(M_A)$, and $P(D)$. We already have $P(D | M_A) = 0.01$ and $P(M_A) = 0.50$ from the given information.

We need to find $P(D)$, the overall probability of producing a defective item. We can use the Law of Total Probability. Since the events $M_A$, $M_B$, and $M_C$ form a partition of the sample space, the Law of Total Probability for event D is:

Substitute the given probabilities into the Law of Total Probability formula:

Calculate the products:

Add these values to find $P(D)$:

Now, substitute the given value $P(D | M_A) = 0.01$, $P(M_A) = 0.50$, and the calculated value $P(D) = 0.0340$ into Bayes' Theorem formula for $P(M_A | D)$:

Calculate the numerator:

Substitute the numerator back into the Bayes' formula:

To simplify this fraction, multiply the numerator and the denominator by 10000 to remove the decimal points:

Simplify the fraction $\frac{50}{340}$ by dividing the numerator and denominator by their greatest common divisor, which is 10.

The probability that the defective item was produced by machine A, given that it is defective, is $\frac{5}{34}$.

---

The final answer is $\mathbf{\frac{5}{34}}$.

Given:

---

Total number of cards in a well-shuffled pack = 52.

Number of diamonds = 13.

Number of non-diamonds = $52 - 13 = 39$.

One card is lost from the pack.

From the remaining 51 cards, two cards are drawn, and both are found to be diamonds.

---

To Find:

---

The probability that the lost card was a diamond, given that the two cards drawn from the remaining pack are both diamonds.

---

Solution:

---

Let $L_D$ be the event that the lost card is a diamond.

Let $L_N$ be the event that the lost card is not a diamond.

The probabilities of the lost card being a diamond or not are:

The events $L_D$ and $L_N$ form a partition of the sample space for the lost card (the lost card is either a diamond or not a diamond).

Let A be the event that two cards drawn from the remaining 51 cards are both diamonds.

We are given that event A has occurred, and we want to find the probability $P(L_D | A)$.

Using Bayes' Theorem, we have:

We need to calculate $P(A | L_D)$ and $P(A)$.

First, consider $P(A | L_D)$: the probability of drawing two diamonds from the remaining 51 cards, given that the lost card was a diamond.

If the lost card was a diamond, the remaining 51 cards contain $13 - 1 = 12$ diamonds and 39 non-diamonds.

The number of ways to choose 2 cards from the remaining 51 is $\binom{51}{2}$.

The number of ways to choose 2 diamonds from the 12 diamonds remaining is $\binom{12}{2}$.

Next, consider $P(A | L_N)$: the probability of drawing two diamonds from the remaining 51 cards, given that the lost card was not a diamond.

If the lost card was not a diamond, the remaining 51 cards contain 13 diamonds and $39 - 1 = 38$ non-diamonds.

The number of ways to choose 2 cards from the remaining 51 is $\binom{51}{2} = 1275$.

The number of ways to choose 2 diamonds from the 13 diamonds remaining is $\binom{13}{2}$.

Now, we find the overall probability $P(A)$ using the Law of Total Probability. Since $L_D$ and $L_N$ form a partition:

Substitute the probabilities $P(A | L_D) = \frac{66}{1275}$, $P(L_D) = \frac{1}{4}$, $P(A | L_N) = \frac{78}{1275}$, and $P(L_N) = \frac{3}{4}$:

Simplify the fraction $\frac{300}{5100}$:

Now, substitute the values $P(A | L_D) = \frac{66}{1275}$, $P(L_D) = \frac{1}{4}$, and $P(A) = \frac{1}{17}$ into Bayes' Theorem formula for $P(L_D | A)$:

To divide by a fraction, multiply by its reciprocal:

Simplify by cancelling common factors. 5100 is $17 \times 300$.

Simplify the fraction $\frac{66}{300}$ by dividing the numerator and denominator by their greatest common divisor, which is 6.

The probability that the lost card was a diamond, given that two diamonds were drawn from the remaining cards, is $\frac{11}{50}$.

---

The final answer is $\mathbf{\frac{11}{50}}$.

Given:

---

Let T be the event that A speaks the truth. $P(T) = \frac{4}{5}$.

Let L be the event that A tells a lie. $P(L) = 1 - P(T) = 1 - \frac{4}{5} = \frac{1}{5}$.

A fair coin is tossed. Let H be the event that a head appears, and T' be the event that a tail appears.

$P(H) = \frac{1}{2}$.

$P(T') = \frac{1}{2}$.

Let R be the event that A reports that a head appears.

---

To Find:

---

The probability that actually there was a head, given that A reports a head. This is the conditional probability $P(H | R)$.

---

Solution:

---

We want to find $P(H | R)$. We can use Bayes' Theorem for this. Bayes' Theorem states:

We need to find $P(R | H)$, $P(H)$, and $P(R)$.

$P(H)$ is given as $\frac{1}{2}$.

$P(R | H)$ is the probability that A reports a head, given that it was actually a head. If the coin shows a head, A reports a head if and only if A speaks the truth. So, $P(R | H) = P(T) = \frac{4}{5}$.

To find $P(R)$, the overall probability that A reports a head, we can use the Law of Total Probability. The event R (reporting a head) can occur in two mutually exclusive ways, based on the actual outcome of the coin toss (Head or Tail):

1. The coin was Head, and A reports Head ($R \cap H$).

2. The coin was Tail, and A reports Head ($R \cap T'$). (This implies A lies).

The events H and T' form a partition of the sample space for the coin toss. The Law of Total Probability for event R is:

We need to find $P(R | T')$. This is the probability that A reports a head, given that it was actually a tail. If the coin shows a tail, A reports a head if and only if A tells a lie. So, $P(R | T') = P(L) = \frac{1}{5}$.

Substitute the probabilities into equation (2): $P(R | H) = \frac{4}{5}$, $P(H) = \frac{1}{2}$, $P(R | T') = \frac{1}{5}$, and $P(T') = \frac{1}{2}$.

Now, substitute the values $P(R | H) = \frac{4}{5}$, $P(H) = \frac{1}{2}$, and the calculated value $P(R) = \frac{1}{2}$ into Bayes' Theorem formula (1):

To divide by a fraction, multiply by its reciprocal:

The probability that there was actually a head, given that A reports a head, is $\frac{4}{5}$.

---

Compare the result with the given options:

(A) $\frac{4}{5}$

(B) $\frac{1}{2}$

(C) $\frac{1}{5}$

(D) $\frac{2}{5}$

The result matches option (A).

---

The final answer is $\mathbf{\frac{4}{5}}$.

Given:

---

A and B are two events such that $A \subset B$ and $P(B) \neq 0$.

---

To Determine:

---

Which of the given statements is correct.

---

Solution:

---

By the definition of conditional probability, for events A and B where $P(B) \neq 0$, the probability of event A given event B is:

We are given that $A \subset B$. This means that every element in set A is also in set B. Consequently, the intersection of events A and B is simply event A itself.

Taking the probability of both sides, we get:

Substitute the result from equation (2) into the conditional probability formula (1):

Now let's evaluate the given options:

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

From (3), we have $P(A|B) = \frac{P(A)}{P(B)}$. This is generally not equal to $\frac{P(B)}{P(A)}$, unless $P(A) = P(B)$ (and they are not zero), which is a specific case of $A \subset B$ (implying $P(B \setminus A) = 0$), not a general rule.

(B) $P(A|B) < P(A)$

Substitute $P(A|B) = \frac{P(A)}{P(B)}$ into the inequality: $\frac{P(A)}{P(B)} < P(A)$.

