Chapter 16 Probability

Solution:

The word ALGORITHM consists of 9 distinct letters: A, L, G, O, R, I, T, H, M.

Total Number of Arrangements:

The total number of ways to arrange 9 distinct letters in a row is given by the factorial of the number of letters.

Total number of arrangements $= 9!$

$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362,880$

Favorable Arrangements (GOR together as a unit):

We want the letters G, O, R to remain together as a single unit. Treat the group (GOR) as one block.

Now, we are arranging the following 7 entities:

(GOR), A, L, I, T, H, M

The number of ways to arrange these 7 distinct entities is $7!$.

Within the unit (GOR), the letters G, O, and R can be arranged among themselves in $3!$ ways.

The possible arrangements of GOR within the unit are GOR, GRO, OGR, ORG, RGO, ROG.

Number of arrangements of the unit (GOR) $= 3! = 3 \times 2 \times 1 = 6$

The total number of favorable arrangements where GOR stay together as a unit is the product of the number of ways to arrange the 7 entities and the number of ways to arrange letters within the GOR unit.

Number of favorable arrangements $= (\text{Number of ways to arrange the 7 entities}) \times (\text{Number of ways to arrange GOR within the unit})$

Number of favorable arrangements $= 7! \times 3!$

$7! = 5,040$

$3! = 6$

Number of favorable arrangements $= 5,040 \times 6 = 30,240$

Probability:

The probability of an event is the ratio of the number of favorable outcomes to the total number of possible outcomes.

Probability (GOR together) = $\frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}}$

Probability (GOR together) $= \frac{7! \times 3!}{9!}$

We can simplify this expression:

Probability $= \frac{7! \times 3!}{9 \times 8 \times 7!}$

Cancel out $7!$ from the numerator and the denominator.

Probability $= \frac{3!}{9 \times 8}$

Substitute the value of $3!$:

$3! = 6$

Probability $= \frac{6}{72}$

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 6.

Probability $= \frac{\cancel{6}^{1}}{\cancel{72}_{12}} = \frac{1}{12}$

The probability that the letters GOR must remain together as a unit when the letters of the word ALGORITHM are arranged at random in a row is $\frac{1}{12}$.

Given:

6 new employees are to be assigned 6 desks that are lined up in a row.

Two of the employees form a married couple.

To Find:

The probability that the married couple will have nonadjacent desks when the assignment of employees to desks is made randomly.

Solution:

Let the six employees be $E_1, E_2, E_3, E_4, E_5, E_6$. Let $E_1$ and $E_2$ represent the married couple.

We are asked to find the probability that $E_1$ and $E_2$ are assigned nonadjacent desks.

Using the hint provided, we will first calculate the probability that the married couple is assigned adjacent desks, and then subtract this probability from 1.

Total Number of Arrangements:

The total number of ways to assign 6 distinct employees to 6 distinct desks in a row is the number of permutations of 6 items.

Total number of arrangements $= 6!$

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

Number of Arrangements where the Married Couple is Adjacent:

To count the arrangements where the married couple is adjacent, we treat the couple ($E_1, E_2$) as a single unit.

Now we have 5 entities to arrange: the coupled unit and the remaining 4 employees ($E_3, E_4, E_5, E_6$).

The number of ways to arrange these 5 distinct entities is $5!$.

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Within the couple unit ($E_1, E_2$), the two individuals can be arranged in $2!$ ways (either $E_1$ is to the left of $E_2$, or $E_2$ is to the left of $E_1$).

$2! = 2 \times 1 = 2$

The total number of arrangements where the married couple is adjacent is the product of the number of ways to arrange the 5 entities and the number of ways to arrange the couple within their unit.

Number of adjacent arrangements $= 5! \times 2! = 120 \times 2 = 240$

Probability that the Married Couple is Adjacent:

The probability that the married couple is adjacent is the ratio of the number of adjacent arrangements to the total number of arrangements.

Probability (Adjacent) = $\frac{\text{Number of adjacent arrangements}}{\text{Total number of arrangements}}$

Probability (Adjacent) $= \frac{5! \times 2!}{6!}$

Probability (Adjacent) $= \frac{240}{720}$

Simplifying the fraction:

Probability (Adjacent) $= \frac{\cancel{240}^{1}}{\cancel{720}_{3}} = \frac{1}{3}$

Probability that the Married Couple is Nonadjacent:

The event that the married couple is nonadjacent is the complement of the event that the married couple is adjacent.

Probability (Nonadjacent) = $1 - $ Probability (Adjacent)

Probability (Nonadjacent) $= 1 - \frac{1}{3}$

Probability (Nonadjacent) $= \frac{3}{3} - \frac{1}{3} = \frac{2}{3}$

The probability that the married couple will have nonadjacent desks is $\frac{2}{3}$.

Given:

An integer is chosen at random from 1 through 1000.

To Find:

The probability that the integer is a multiple of 2 or a multiple of 9.

Solution:

The total number of integers from 1 through 1000 is 1000.

Total number of possible outcomes = 1000.

Let A be the event that the integer is a multiple of 2.

Let B be the event that the integer is a multiple of 9.

We want to find the probability of the event A or B, which is $P(A \cup B)$.

Using the Principle of Inclusion-Exclusion:

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

First, find the number of integers that are multiples of 2 in the range [1, 1000].

Number of multiples of 2 = $\lfloor \frac{1000}{2} \rfloor = 500$

So, the number of outcomes in event A is 500.

Next, find the number of integers that are multiples of 9 in the range [1, 1000].

Number of multiples of 9 = $\lfloor \frac{1000}{9} \rfloor = 111$

So, the number of outcomes in event B is 111.

Now, find the number of integers that are multiples of both 2 and 9. An integer that is a multiple of both 2 and 9 must be a multiple of their least common multiple (LCM).

LCM(2, 9) = 18

So, we need to find the number of multiples of 18 in the range [1, 1000].

Number of multiples of 18 = $\lfloor \frac{1000}{18} \rfloor = 55$

So, the number of outcomes in event $A \cap B$ is 55.

Now, apply the Principle of Inclusion-Exclusion to find the number of integers that are multiples of 2 or 9.

Number of (multiples of 2 or 9) = (Number of multiples of 2) + (Number of multiples of 9) - (Number of multiples of 18)

Number of (multiples of 2 or 9) $= 500 + 111 - 55$

Number of (multiples of 2 or 9) $= 611 - 55 = 556$

This is the number of favorable outcomes.

The probability is the ratio of the number of favorable outcomes to the total number of possible outcomes.

Probability (multiple of 2 or 9) $= \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

Probability $= \frac{556}{1000}$

We can simplify this fraction by dividing the numerator and denominator by their greatest common divisor. Both are divisible by 4.

$556 \div 4 = 139$

$1000 \div 4 = 250$

Probability $= \frac{139}{250}$

The probability that the integer chosen is a multiple of 2 or a multiple of 9 is $\frac{139}{250}$.

Solution:

The experiment involves rolling a standard six-sided die until a 2 appears for the first time. The possible outcomes on each roll are {1, 2, 3, 4, 5, 6}.

(i) Number of elements when 2 appears on the k^th roll:

For the 2 to appear for the first time on the k^th roll, the following conditions must be met:

The first $k-1$ rolls must not result in a 2. There are 5 possible outcomes for each of these rolls (any number except 2).

The k^th roll must result in a 2. There is only 1 possible outcome for this roll (the number 2).

Since the rolls are independent events, the total number of elements in the sample space corresponding to this event is the product of the number of outcomes for each roll.

Number of elements = (Number of outcomes for roll 1) $\times$ (Number of outcomes for roll 2) $\times$ ... $\times$ (Number of outcomes for roll k-1) $\times$ (Number of outcome for roll k)

Number of elements = $5 \times 5 \times \dots \times 5$ (for the first $k-1$ rolls) $\times 1$ (for the k^th roll)

Number of elements $= 5^{k-1} \times 1 = 5^{k-1}$

So, there are $5^{k-1}$ elements in the sample space where the 2 appears exactly on the k^th roll for the first time.