If $P(A) > 0$, we can divide both sides by $P(A)$: $\frac{1}{P(B)} < 1$. This implies $1 < P(B)$. However, the probability of any event B cannot be greater than 1 ($P(B) \le 1$). If $P(A) = 0$, the inequality $0 < 0$ is false. Thus, this option is generally incorrect.

(C) $P(A|B) \ge P(A)$

Substitute $P(A|B) = \frac{P(A)}{P(B)}$ into the inequality: $\frac{P(A)}{P(B)} \ge P(A)$.

Let's analyze this inequality:

We know the following:

• $P(A) \ge 0$ (Probability is non-negative).
• $P(B) > 0$ (Given).
• Since B is an event in a probability space, $P(B) \le 1$. This implies $1 - P(B) \ge 0$.

So, the term $\frac{1 - P(B)}{P(B)}$ is non-negative (since $1 - P(B) \ge 0$ and $P(B) > 0$).

The product of a non-negative number ($P(A)$) and a non-negative number ($\frac{1 - P(B)}{P(B)}$) must be non-negative.

This confirms that $\frac{P(A)}{P(B)} \ge P(A)$, which means $P(A|B) \ge P(A)$. This inequality holds true whenever $A \subset B$ and $P(B) \neq 0$.

(D) None of these

Since option (C) is correct, this option is incorrect.

---

The condition $A \subset B$ implies that the occurrence of A guarantees the occurrence of B. When we consider the probability of A given B has occurred, we are looking at the probability of A within the reduced sample space B. Since A is entirely contained within B, the probability of A relative to B ($P(A|B)$) is $P(A)/P(B)$. Because $P(B) \le 1$ (and $P(B) > 0$), dividing $P(A)$ by $P(B)$ results in a value greater than or equal to $P(A)$. The equality holds if $P(B) = 1$ or $P(A) = 0$. If $P(A)>0$ and $P(B)<1$, then $P(A|B) > P(A)$.

---

The final answer is $\mathbf{P(A|B) \ge P(A)}$.

Given:

The distribution of coloured balls in four boxes is provided in the table below. A box is selected at random, and then a ball is randomly drawn from the selected box.

Box	Colour
	Black	White	Red	Blue
I	3	4	5	6
II	2	2	2	2
III	1	2	3	1
IV	4	3	1	5

---

To Find:

The probability that the ball drawn is from Box III, given that the ball is black.

---

Solution:

Let $B_i$ denote the event that Box $i$ is selected, for $i = 1, 2, 3, 4$.

Let $A$ denote the event that a black ball is drawn.

Since a box is selected at random, the probability of selecting any box is equal.

$$P(B_1) = P(B_2) = P(B_3) = P(B_4) = \frac{1}{4}$$

Now, we calculate the probability of drawing a black ball given that a specific box is selected, i.e., $P(A | B_i)$.

For Box I:

Total number of balls in Box I = $3 + 4 + 5 + 6 = 18$

Number of black balls in Box I = $3$

$$P(A | B_1) = \frac{3}{18} = \frac{1}{6}$$

For Box II:

Total number of balls in Box II = $2 + 2 + 2 + 2 = 8$

Number of black balls in Box II = $2$

$$P(A | B_2) = \frac{2}{8} = \frac{1}{4}$$

For Box III:

Total number of balls in Box III = $1 + 2 + 3 + 1 = 7$

Number of black balls in Box III = $1$

$$P(A | B_3) = \frac{1}{7}$$

For Box IV:

Total number of balls in Box IV = $4 + 3 + 1 + 5 = 13$

Number of black balls in Box IV = $4$

$$P(A | B_4) = \frac{4}{13}$$

We want to find the probability that the ball is from Box III given that it is black, i.e., $P(B_3 | A)$. We use Bayes' Theorem:

To find $P(A)$, the total probability of drawing a black ball, we use the Law of Total Probability:

Substituting the probabilities we found:

$$P(A) = P(A | B_1)P(B_1) + P(A | B_2)P(B_2) + P(A | B_3)P(B_3) + P(A | B_4)P(B_4)$$

$$P(A) = \left(\frac{1}{6}\right)\left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)\left(\frac{1}{4}\right) + \left(\frac{1}{7}\right)\left(\frac{1}{4}\right) + \left(\frac{4}{13}\right)\left(\frac{1}{4}\right)$$

Factor out the common term $\frac{1}{4}$:

$$P(A) = \frac{1}{4} \left(\frac{1}{6} + \frac{1}{4} + \frac{1}{7} + \frac{4}{13}\right)$$

To sum the fractions inside the parenthesis, we find a common denominator, which is $LCM(6, 4, 7, 13) = 1092$.

$$\frac{1}{6} = \frac{1 \times 182}{6 \times 182} = \frac{182}{1092}$$

$$\frac{1}{4} = \frac{1 \times 273}{4 \times 273} = \frac{273}{1092}$$

$$\frac{1}{7} = \frac{1 \times 156}{7 \times 156} = \frac{156}{1092}$$

$$\frac{4}{13} = \frac{4 \times 84}{13 \times 84} = \frac{336}{1092}$$

Sum of fractions = $\frac{182 + 273 + 156 + 336}{1092} = \frac{947}{1092}$

Now, substitute this back into the expression for $P(A)$:

$$P(A) = \frac{1}{4} \times \frac{947}{1092} = \frac{947}{4368}$$

Finally, substitute the values of $P(A | B_3)$, $P(B_3)$, and $P(A)$ into Bayes' Theorem (equation i):

$$P(B_3 | A) = \frac{\left(\frac{1}{7}\right) \left(\frac{1}{4}\right)}{\frac{947}{4368}}$$

$$P(B_3 | A) = \frac{\frac{1}{28}}{\frac{947}{4368}}$$

$$P(B_3 | A) = \frac{1}{28} \times \frac{4368}{947}$$

Simplify the fraction $\frac{4368}{28}$:

$$\frac{4368}{28} = 156$$

Thus,

$$P(B_3 | A) = \frac{156}{947}$$

---

The probability that the ball drawn is from Box III, given that it is black, is $\frac{156}{947}$.

Given:

A and B throw a fair die alternatively until one gets a '6'. The one who gets a '6' first wins. A starts the game.

---

To Find:

The respective probabilities of A winning and B winning.

---

Solution:

Let $S$ be the event of getting a '6' in a single throw of a die.

Let $F$ be the event of not getting a '6' in a single throw.

The probability of getting a '6' is $p = P(S)$. Since the die is fair, there is one favorable outcome (6) out of six possible outcomes (1, 2, 3, 4, 5, 6).

$$p = \frac{1}{6}$$

The probability of not getting a '6' is $q = P(F)$.

$$q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$$

A starts the game. A can win on their first throw, or on their second throw (which is the 3rd throw of the game), or on their third throw (which is the 5th throw of the game), and so on.

A wins on the 1st throw if A gets a '6'. The probability is $p$.

A wins on the 3rd throw if A fails, B fails, and then A gets a '6'. The probability is $q \times q \times p = q^2 p$.

A wins on the 5th throw if A fails, B fails, A fails, B fails, and then A gets a '6'. The probability is $q \times q \times q \times q \times p = q^4 p$.

This pattern continues. The probability of A winning is the sum of the probabilities of A winning on any of their turns:

$$P(\text{A wins}) = p + q^2 p + q^4 p + q^6 p + ...$$

This is an infinite geometric series with first term $a = p$ and common ratio $r = q^2$. Since $|r| = \left(\frac{5}{6}\right)^2 = \frac{25}{36} < 1$, the series converges. The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.

Substitute the values $p = \frac{1}{6}$ and $q = \frac{5}{6}$ into equation (i):

$$P(\text{A wins}) = \frac{\frac{1}{6}}{1 - \left(\frac{5}{6}\right)^2} = \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{\frac{1}{6}}{\frac{36 - 25}{36}} = \frac{\frac{1}{6}}{\frac{11}{36}}$$

$$P(\text{A wins}) = \frac{1}{6} \times \frac{36}{11} = \frac{36}{6 \times 11} = \frac{6}{11}$$

Now, consider the probability of B winning. B can win on their first throw (which is the 2nd throw of the game), or on their second throw (which is the 4th throw of the game), and so on.

B wins on the 2nd throw (B's 1st) if A fails and B gets a '6'. The probability is $q \times p = qp$.

B wins on the 4th throw (B's 2nd) if A fails, B fails, A fails, and then B gets a '6'. The probability is $q \times q \times q \times p = q^3 p$.

B wins on the 6th throw (B's 3rd) if A fails, B fails, A fails, B fails, A fails, and then B gets a '6'. The probability is $q \times q \times q \times q \times q \times p = q^5 p$.

This pattern continues. The probability of B winning is the sum of the probabilities of B winning on any of their turns:

$$P(\text{B wins}) = q p + q^3 p + q^5 p + q^7 p + ...$$

This is an infinite geometric series with first term $a = qp$ and common ratio $r = q^2$. The sum is given by $S = \frac{a}{1-r}$.

Substitute the values $p = \frac{1}{6}$ and $q = \frac{5}{6}$ into equation (ii):

$$P(\text{B wins}) = \frac{\left(\frac{5}{6}\right) \left(\frac{1}{6}\right)}{1 - \left(\frac{5}{6}\right)^2} = \frac{\frac{5}{36}}{1 - \frac{25}{36}} = \frac{\frac{5}{36}}{\frac{11}{36}}$$

$$P(\text{B wins}) = \frac{5}{36} \times \frac{36}{11} = \frac{5}{11}$$

---

Alternatively:

Since the game continues until one person wins, and the probability of success $p > 0$, the game is certain to end. Thus, either A wins or B wins.

Using the probability of A winning calculated previously, we can find the probability of B winning:

$$P(\text{B wins}) = 1 - P(\text{A wins}) = 1 - \frac{6}{11} = \frac{11}{11} - \frac{6}{11} = \frac{11 - 6}{11} = \frac{5}{11}$$

This confirms the result obtained using the geometric series for B's probability.

---

The probability that A wins is $\frac{6}{11}$.

The probability that B wins is $\frac{5}{11}$.

Given:

Probability of producing an acceptable item when the machine is correctly set up = $90\% = 0.90$.

Probability of producing an acceptable item when the machine is incorrectly set up = $40\% = 0.40$.

Probability that the machine is correctly set up (based on past experience) = $80\% = 0.80$.

Observation: The machine produces 2 acceptable items.

---

To Find:

The probability that the machine is correctly set up, given that it produced 2 acceptable items.

---

Solution:

Let $C$ be the event that the machine is correctly set up.

Let $I$ be the event that the machine is incorrectly set up.

Let $A$ be the event that the machine produces 2 acceptable items.

We are given the prior probabilities:

Since the machine is either correctly or incorrectly set up, $I$ is the complement of $C$.

We are given the conditional probabilities of producing a single acceptable item:

$$P(\text{acceptable item} | C) = 0.90$$

$$P(\text{acceptable item} | I) = 0.40$$

The event $A$ is producing 2 acceptable items. Assuming the outcome of producing each item is independent given the setup state, the probability of producing 2 consecutive acceptable items is the product of the probabilities of producing acceptable items for each item.

Probability of producing 2 acceptable items given correctly set up:

$$P(A | C) = (0.90) \times (0.90) = 0.81$$

Probability of producing 2 acceptable items given incorrectly set up:

$$P(A | I) = (0.40) \times (0.40) = 0.16$$

We want to find the posterior probability $P(C | A)$, which is the probability that the machine is correctly set up given that 2 acceptable items were produced. We can use Bayes' Theorem:

To find $P(A)$, the total probability of producing 2 acceptable items, we use the Law of Total Probability:

Substitute the known values into equation (ii):

$$P(A) = (0.81)(0.80) + (0.16)(0.20)$$

$$P(A) = 0.648 + 0.032$$

$$P(A) = 0.680$$

Now, substitute the values of $P(A | C)$, $P(C)$, and $P(A)$ into Bayes' Theorem (equation i):

$$P(C | A) = \frac{(0.81)(0.80)}{0.680}$$

$$P(C | A) = \frac{0.648}{0.680}$$

To simplify the fraction, we can multiply the numerator and denominator by 1000 to remove decimals:

$$P(C | A) = \frac{648}{680}$$

We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor. Both are divisible by 8:

$$648 \div 8 = 81$$

$$680 \div 8 = 85$$

$$P(C | A) = \frac{81}{85}$$

---

The probability that the machine is correctly set up, given that it produced 2 acceptable items, is $\frac{81}{85}$.

Given:

A and B are two events such that $P(A) \neq 0$.

---

To Find:

$P(B|A)$ under the conditions:

(i) A is a subset of B

(ii) A $\cap$ B = $\phi$ (A and B are disjoint events)

---

Solution:

The formula for the conditional probability of event B occurring given that event A has occurred is:

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

This formula is valid since it is given that $P(A) \neq 0$.

(i) A is a subset of B ($A \subset B$)

If A is a subset of B, it means that whenever event A occurs, event B must also occur. In terms of set theory, the intersection of A and B is equal to A.

Therefore, the probability of the intersection is equal to the probability of A.

Substitute this into the formula for $P(B|A)$:

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

Since $P(A) \neq 0$, we can simplify this expression:

$$P(B|A) = 1$$

(ii) A $\cap$ B = $\phi$ (A and B are disjoint events)

If A and B are disjoint events, their intersection is the empty set ($\phi$). This means that events A and B cannot occur at the same time.

The probability of the empty set is 0.

Substitute this into the formula for $P(B|A)$:

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

Since $P(A) \neq 0$, the fraction is equal to 0.

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

---

Final Answer:

(i) If A is a subset of B, $P(B|A) = 1$.

(ii) If A $\cap$ B = $\phi$, $P(B|A) = 0$.

Given:

A couple has two children.

We assume that the gender of each child is independent and the probability of having a male (M) child is equal to the probability of having a female (F) child, i.e., $P(M) = P(F) = \frac{1}{2}$.

The sample space for the genders of two children, in order of birth (Elder, Younger), is:

$$S = \{MM, MF, FM, FF\}$$

Each outcome in the sample space is equally likely, with a probability of $\frac{1}{4}$.

$$P(MM) = \frac{1}{4}$$

$$P(MF) = \frac{1}{4}$$

$$P(FM) = \frac{1}{4}$$

$$P(FF) = \frac{1}{4}$$

---

To Find:

(i) The probability that both children are males, given that at least one child is male.

(ii) The probability that both children are females, given that the elder child is a female.

---

Solution:

(i) Probability that both children are males, if at least one is male.

Let $A$ be the event that at least one child is male.

The outcomes where at least one child is male are: $\{MM, MF, FM\}$.

$$A = \{MM, MF, FM\}$$

$$P(A) = P(MM) + P(MF) + P(FM) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$$

Let $B$ be the event that both children are males.