(ii) Number of elements when 2 appears not later than the k^th roll:

The event that the 2 appears not later than the k^th roll means that the first 2 appears on the 1^st roll OR the 2^nd roll OR the 3^rd roll OR ... OR the k^th roll.

Let $E_i$ be the event that the 2 appears for the first time on the i^th roll.

We want to find the number of elements in the event $E_1 \cup E_2 \cup \dots \cup E_k$.

Since these events ($E_i$) are mutually exclusive (the first 2 cannot appear for the first time on both the i^th and j^th roll if $i \neq j$), the number of elements in their union is the sum of the number of elements in each event.

Number of elements = Number of elements in $E_1$ + Number of elements in $E_2$ + ... + Number of elements in $E_k$

Using the result from part (i), the number of elements in $E_i$ is $5^{i-1}$.

Number of elements = $5^{1-1} + 5^{2-1} + 5^{3-1} + \dots + 5^{k-1}$

Number of elements = $5^0 + 5^1 + 5^2 + \dots + 5^{k-1}$

Number of elements = $1 + 5 + 5^2 + \dots + 5^{k-1}$

This is a finite geometric series with first term $a = 1$, common ratio $r = 5$, and number of terms $n = k$.

The sum of a finite geometric series is given by the formula $S_n = a \frac{r^n - 1}{r - 1}$.

In this case, $S_k = 1 \times \frac{5^k - 1}{5 - 1} = \frac{5^k - 1}{4}$.

So, there are $\frac{5^k - 1}{4}$ elements in the sample space where the 2 appears not later than the k^th roll.

Given:

A loaded die where the probability of an odd number is twice the probability of an even number.

Event G is that a number greater than 3 occurs on a single roll.

To Find:

The probability of event G, P(G).

Solution:

The possible outcomes of a single roll of a standard die are {1, 2, 3, 4, 5, 6}.

Let $P(\text{even})$ be the probability of rolling an even number (2, 4, or 6).

Let $P(\text{odd})$ be the probability of rolling an odd number (1, 3, or 5).

According to the problem statement, $P(\text{odd}) = 2 \times P(\text{even})$.

Since each even number is equally likely, the probability of rolling a specific even number (2, 4, or 6) is the same. Let this probability be $p$.

So, $P(2) = P(4) = P(6) = p$.

Since each odd number is equally likely and its probability is twice that of an even number, the probability of rolling a specific odd number (1, 3, or 5) is $2p$.

So, $P(1) = P(3) = P(5) = 2p$.

The sum of probabilities of all possible outcomes must equal 1.

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

$2p + p + 2p + p + 2p + p = 1$

Combine the terms with $p$:

$(2+1+2+1+2+1)p = 1$

$9p = 1$

Solve for $p$:

$p = \frac{1}{9}$

Now we know the probability of each outcome:

$P(2) = P(4) = P(6) = \frac{1}{9}$

$P(1) = P(3) = P(5) = 2 \times \frac{1}{9} = \frac{2}{9}$

The event G is that a number greater than 3 occurs. The outcomes in event G are {4, 5, 6}.

To find the probability of event G, we sum the probabilities of the outcomes in G.

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

$P(G) = \frac{1}{9} + \frac{2}{9} + \frac{1}{9}$

$P(G) = \frac{1+2+1}{9} = \frac{4}{9}$

The probability of event G is $\frac{4}{9}$.

Given:

Let C be the event that a family owns a colour television set.

Let B be the event that a family owns a black and white television set.

We are given the following probabilities:

$P(C) = 0.87$

$P(B) = 0.36$

The probability that a family owns both kinds of sets is given as:

$P(C \cap B) = 0.30$

To Find:

The probability that a family owns either one or both kinds of sets. This corresponds to the probability of the union of events C and B, denoted as $P(C \cup B)$.