The outcome where both children are males is: $\{MM\}$.

$$B = \{MM\}$$

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

We need to find the probability of event B occurring given that event A has occurred, i.e., $P(B|A)$.

The intersection of events A and B, $A \cap B$, contains the outcomes that are in both A and B.

$$A \cap B = \{MM, MF, FM\} \cap \{MM\} = \{MM\}$$

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

Using the formula for conditional probability:

Substitute the values of $P(A \cap B)$ and $P(A)$ into equation (i):

$$P(\text{both males} | \text{at least one male}) = \frac{\frac{1}{4}}{\frac{3}{4}}$$

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

---

(ii) Probability that both children are females, if the elder child is a female.

Let $C$ be the event that the elder child is a female.

The outcomes where the elder child is female are: $\{FM, FF\}$.

$$C = \{FM, FF\}$$

$$P(C) = P(FM) + P(FF) = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$$

Let $D$ be the event that both children are females.

The outcome where both children are females is: $\{FF\}$.

$$D = \{FF\}$$

$$P(D) = P(FF) = \frac{1}{4}$$

We need to find the probability of event D occurring given that event C has occurred, i.e., $P(D|C)$.

The intersection of events C and D, $C \cap D$, contains the outcomes that are in both C and D.

$$C \cap D = \{FM, FF\} \cap \{FF\} = \{FF\}$$

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

Using the formula for conditional probability:

Substitute the values of $P(C \cap D)$ and $P(C)$ into equation (ii):

$$P(\text{both females} | \text{elder child is female}) = \frac{\frac{1}{4}}{\frac{1}{2}}$$

$$P(D|C) = \frac{1}{4} \times \frac{2}{1} = \frac{2}{4} = \frac{1}{2}$$

---

Final Answer:

(i) The probability that both children are males, given that at least one of the children is male, is $\frac{1}{3}$.

(ii) The probability that both children are females, given that the elder child is a female, is $\frac{1}{2}$.

Given:

Let $M$ be the event that the selected person is male.

Let $F$ be the event that the selected person is female.

Let $G$ be the event that the selected person has grey hair.

We are given the following probabilities:

The probability that a man has grey hair is $P(G | M) = 5\% = 0.05$.

The probability that a woman has grey hair is $P(G | F) = 0.25\% = 0.0025$.

It is stated that there are equal numbers of males and females in the population. Therefore, the prior probabilities of selecting a male or a female are:

$$P(M) = 0.5$$

$$P(F) = 0.5$$

---

To Find:

The probability that a randomly selected grey-haired person is male, i.e., $P(M | G)$.

---

Solution:

We can use Bayes' Theorem to find the required probability $P(M | G)$. Bayes' Theorem is given by:

$$P(M | G) = \frac{P(G | M) P(M)}{P(G)}$$

To use this formula, we first need to find the total probability of a randomly selected person having grey hair, $P(G)$. We can find this using the Law of Total Probability. Since a person is either male or female (and these events are mutually exclusive and exhaustive), we have:

$$P(G) = P(G \cap M) + P(G \cap F)$$

Using the definition of conditional probability, $P(G \cap M) = P(G | M) P(M)$ and $P(G \cap F) = P(G | F) P(F)$.

$$P(G) = P(G | M) P(M) + P(G | F) P(F)$$

Substitute the given probability values into this equation:

$$P(G) = (0.05)(0.5) + (0.0025)(0.5)$$

$$P(G) = 0.025 + 0.00125$$

$$P(G) = 0.02625$$

Now, substitute the values of $P(G | M)$, $P(M)$, and $P(G)$ into Bayes' Theorem to find $P(M | G)$:

$$P(M | G) = \frac{(0.05)(0.5)}{0.02625}$$

$$P(M | G) = \frac{0.025}{0.02625}$$

To simplify the fraction, we can eliminate the decimals by multiplying the numerator and denominator by 10000:

$$P(M | G) = \frac{0.025 \times 10000}{0.02625 \times 10000} = \frac{250}{262.5} \times \frac{10}{10} = \frac{2500}{2625}$$

Now, simplify the fraction $\frac{2500}{2625}$ by dividing both the numerator and the denominator by their greatest common divisor. Both are divisible by 25:

$$2500 \div 25 = 100$$

$$2625 \div 25 = 105$$

So, $P(M | G) = \frac{100}{105}$.

This fraction can be simplified further by dividing both numerator and denominator by 5:

$$100 \div 5 = 20$$

$$105 \div 5 = 21$$

Thus,

$$P(M | G) = \frac{20}{21}$$

---

The probability that the grey haired person selected at random is male is $\frac{20}{21}$.

Given:

Percentage of people who are right-handed = $90\%$

Size of the random sample = $10$ people

---

To Find:

The probability that at most 6 of the 10 people in the sample are right-handed.

---

Solution:

This problem can be modeled using a binomial distribution because we have a fixed number of independent trials (selecting 10 people), each trial has two possible outcomes (right-handed or not right-handed), and the probability of success (being right-handed) is constant for each trial.

Let $n$ be the number of people in the sample, so $n = 10$.

Let $p$ be the probability that a randomly selected person is right-handed, so $p = 90\% = 0.90$.

Let $q$ be the probability that a randomly selected person is not right-handed, so $q = 1 - p = 1 - 0.90 = 0.10$.

Let $X$ be the random variable representing the number of right-handed people in the sample of 10. $X$ follows a binomial distribution with parameters $n=10$ and $p=0.90$, denoted as $X \sim B(10, 0.90)$.

The probability mass function of a binomial distribution is given by:

$$P(X=k) = \binom{n}{k} p^k q^{n-k}$$

For this problem, $P(X=k) = \binom{10}{k} (0.9)^k (0.1)^{10-k}$, where $k$ is the number of right-handed people.

We need to find the probability that at most 6 people are right-handed, which means $X \leq 6$. This is given by:

$$P(X \leq 6) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)$$

$$P(X \leq 6) = \sum_{k=0}^{6} P(X=k) = \sum_{k=0}^{6} \binom{10}{k} (0.9)^k (0.1)^{10-k}$$

Calculating the sum of 7 terms can be tedious. Alternatively, we can use the complement rule:

$$P(X \leq 6) = 1 - P(X > 6)$$

The event $X > 6$ means that the number of right-handed people is 7, 8, 9, or 10.

$$P(X > 6) = P(X=7) + P(X=8) + P(X=9) + P(X=10)$$

Let's calculate each of these probabilities:

$$P(X=7) = \binom{10}{7} (0.9)^7 (0.1)^{10-7} = \binom{10}{7} (0.9)^7 (0.1)^3$$

$$P(X=8) = \binom{10}{8} (0.9)^8 (0.1)^{10-8} = \binom{10}{8} (0.9)^8 (0.1)^2$$

$$P(X=9) = \binom{10}{9} (0.9)^9 (0.1)^{10-9} = \binom{10}{9} (0.9)^9 (0.1)^1$$

$$P(X=10) = \binom{10}{10} (0.9)^{10} (0.1)^{10-10} = \binom{10}{10} (0.9)^{10} (0.1)^0$$

Now, calculate the binomial coefficients:

$$\binom{10}{7} = \binom{10}{10-7} = \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$$

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

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

$$\binom{10}{10} = \binom{10}{0} = 1$$

So, the probabilities are:

$$P(X=7) = 120 \times (0.9)^7 \times (0.1)^3$$

$$P(X=8) = 45 \times (0.9)^8 \times (0.1)^2$$

$$P(X=9) = 10 \times (0.9)^9 \times (0.1)^1$$

$$P(X=10) = 1 \times (0.9)^{10} \times 1 = (0.9)^{10}$$

The probability $P(X > 6)$ is the sum of these values:

$$P(X > 6) = 120(0.9)^7(0.1)^3 + 45(0.9)^8(0.1)^2 + 10(0.9)^9(0.1)^1 + (0.9)^{10}$$

Finally, the probability that at most 6 people are right-handed is:

$$P(X \leq 6) = 1 - P(X > 6) = 1 - [120(0.9)^7(0.1)^3 + 45(0.9)^8(0.1)^2 + 10(0.9)^9(0.1) + (0.9)^{10}]$$

Calculating the numerical value requires computation. Using a calculator:

$(0.9)^7 \approx 0.4782969$, $(0.9)^8 \approx 0.43046721$, $(0.9)^9 \approx 0.387420489$, $(0.9)^{10} \approx 0.3486784401$

$(0.1)^3 = 0.001$, $(0.1)^2 = 0.01$, $(0.1)^1 = 0.1$

$P(X=7) \approx 120 \times 0.4782969 \times 0.001 \approx 0.057395628$

$P(X=8) \approx 45 \times 0.43046721 \times 0.01 \approx 0.1937102445$

$P(X=9) \approx 10 \times 0.387420489 \times 0.1 \approx 0.387420489$

$P(X=10) \approx 0.3486784401$

$P(X > 6) \approx 0.057395628 + 0.1937102445 + 0.387420489 + 0.3486784401 \approx 0.9872048016$

$P(X \leq 6) = 1 - P(X > 6) \approx 1 - 0.9872048016 \approx 0.0127951984$

---

The probability that at most 6 of a random sample of 10 people are right-handed is $1 - [120(0.9)^7(0.1)^3 + 45(0.9)^8(0.1)^2 + 10(0.9)^9(0.1) + (0.9)^{10}]$. Numerically, this probability is approximately $0.0128$ (rounded to four decimal places).

Given:

A leap year is selected at random.

---

To Find:

The probability that the leap year contains 53 Tuesdays.

---

Solution:

A leap year has 366 days.

We can express the number of days in a leap year in terms of weeks and extra days.

$$366 \text{ days} = 52 \times 7 \text{ days} + 2 \text{ days}$$

This means a leap year has exactly 52 full weeks and 2 extra days.

The 52 full weeks guarantee that there will be 52 Tuesdays.

For the leap year to have 53 Tuesdays, one of the 2 extra days must be a Tuesday.

The 2 extra days must be consecutive days of the week.

The possible pairs for these 2 consecutive extra days are:

1. Sunday and Monday (SM)

2. Monday and Tuesday (MT)

3. Tuesday and Wednesday (TW)

4. Wednesday and Thursday (WT)

5. Thursday and Friday (TF)

6. Friday and Saturday (FS)

7. Saturday and Sunday (SS)

There are 7 possible outcomes for the pair of extra days, and each outcome is equally likely as the leap year is selected at random.

We need to identify the outcomes where one of the extra days is a Tuesday.

The pairs containing a Tuesday are (Monday, Tuesday) and (Tuesday, Wednesday).

These are 2 favorable outcomes out of the 7 possible outcomes.

The probability of the leap year containing 53 Tuesdays is the ratio of the number of favorable outcomes to the total number of possible outcomes.

$$P(\text{53 Tuesdays}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$$

$$P(\text{53 Tuesdays}) = \frac{2}{7}$$

---

The chance that a randomly selected leap year will contain 53 Tuesdays is $\frac{2}{7}$.

Given:

Four boxes A, B, C, and D containing marbles of different colours as per the table:

Box	Marble colour
	Red	White	Black
A	1	6	3
B	6	2	2
C	8	1	1
D	0	6	4

A box is selected at random, and a single marble is drawn from it.

The drawn marble is red.

---

To Find:

The probability that the red marble was drawn from Box A, Box B, and Box C.

---

Solution:

Let $B_A, B_B, B_C, B_D$ be the events that Box A, B, C, or D is selected, respectively.

Let $R$ be the event that a red marble is drawn.

Since one of the four boxes is selected at random, the probability of selecting any box is equal:

$$P(B_A) = P(B_B) = P(B_C) = P(B_D) = \frac{1}{4}$$

Next, we find the total number of marbles in each box and the number of red marbles in each box to calculate the conditional probabilities of drawing a red marble given the selected box, $P(R | B_i)$.

Total marbles in Box A = $1 + 6 + 3 = 10$. Number of red marbles = 1.

$$P(R | B_A) = \frac{1}{10}$$

Total marbles in Box B = $6 + 2 + 2 = 10$. Number of red marbles = 6.

$$P(R | B_B) = \frac{6}{10} = \frac{3}{5}$$

Total marbles in Box C = $8 + 1 + 1 = 10$. Number of red marbles = 8.

$$P(R | B_C) = \frac{8}{10} = \frac{4}{5}$$

Total marbles in Box D = $0 + 6 + 4 = 10$. Number of red marbles = 0.

$$P(R | B_D) = \frac{0}{10} = 0$$

Now, we need to find the total probability of drawing a red marble, $P(R)$. We use the Law of Total Probability:

$$P(R) = P(R | B_A) P(B_A) + P(R | B_B) P(B_B) + P(R | B_C) P(B_C) + P(R | B_D) P(B_D)$$

Substitute the calculated probabilities:

$$P(R) = \left(\frac{1}{10}\right)\left(\frac{1}{4}\right) + \left(\frac{6}{10}\right)\left(\frac{1}{4}\right) + \left(\frac{8}{10}\right)\left(\frac{1}{4}\right) + \left(\frac{0}{10}\right)\left(\frac{1}{4}\right)$$

$$P(R) = \frac{1}{40} + \frac{6}{40} + \frac{8}{40} + 0$$

$$P(R) = \frac{1 + 6 + 8}{40} = \frac{15}{40}$$

Simplify the fraction $\frac{15}{40}$ by dividing numerator and denominator by 5:

$$\frac{\cancel{15}^{3}}{\cancel{40}_{8}} = \frac{3}{8}$$

$$P(R) = \frac{3}{8}$$

We want to find the probability that the red marble was drawn from a specific box, given that the marble is red. We use Bayes' Theorem:

$$P(B_i | R) = \frac{P(R | B_i) P(B_i)}{P(R)}$$

Probability that the red marble was drawn from Box A ($P(B_A | R)$):

$$P(B_A | R) = \frac{P(R | B_A) P(B_A)}{P(R)}$$

$$P(B_A | R) = \frac{\left(\frac{1}{10}\right)\left(\frac{1}{4}\right)}{\frac{3}{8}}$$

$$P(B_A | R) = \frac{\frac{1}{40}}{\frac{3}{8}}$$

$$P(B_A | R) = \frac{1}{40} \times \frac{8}{3} = \frac{8}{120}$$

Simplify the fraction $\frac{8}{120}$ by dividing numerator and denominator by 8:

$$\frac{\cancel{8}^{1}}{\cancel{120}_{15}} = \frac{1}{15}$$

$$P(B_A | R) = \frac{1}{15}$$

Probability that the red marble was drawn from Box B ($P(B_B | R)$):

$$P(B_B | R) = \frac{P(R | B_B) P(B_B)}{P(R)}$$

$$P(B_B | R) = \frac{\left(\frac{6}{10}\right)\left(\frac{1}{4}\right)}{\frac{3}{8}}$$