Solution:

We use the formula for the probability of the union of two events, which is given by:

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

Substitute the given values into the formula:

$P(C \cup B) = 0.87 + 0.36 - 0.30$

First, add $P(C)$ and $P(B)$:

$0.87 + 0.36 = 1.23$

Now, subtract $P(C \cap B)$ from the result:

$1.23 - 0.30 = 0.93$

So, the probability that a family owns either a colour television set or a black and white television set or both is 0.93.

The probability that a family owns either anyone or both kinds of sets is $0.93$.

Given:

A and B are mutually exclusive events.

$P(A) = 0.35$

$P(B) = 0.45$

To Find:

(a) $P(A')$

(b) $P(B')$

(c) $P(A \cup B)$

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

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

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

Solution:

Since A and B are mutually exclusive events, they cannot occur simultaneously. This means the probability of their intersection is 0, i.e., $P(A \cap B) = 0$.

(a) $P(A')$

The probability of the complement of an event A is given by $P(A') = 1 - P(A)$.

$P(A') = 1 - 0.35$

$P(A') = 0.65$

(b) $P(B')$

The probability of the complement of an event B is given by $P(B') = 1 - P(B)$.

$P(B') = 1 - 0.45$

$P(B') = 0.55$

(c) $P(A \cup B)$

For any two events A and B, the probability of their union is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.

Since A and B are mutually exclusive, $P(A \cap B) = 0$.

Therefore, $P(A \cup B) = P(A) + P(B)$.

$P(A \cup B) = 0.35 + 0.45$

$P(A \cup B) = 0.80$

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

Since A and B are mutually exclusive events, their intersection is the empty set ($\emptyset$). The probability of an empty set is 0.

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

$P(A \cap B) = 0$

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

The intersection $A \cap B'$ represents the event where A occurs and B does not occur. Since A and B are mutually exclusive, if A occurs, B cannot occur. Therefore, the event A occurring implies that B must not occur, which means the outcome is in $B'$. Thus, the event $A \cap B'$ is the same as event A.

$A \cap B' = A$

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

$P(A \cap B') = 0.35$

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

The intersection $A' \cap B'$ represents the event where neither A occurs nor B occurs. Using De Morgan's Law, this is equivalent to the complement of the union of A and B.

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

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

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

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

From part (c), we found $P(A \cup B) = 0.80$.

$P(A' \cap B') = 1 - 0.80$

$P(A' \cap B') = 0.20$

Given:

Let the events be:

VC: Surgery is rated Very Complex, $P(VC) = 0.15$

C: Surgery is rated Complex, $P(C) = 0.20$

R: Surgery is rated Routine, $P(R) = 0.31$

S: Surgery is rated Simple, $P(S) = 0.26$

VS: Surgery is rated Very Simple, $P(VS) = 0.08$

The categories (Very Complex, Complex, Routine, Simple, Very Simple) are mutually exclusive, meaning a surgery cannot belong to more than one category simultaneously.

To Find:

(a) $P(\text{Complex or Very Complex})$

(b) $P(\text{Neither Very Complex nor Very Simple})$

(c) $P(\text{Routine or Complex})$

(d) $P(\text{Routine or Simple})$

Solution:

Since the different ratings are mutually exclusive events, the probability of the union of any two (or more) ratings is simply the sum of their individual probabilities.

(a) Probability of being Complex or Very Complex:

This is the probability of the event C or VC, i.e., $P(C \cup VC)$.

$P(C \cup VC) = P(C) + P(VC)$ (Since C and VC are mutually exclusive)

$P(C \cup VC) = 0.20 + 0.15$

$P(C \cup VC) = 0.35$

(b) Probability of being Neither Very Complex nor Very Simple:

This means the surgery is not in the VC category and not in the VS category. This is the complement of the event (VC or VS). So we want $P((VC \cup VS)')$.

$P((VC \cup VS)') = 1 - P(VC \cup VS)$

Since VC and VS are mutually exclusive, $P(VC \cup VS) = P(VC) + P(VS)$.