$$P(B_B | R) = \frac{\frac{6}{40}}{\frac{3}{8}}$$

$$P(B_B | R) = \frac{6}{40} \times \frac{8}{3} = \frac{48}{120}$$

Simplify the fraction $\frac{48}{120}$ by dividing numerator and denominator by 24:

$$\frac{\cancel{48}^{2}}{\cancel{120}_{5}} = \frac{2}{5}$$

$$P(B_B | R) = \frac{2}{5}$$

Probability that the red marble was drawn from Box C ($P(B_C | R)$):

$$P(B_C | R) = \frac{P(R | B_C) P(B_C)}{P(R)}$$

$$P(B_C | R) = \frac{\left(\frac{8}{10}\right)\left(\frac{1}{4}\right)}{\frac{3}{8}}$$

$$P(B_C | R) = \frac{\frac{8}{40}}{\frac{3}{8}}$$

$$P(B_C | R) = \frac{8}{40} \times \frac{8}{3} = \frac{64}{120}$$

Simplify the fraction $\frac{64}{120}$ by dividing numerator and denominator by 8:

$$\frac{\cancel{64}^{8}}{\cancel{120}_{15}} = \frac{8}{15}$$

$$P(B_C | R) = \frac{8}{15}$$

Note: The question did not ask for the probability it was drawn from Box D. $P(B_D | R) = \frac{P(R | B_D) P(B_D)}{P(R)} = \frac{(0)(\frac{1}{4})}{\frac{3}{8}} = 0$, which makes sense as Box D has no red marbles.

---

The probability that the red marble was drawn from Box A is $\frac{1}{15}$.

The probability that the red marble was drawn from Box B is $\frac{2}{5}$.

The probability that the red marble was drawn from Box C is $\frac{8}{15}$.

Given:

Let $H_0$ be the baseline event of having a heart attack. The initial probability of a heart attack is $P(H_0) = 40\% = 0.40$.

Let $M$ be the event that the patient chooses the meditation and yoga course.

Let $D$ be the event that the patient chooses the prescription drug.

Let $H$ be the event that the patient suffers a heart attack after choosing one of the options.

The patient can choose either option with equal probabilities:

$$P(M) = P(D) = \frac{1}{2} = 0.5$$

The meditation and yoga course reduces the risk by 30%. This means the probability of a heart attack given meditation/yoga is 70% of the baseline risk:

$$P(H | M) = P(H_0) \times (1 - 0.30) = 0.40 \times 0.70 = 0.28$$

The prescription drug reduces the risk by 25%. This means the probability of a heart attack given the drug is 75% of the baseline risk:

$$P(H | D) = P(H_0) \times (1 - 0.25) = 0.40 \times 0.75 = 0.30$$

---

To Find:

The probability that the patient followed a course of meditation and yoga, given that the patient suffered a heart attack, i.e., $P(M | H)$.

---

Solution:

We need to find $P(M | H)$. We can use Bayes' Theorem for this:

First, we need to find the total probability of a patient suffering a heart attack after choosing one of the options, $P(H)$. Since the patient must choose either M or D (and these are mutually exclusive and exhaustive events), we can use the Law of Total Probability:

Substitute the known probability values into equation (ii):

$$P(H) = (0.28)(0.5) + (0.30)(0.5)$$

$$P(H) = 0.14 + 0.15$$

$$P(H) = 0.29$$

Now, substitute the values of $P(H | M)$, $P(M)$, and $P(H)$ into Bayes' Theorem (equation i):

$$P(M | H) = \frac{(0.28)(0.5)}{0.29}$$

$$P(M | H) = \frac{0.14}{0.29}$$

To express this as a fraction, we can write it as $\frac{14}{29}$. This fraction cannot be simplified further as 14 and 29 have no common factors other than 1.

---

The probability that the patient followed a course of meditation and yoga, given that they suffered a heart attack, is $\frac{14}{29}$.

Given:

A second order determinant is given by $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$.

Each element $a, b, c, d$ can take a value from $\{0, 1\}$.

The values of $a, b, c, d$ are chosen independently, with $P(\text{element}=0) = \frac{1}{2}$ and $P(\text{element}=1) = \frac{1}{2}$ for each element.

---

To Find:

The probability that the value of the determinant is positive.

---

Solution:

The value of a second order determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is given by the formula:

$$ \text{Value} = ad - bc $$

Since each element $a, b, c, d$ can be either 0 or 1, and they are chosen independently, the total number of possible determinants is $2 \times 2 \times 2 \times 2 = 2^4 = 16$.

Each specific determinant (defined by the values of $a, b, c, d$) has a probability of $\left(\frac{1}{2}\right)^4 = \frac{1}{16}$ of being chosen.

We want to find the probability that the value of the determinant is positive, i.e., $ad - bc > 0$.

This inequality is equivalent to $ad > bc$.

Since $a, b, c, d \in \{0, 1\}$, the possible values for the products $ad$ and $bc$ are $0 \times 0 = 0$, $0 \times 1 = 0$, $1 \times 0 = 0$, and $1 \times 1 = 1$. So, $ad \in \{0, 1\}$ and $bc \in \{0, 1\}$.

The inequality $ad > bc$ can only be satisfied if $ad=1$ and $bc=0$.

Let's analyze the condition $ad=1$ and $bc=0$:

For $ad=1$, both $a$ and $d$ must be 1. The probability of this occurring is $P(a=1 \text{ and } d=1) = P(a=1) \times P(d=1)$ (due to independence).

$$P(a=1 \text{ and } d=1) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$

For $bc=0$, at least one of $b$ or $c$ must be 0. The opposite event is $bc=1$, which occurs only when both $b=1$ and $c=1$. The probability of $bc=1$ is $P(b=1 \text{ and } c=1) = P(b=1) \times P(c=1) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

The probability of $bc=0$ is the complement of $bc=1$:

$$P(bc=0) = 1 - P(bc=1) = 1 - \frac{1}{4} = \frac{3}{4}$$

Since the choices of $a, d$ are independent of the choices of $b, c$, the event ($ad=1$) and the event ($bc=0$) are independent.

The probability that $ad=1$ AND $bc=0$ is the product of their individual probabilities:

$$P(ad=1 \text{ and } bc=0) = P(ad=1) \times P(bc=0)$$

$$P(ad - bc > 0) = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$$

Alternatively, we can list the specific determinants that result in a positive value ($ad-bc > 0$, i.e., $ad=1$ and $bc=0$). For $ad=1$, we must have $a=1$ and $d=1$. For $bc=0$, the pair $(b,c)$ can be $(0,0)$, $(0,1)$, or $(1,0)$.

The determinants with positive values are those where $(a,b,c,d)$ are:

1. $(1, 0, 0, 1)$: Value $1 \times 1 - 0 \times 0 = 1 > 0$

2. $(1, 0, 1, 1)$: Value $1 \times 1 - 0 \times 1 = 1 > 0$

3. $(1, 1, 0, 1)$: Value $1 \times 1 - 1 \times 0 = 1 > 0$

There are 3 such determinants. Each of these determinants has a probability of $\frac{1}{16}$ of being chosen.

The probability of getting a positive determinant is the sum of the probabilities of these 3 outcomes:

$$P(\text{positive determinant}) = P(1,0,0,1) + P(1,0,1,1) + P(1,1,0,1)$$

$$P(\text{positive determinant}) = \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{3}{16}$$

---

The probability that the value of the determinant is positive is $\frac{3}{16}$.

Given:

Let $F_A$ be the event that subsystem A fails.

Let $F_B$ be the event that subsystem B fails.