$P(VC \cup VS) = 0.15 + 0.08 = 0.23$

$P((VC \cup VS)') = 1 - 0.23 = 0.77$

Alternatively, "neither Very Complex nor Very Simple" means the surgery is Complex, Routine, or Simple. Since these events are mutually exclusive:

$P(\text{C or R or S}) = P(C) + P(R) + P(S)$

$P(\text{C or R or S}) = 0.20 + 0.31 + 0.26 = 0.77$

(c) Probability of being Routine or Complex:

This is the probability of the event R or C, i.e., $P(R \cup C)$.

$P(R \cup C) = P(R) + P(C)$ (Since R and C are mutually exclusive)

$P(R \cup C) = 0.31 + 0.20$

$P(R \cup C) = 0.51$

(d) Probability of being Routine or Simple:

This is the probability of the event R or S, i.e., $P(R \cup S)$.

$P(R \cup S) = P(R) + P(S)$ (Since R and S are mutually exclusive)

$P(R \cup S) = 0.31 + 0.26$

$P(R \cup S) = 0.57$

Given:

Four candidates A, B, C, and D.

The probabilities of their selection are related as follows:

A is twice as likely as B: $P(A) = 2 P(B)$

B and C are given about the same chance: Assuming "about the same chance" means equal chance, $P(B) = P(C)$.

C is twice as likely as D: $P(C) = 2 P(D)$

Since A, B, C, and D are the only candidates, the sum of their probabilities of being selected must be 1.

$P(A) + P(B) + P(C) + P(D) = 1$

To Find:

(a) The probability that C will be selected, $P(C)$.

(b) The probability that A will not be selected, $P(A')$.

Solution:

Let's express the probabilities of A, B, and C in terms of the probability of D.

From the given information, we have:

$P(C) = 2 P(D)$

$P(B) = P(C)$

Substituting the first equation into the second:

$P(B) = 2 P(D)$

Also, we are given $P(A) = 2 P(B)$.

Substituting the expression for $P(B)$:

$P(A) = 2 \times (2 P(D))$

$P(A) = 4 P(D)$

Now, substitute these expressions into the equation for the sum of probabilities:

$P(A) + P(B) + P(C) + P(D) = 1$

$4 P(D) + 2 P(D) + 2 P(D) + P(D) = 1$

Combine the terms:

$(4 + 2 + 2 + 1) P(D) = 1$

$9 P(D) = 1$

Solving for $P(D)$:

$P(D) = \frac{1}{9}$

Now we can find the probabilities for the other candidates:

$P(C) = 2 P(D) = 2 \times \frac{1}{9} = \frac{2}{9}$

$P(B) = P(C) = \frac{2}{9}$

$P(A) = 4 P(D) = 4 \times \frac{1}{9} = \frac{4}{9}$

Let's verify that the probabilities sum to 1:

$P(A) + P(B) + P(C) + P(D) = \frac{4}{9} + \frac{2}{9} + \frac{2}{9} + \frac{1}{9} = \frac{4+2+2+1}{9} = \frac{9}{9} = 1$. The probabilities are consistent.

(a) Probability that C will be selected:

We found $P(C) = \frac{2}{9}$.

(b) Probability that A will not be selected:

The event that A will not be selected is the complement of event A. The probability of the complement $A'$ is $P(A') = 1 - P(A)$.

$P(A') = 1 - \frac{4}{9}$

$P(A') = \frac{9}{9} - \frac{4}{9} = \frac{5}{9}$

Given:

The sample space $S = \{\text{John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}\}$.

Let J, R, A, G represent the events that John, Rita, Aslam, or Gurpreet is promoted, respectively.

The given probability relationships are:

The chances of John’s promotion is same as that of Gurpreet: $P(J) = P(G)$.

Rita’s chances of promotion are twice as likely as Johns: $P(R) = 2 P(J)$.

Aslam’s chances are four times that of John: $P(A) = 4 P(J)$.

To Find:

(a) $P(\text{John promoted})$, $P(\text{Rita promoted})$, $P(\text{Aslam promoted})$, $P(\text{Gurpreet promoted})$.

(b) $P(A)$ where $A = \{\text{John promoted or Gurpreet promoted}\}$.

Solution:

The events J, R, A, and G are mutually exclusive and exhaustive (since exactly one person will be promoted). Therefore, the sum of their probabilities is 1.

$P(J) + P(R) + P(A) + P(G) = 1$

We can express the probabilities of R, A, and G in terms of $P(J)$ using the given relationships:

$P(R) = 2 P(J)$

$P(A) = 4 P(J)$

$P(G) = P(J)$

Substitute these expressions into the sum of probabilities equation:

$P(J) + 2 P(J) + 4 P(J) + P(J) = 1$

Combine the terms with $P(J)$:

$(1 + 2 + 4 + 1) P(J) = 1$

$8 P(J) = 1$

Solving for $P(J)$:

$P(J) = \frac{1}{8}$

(a) Determine individual probabilities:

$P(\text{John promoted}) = P(J) = \frac{1}{8}$

$P(\text{Rita promoted}) = P(R) = 2 P(J) = 2 \times \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$

$P(\text{Aslam promoted}) = P(A) = 4 P(J) = 4 \times \frac{1}{8} = \frac{4}{8} = \frac{1}{2}$

$P(\text{Gurpreet promoted}) = P(G) = P(J) = \frac{1}{8}$

(b) Find P(A) where A = {John promoted or Gurpreet promoted}:

The event A is the union of events J and G, i.e., $A = J \cup G$.

Since J and G are mutually exclusive events (only one person can be promoted), the probability of their union is the sum of their individual probabilities.

$P(A) = P(J \cup G) = P(J) + P(G)$

Substitute the probabilities found in part (a):

$P(A) = \frac{1}{8} + \frac{1}{8}$

$P(A) = \frac{1+1}{8} = \frac{2}{8} = \frac{1}{4}$

Solution:

Based on the given Venn diagram, the probabilities of the disjoint regions are:

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

$P(A \cap B \cap C') = 0.06$

$P(A \cap B' \cap C) = 0.04$

$P(A' \cap B \cap C) = 0.05$

$P(A \cap B' \cap C') = 0.10$

$P(A' \cap B \cap C') = 0.13$

$P(A' \cap B' \cap C) = 0.17$

$P(A' \cap B' \cap C') = 0.44$

Let's verify the sum of probabilities: $0.01+0.06+0.04+0.05+0.10+0.13+0.17+0.44 = 1.00$.

(a) Determine P(A):

The probability of event A is the sum of the probabilities of all disjoint regions within A.

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

$P(A) = 0.01 + 0.06 + 0.04 + 0.10$

$P(A) = 0.21$

(b) Determine P(B $\cap$ $\overline{C}$):

$\overline{C}$ is the complement of C, denoted as $C'$. $B \cap C'$ represents the region that is in B but not in C. This consists of the disjoint regions $A \cap B \cap C'$ and $A' \cap B \cap C'$.

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

$P(B \cap C') = 0.06 + 0.13$

$P(B \cap C') = 0.19$

(c) Determine P(A ∪ B):

The probability of the union of A and B is the sum of the probabilities of all disjoint regions within A or B or both.

$P(A \cup B) = P(A \cap B \cap C) + P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C) + P(A \cap B' \cap C') + P(A' \cap B \cap C')$

$P(A \cup B) = 0.01 + 0.06 + 0.04 + 0.05 + 0.10 + 0.13$

$P(A \cup B) = 0.39$

Alternatively, using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:

First, calculate P(B):

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

$P(B) = 0.01 + 0.06 + 0.05 + 0.13 = 0.25$

Next, calculate $P(A \cap B)$. This is the region common to A and B.

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

$P(A \cap B) = 0.01 + 0.06 = 0.07$ (This matches the example given in the question).

Now, apply the formula:

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

$P(A \cup B) = 0.21 + 0.25 - 0.07$

$P(A \cup B) = 0.46 - 0.07 = 0.39$

Both methods yield the same result.