We are given the following probabilities:

$$P(F_A) = 0.2$$

The event "B fails alone" means B fails and A does not fail. This can be represented as $F_B \cap F_A'$.

$$P(F_B \cap F_A') = 0.15$$

The event "A and B fail" means both A and B fail. This can be represented as $F_A \cap F_B$.

$$P(F_A \cap F_B) = 0.15$$

---

To Find:

(i) $P(F_A | F_B)$, the probability that A fails given that B has failed.

(ii) $P(F_A \text{ fails alone})$, the probability that A fails and B does not fail.

---

Solution:

First, let's find the probability that B fails, $P(F_B)$. The event $F_B$ can be partitioned into two mutually exclusive events: ($F_B$ and $F_A$) and ($F_B$ and not $F_A$). That is, $F_B = (F_B \cap F_A) \cup (F_B \cap F_A')$.

Therefore,

Substitute the given values:

$$P(F_B) = P(F_A \cap F_B) + P(F_B \text{ fails alone})$$

$$P(F_B) = 0.15 + 0.15$$

$$P(F_B) = 0.30$$

(i) Find P(A fails|B has failed), which is $P(F_A | F_B)$.

Using the definition of conditional probability:

Substitute the known values $P(F_A \cap F_B) = 0.15$ and $P(F_B) = 0.30$:

$$P(F_A | F_B) = \frac{0.15}{0.30}$$

$$P(F_A | F_B) = \frac{15}{30} = \frac{1}{2}$$

---

(ii) Find P(A fails alone), which is $P(F_A \cap F_B')$.

The event $F_A$ can be partitioned into two mutually exclusive events: ($F_A$ and $F_B$) and ($F_A$ and not $F_B$). That is, $F_A = (F_A \cap F_B) \cup (F_A \cap F_B')$.

Therefore,

We are looking for $P(F_A \cap F_B')$. Rearranging the equation:

Substitute the given values $P(F_A) = 0.2$ and $P(F_A \cap F_B) = 0.15$:

$$P(F_A \text{ fails alone}) = 0.2 - 0.15$$

$$P(F_A \text{ fails alone}) = 0.05$$

---

Final Answer:

(i) The probability that A fails given B has failed is $0.5$ or $\frac{1}{2}$.

(ii) The probability that A fails alone is $0.05$.

Given:

Bag I contains 3 red balls and 4 black balls.

Bag II contains 4 red balls and 5 black balls.

One ball is transferred from Bag I to Bag II.

Then a ball is drawn from Bag II, and it is found to be red.

---

To Find:

The probability that the transferred ball was black, given that the ball drawn from Bag II was red.

---

Solution:

Let $T_R$ be the event that a red ball is transferred from Bag I to Bag II.

Let $T_B$ be the event that a black ball is transferred from Bag I to Bag II.

Let $E$ be the event that a red ball is drawn from Bag II after the transfer.

Initially, Bag I has $3$ red and $4$ black balls, for a total of $3 + 4 = 7$ balls.

The probability of transferring a red ball from Bag I is:

The probability of transferring a black ball from Bag I is:

Note that $P(T_R) + P(T_B) = \frac{3}{7} + \frac{4}{7} = 1$.

Now, we consider the composition of Bag II after the transfer and the probability of drawing a red ball (event $E$).

Case 1: A red ball is transferred from Bag I ($T_R$ occurs).

Bag II initially has 4 red and 5 black balls. After transferring one red ball, Bag II will have $4+1 = 5$ red balls and $5$ black balls, for a total of $5+5 = 10$ balls.

The probability of drawing a red ball from Bag II in this case is $P(E | T_R)$:

Case 2: A black ball is transferred from Bag I ($T_B$ occurs).

Bag II initially has 4 red and 5 black balls. After transferring one black ball, Bag II will have $4$ red balls and $5+1 = 6$ black balls, for a total of $4+6 = 10$ balls.

The probability of drawing a red ball from Bag II in this case is $P(E | T_B)$:

We are given that the ball drawn from Bag II is red (event $E$ has occurred), and we want to find the probability that the transferred ball was black, i.e., $P(T_B | E)$. We can use Bayes' Theorem:

To find $P(E)$, the total probability of drawing a red ball from Bag II, we use the Law of Total Probability:

Substitute the calculated probabilities into equation (ii):

$$P(E) = \left(\frac{1}{2}\right)\left(\frac{3}{7}\right) + \left(\frac{2}{5}\right)\left(\frac{4}{7}\right)$$

$$P(E) = \frac{3}{14} + \frac{8}{35}$$

To add these fractions, find a common denominator. The least common multiple of 14 and 35 is 70.

$$P(E) = \frac{3 \times 5}{14 \times 5} + \frac{8 \times 2}{35 \times 2} = \frac{15}{70} + \frac{16}{70} = \frac{15 + 16}{70} = \frac{31}{70}$$

Now, substitute the values of $P(E | T_B)$, $P(T_B)$, and $P(E)$ into Bayes' Theorem (equation i):

$$P(T_B | E) = \frac{\left(\frac{2}{5}\right) \left(\frac{4}{7}\right)}{\frac{31}{70}}$$

$$P(T_B | E) = \frac{\frac{8}{35}}{\frac{31}{70}}$$

$$P(T_B | E) = \frac{8}{35} \times \frac{70}{31}$$

Simplify the expression:

$$P(T_B | E) = \frac{8}{\cancel{35}_{1}} \times \frac{\cancel{70}^{2}}{31}$$

$$P(T_B | E) = \frac{8 \times 2}{1 \times 31} = \frac{16}{31}$$

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The probability that the transferred ball is black, given that the drawn ball is red, is $\frac{16}{31}$.

Given:

A and B are two events.

$P(A) \neq 0$

$P(B | A) = 1$

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To Find:

The relationship between events A and B.

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Solution:

The definition of the conditional probability of event B given event A is:

This formula is valid since $P(A) \neq 0$.

We are given that $P(B | A) = 1$. Substitute this into the formula (i):

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

Since $P(A) \neq 0$, we can multiply both sides by $P(A)$:

Now we need to interpret what the condition $P(A \cap B) = P(A)$ means about the relationship between events A and B.

Recall that the event $A \cap B$ occurs if and only if both event A and event B occur.

The condition $P(A \cap B) = P(A)$ implies that the probability of A and B occurring together is equal to the probability of A occurring alone.

Consider the relationship between the sets of outcomes for events A and $A \cap B$. The set of outcomes for $A \cap B$ is a subset of the set of outcomes for A. If their probabilities are equal, it means that every outcome in A must also be in $A \cap B$.

If an outcome $\omega$ is in A, and $P(A \cap B) = P(A)$, then $\omega$ must also be in $A \cap B$. If $\omega$ is in $A \cap B$, then by definition, $\omega$ must be in A AND $\omega$ must be in B. Thus, if $\omega \in A$, then $\omega \in B$.

This is the definition of A being a subset of B.

Let's examine the given options:

(A) A ⊂ B: If A is a subset of B, then $A \cap B = A$. This leads to $P(A \cap B) = P(A)$, which matches our derived condition.

(B) B ⊂ A: If B is a subset of A, then $A \cap B = B$. This leads to $P(A \cap B) = P(B)$. So, $P(B) = P(A)$. This does not necessarily mean B is a subset of A (unless $P(A)=P(B)=0$, which is ruled out by $P(A) \neq 0$ in the context of the given options). For example, if A is the event of rolling an even number on a die {2, 4, 6} and B is the event of rolling a number greater than 3 {4, 5, 6}, then $P(A)=3/6=1/2$, $P(B)=3/6=1/2$. $A \cap B = \{4, 6\}$, $P(A \cap B) = 2/6 = 1/3$. Here $P(A)=P(B)$ but $P(A \cap B) \neq P(A)$, and neither is a subset of the other. However, if $A \subset B$, then $P(A \cap B) = P(A)$. If $P(A \cap B) = P(A)$, it forces $A$ to be contained within $B$ in terms of probability measure, which implies set inclusion unless $P(A)=0$.

(C) B = $\phi$: If B is the empty set, $A \cap B = \phi$. Then $P(A \cap B) = P(\phi) = 0$. From equation (ii), $P(A) = 0$. But we are given $P(A) \neq 0$. So, this is incorrect.

(D) A = $\phi$: If A is the empty set, $P(A) = P(\phi) = 0$. But we are given $P(A) \neq 0$. So, this is incorrect.

The condition $P(A \cap B) = P(A)$ with $P(A) \neq 0$ precisely means that the event A can only occur when event B also occurs, which is the definition of A being a subset of B ($A \subseteq B$). Since A is a proper subset (A $\subset$ B) is an option, and implies $A \subseteq B$, this is the correct choice.

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The correct answer is (A) A ⊂ B.

Given:

A and B are two events.

$$P(A|B) > P(A)$$

For $P(A|B)$ to be defined, we must have $P(B) \neq 0$. Also, from the given inequality $P(A|B) > P(A)$, if $P(A) = 0$, then $P(A|B) > 0$. However, if $P(A) = 0$, then $P(A \cap B) \le P(A) = 0$, so $P(A \cap B) = 0$. Then $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{P(B)} = 0$ (assuming $P(B) \neq 0$), which contradicts $P(A|B) > 0$. Therefore, we must have $P(A) \neq 0$. Since options involve $P(B|A)$, we must also have $P(A) \neq 0$ for $P(B|A)$ to be defined. So, we assume $P(A) \neq 0$ and $P(B) \neq 0$.

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To Find:

Which of the given relationships is correct.

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Solution:

The definition of conditional probability $P(A|B)$ is:

We are given the inequality $P(A|B) > P(A)$. Substituting the definition of $P(A|B)$:

Since we are assuming $P(B) > 0$, we can multiply both sides of the inequality (i) by $P(B)$:

This inequality tells us that the probability of both A and B occurring is greater than the product of their individual probabilities. This means that events A and B are positively correlated; the occurrence of one event makes the other event more likely.

Now consider the probability $P(B|A)$. The definition of conditional probability $P(B|A)$ is:

Substitute the inequality $P(A \cap B) > P(A) \cdot P(B)$ from (ii) into the numerator of the expression for $P(B|A)$:

Since we are assuming $P(A) > 0$, we can cancel $P(A)$ from the numerator and denominator on the right side of the inequality (iii):

$$P(B|A) > P(B)$$

This result shows that if $P(A|B) > P(A)$, then it implies $P(B|A) > P(B)$. This corresponds to option (C).

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The correct answer is (C) P(B|A) > P(B).

Given:

A and B are two events such that $P(A) + P(B) – P(A \cap B) = P(A)$.

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To Find:

The correct relationship among the given options.

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Solution:

The given equation is $P(A) + P(B) – P(A \cap B) = P(A)$.

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

Substitute the given equation into the formula for $P(A \cup B)$:

From $P(A \cup B) = P(A)$, we can substitute the expression from equation (i):

$$P(A) = P(A) + P(B) – P(A \cap B)$$

Subtract $P(A)$ from both sides of the equation:

$$0 = P(B) – P(A \cap B)$$

Rearranging the terms, we get:

Now let's consider the options involving conditional probabilities. The formula for conditional probability $P(A|B)$ is $\frac{P(A \cap B)}{P(B)}$, provided $P(B) \neq 0$. The formula for conditional probability $P(B|A)$ is $\frac{P(A \cap B)}{P(A)}$, provided $P(A) \neq 0$.

We are given $P(A) \neq 0$ implicitly by the options (A) and (C) involving $P(B|A)$. Let's examine the option (B): $P(A|B) = 1$.

If $P(B) \neq 0$, we can use the definition of $P(A|B)$ and equation (ii):

So, if $P(B) \neq 0$, the given condition implies $P(A|B) = 1$.

What if $P(B) = 0$? If $P(B) = 0$, then from equation (ii), $P(A \cap B) = 0$. The original given equation becomes $P(A) + 0 - 0 = P(A)$, which simplifies to $P(A) = P(A)$, which is always true for any event A.

If $P(B) = 0$, the conditional probability $P(A|B)$ is typically considered undefined by the formula $\frac{P(A \cap B)}{P(B)}$. However, in the context of multiple-choice questions like this, the options involving conditional probabilities imply that the denominator probability is non-zero where the conditional probability is stated to be equal to a specific value (like 0 or 1).

Given that $P(A|B) = 1$ is one of the options, and we derived that this holds true whenever $P(A|B)$ is defined (i.e., when $P(B) \neq 0$), this is the most likely intended correct answer.

Let's check the other options using $P(A \cap B) = P(B)$ and the fact that $P(A) \neq 0$.

(A) $P(B|A) = 1$: $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{P(B)}{P(A)}$. This is equal to 1 only if $P(B) = P(A)$. The given condition $P(A \cap B) = P(B)$ does not imply $P(A) = P(B)$ necessarily (e.g., let A be rolling a number $\le 4$ on a die, B be rolling a prime number $\{2,3,5\}$. $A=\{1,2,3,4\}, P(A)=4/6$. $B=\{2,3,5\}, P(B)=3/6$. $A \cap B = \{2,3\}, P(A \cap B)=2/6$. $P(A \cup B) = P(A)+P(B)-P(A \cap B) = 4/6 + 3/6 - 2/6 = 5/6$. $P(A)=4/6$. Here $P(A \cup B) \neq P(A)$ and $P(A \cap B) \neq P(B)$. Let's try another example where $P(A \cap B) = P(B)$. Let A be $\{1,2,3,4\}$ and B be $\{1,2\}$. $P(A)=4/6$, $P(B)=2/6$. $A \cap B = \{1,2\}, P(A \cap B)=2/6$. Here $P(A \cap B) = P(B)$ is true. Let's check the original condition: $P(A) + P(B) - P(A \cap B) = 4/6 + 2/6 - 2/6 = 4/6 = P(A)$. The original condition holds. Now check $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{2/6}{4/6} = \frac{2}{4} = \frac{1}{2}$. This is not 1. So option (A) is not always true.)

(C) $P(B|A) = 0$: This would mean $P(A \cap B) = 0$. From equation (ii), this implies $P(B) = 0$. As discussed, the given condition holds if $P(B)=0$, but it also holds for other cases where $P(B) \neq 0$ (as shown by the example above where $P(B)=2/6$). So option (C) is not always true.

(D) $P(A|B) = 0$: This would mean $P(A \cap B) = 0$. From equation (ii), this implies $P(B) = 0$. As discussed, the given condition does not require $P(B)=0$. So option (D) is not always true.

Therefore, the only option that is a direct consequence of the given condition $P(A) + P(B) – P(A \cap B) = P(A)$, assuming the conditional probabilities are well-defined, is $P(A|B) = 1$. The condition $P(A \cap B) = P(B)$ means that event B can only occur when event A also occurs (if $P(B) \neq 0$), which is exactly what $P(A|B)=1$ means.

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The correct answer is (B) P(A|B) = 1.