(d) Determine P(A $\cap$ $\overline{B}$):

$\overline{B}$ is the complement of B, denoted as $B'$. $A \cap B'$ represents the region that is in A but not in B. This consists of the disjoint regions $A \cap B' \cap C$ and $A \cap B' \cap C'$.

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

$P(A \cap B') = 0.04 + 0.10$

$P(A \cap B') = 0.14$

(e) Determine P(B $\cap$ C):

The intersection $B \cap C$ is the region common to B and C. This consists of the disjoint regions $A \cap B \cap C$ and $A' \cap B \cap C$.

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

$P(B \cap C) = 0.01 + 0.05$

$P(B \cap C) = 0.06$

(f) Probability of exactly one of the three occurs:

This event corresponds to the union of the regions where only one of the events occurs. These disjoint regions are A only ($A \cap B' \cap C'$), B only ($A' \cap B \cap C'$), and C only ($A' \cap B' \cap C$).

Probability (exactly one) $= P(A \cap B' \cap C') + P(A' \cap B \cap C') + P(A' \cap B' \cap C)$

Probability (exactly one) $= 0.10 + 0.13 + 0.17$

Probability (exactly one) $= 0.40$

Given:

Urn 1 contains: 2 Black balls (B1, B2) and 1 White ball (W). Total 3 balls.

Urn 2 contains: 1 Black ball (B) and 2 White balls (W1, W2). Total 3 balls.

An experiment is performed: Choose an urn at random, then choose two balls without replacement from the chosen urn.

To Find:

(a) The sample space of all possible outcomes.

(b) The probability of choosing two black balls.

(c) The probability of choosing two balls of opposite colour.

Solution:

The experiment has two stages: selecting an urn and then drawing two balls from it without replacement. Since an urn is chosen at random, the probability of selecting Urn 1 is $\frac{1}{2}$, and the probability of selecting Urn 2 is $\frac{1}{2}$.

(a) Write the sample space showing all possible outcomes:

An outcome is represented by the sequence of choices: (Urn chosen, first ball drawn, second ball drawn).

Consider the outcomes when Urn 1 is chosen (containing B1, B2, W). There are 3 choices for the first ball and 2 choices for the second ball, giving $3 \times 2 = 6$ possible ordered draws:

(Urn 1, B1, B2)

(Urn 1, B1, W)

(Urn 1, B2, B1)

(Urn 1, B2, W)

(Urn 1, W, B1)

(Urn 1, W, B2)

Consider the outcomes when Urn 2 is chosen (containing B, W1, W2). There are 3 choices for the first ball and 2 choices for the second ball, giving $3 \times 2 = 6$ possible ordered draws:

(Urn 2, B, W1)

(Urn 2, B, W2)

(Urn 2, W1, B)

(Urn 2, W1, W2)

(Urn 2, W2, B)

(Urn 2, W2, W1)

The complete sample space S is the collection of all these 12 distinct outcomes:

$S = \{(Urn \; 1, B1, B2), (Urn \; 1, B1, W), (Urn \; 1, B2, B1), (Urn \; 1, B2, W), (Urn \; 1, W, B1), (Urn \; 1, W, B2), (Urn \; 2, B, W1), (Urn \; 2, B, W2), (Urn \; 2, W1, B), (Urn \; 2, W1, W2), (Urn \; 2, W2, B), (Urn \; 2, W2, W1)\}$

(b) What is the probability that two black balls are chosen?

Let $E_{BB}$ be the event that two black balls are chosen.

This event occurs if we choose Urn 1 and draw two black balls, OR choose Urn 2 and draw two black balls.

$P(E_{BB}) = P(\text{Urn 1 and 2 Black}) + P(\text{Urn 2 and 2 Black})$

Probability of choosing Urn 1 and drawing 2 black balls:

$P(\text{Urn 1 and 2 Black}) = P(\text{Choose Urn 1}) \times P(\text{Draw 2 Black | Urn 1})$

$P(\text{Choose Urn 1}) = \frac{1}{2}$

In Urn 1 (B1, B2, W), the probability of drawing a black ball first is $\frac{2}{3}$. Given the first is black, the probability of drawing a second black ball from the remaining 2 balls (1 black, 1 white) is $\frac{1}{2}$.

$P(\text{Draw 2 Black | Urn 1}) = P(\text{1st B}) \times P(\text{2nd B | 1st B}) = \frac{2}{3} \times \frac{1}{2} = \frac{2}{6} = \frac{1}{3}$

$P(\text{Urn 1 and 2 Black}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$

Probability of choosing Urn 2 and drawing 2 black balls:

$P(\text{Urn 2 and 2 Black}) = P(\text{Choose Urn 2}) \times P(\text{Draw 2 Black | Urn 2})$

$P(\text{Choose Urn 2}) = \frac{1}{2}$

In Urn 2 (B, W1, W2), there is only 1 black ball. It is impossible to draw two black balls without replacement.

$P(\text{Draw 2 Black | Urn 2}) = 0$

$P(\text{Urn 2 and 2 Black}) = \frac{1}{2} \times 0 = 0$

Summing the probabilities:

$P(E_{BB}) = \frac{1}{6} + 0 = \frac{1}{6}$

(c) What is the probability that two balls of opposite colour are chosen?

Let $E_{Opp}$ be the event that two balls of opposite colour (one black and one white) are chosen.

This can happen by drawing a Black ball then a White ball (BW) or a White ball then a Black ball (WB).

$P(E_{Opp}) = P(\text{Urn 1 and Opposite}) + P(\text{Urn 2 and Opposite})$

Probability of choosing Urn 1 and drawing opposite colours:

$P(\text{Urn 1 and Opposite}) = P(\text{Choose Urn 1}) \times P(\text{Opposite | Urn 1})$

$P(\text{Choose Urn 1}) = \frac{1}{2}$

In Urn 1 (B1, B2, W), $P(\text{Opposite | Urn 1}) = P(\text{BW | Urn 1}) + P(\text{WB | Urn 1})$

$P(\text{BW | Urn 1}) = P(\text{1st B}) \times P(\text{2nd W | 1st B}) = \frac{2}{3} \times \frac{1}{2} = \frac{2}{6} = \frac{1}{3}$

$P(\text{WB | Urn 1}) = P(\text{1st W}) \times P(\text{2nd B | 1st W}) = \frac{1}{3} \times \frac{2}{2} = \frac{2}{6} = \frac{1}{3}$

$P(\text{Opposite | Urn 1}) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

$P(\text{Urn 1 and Opposite}) = \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3}$

Probability of choosing Urn 2 and drawing opposite colours:

$P(\text{Urn 2 and Opposite}) = P(\text{Choose Urn 2}) \times P(\text{Opposite | Urn 2})$

$P(\text{Choose Urn 2}) = \frac{1}{2}$

In Urn 2 (B, W1, W2), $P(\text{Opposite | Urn 2}) = P(\text{BW | Urn 2}) + P(\text{WB | Urn 2})$

$P(\text{BW | Urn 2}) = P(\text{1st B}) \times P(\text{2nd W | 1st B}) = \frac{1}{3} \times \frac{2}{2} = \frac{2}{6} = \frac{1}{3}$

$P(\text{WB | Urn 2}) = P(\text{1st W}) \times P(\text{2nd B | 1st W}) = \frac{2}{3} \times \frac{1}{2} = \frac{2}{6} = \frac{1}{3}$

$P(\text{Opposite | Urn 2}) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

$P(\text{Urn 2 and Opposite}) = \frac{1}{2} \times \frac{2}{3} = \frac{2}{6} = \frac{1}{3}$

Summing the probabilities:

$P(E_{Opp}) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

Final Answers:

(a) The sample space S is:

\{(Urn 1, B1, B2), (Urn 1, B1, W), (Urn 1, B2, B1), (Urn 1, B2, W), (Urn 1, W, B1), (Urn 1, W, B2), (Urn 2, B, W1), (Urn 2, B, W2), (Urn 2, W1, B), (Urn 2, W1, W2), (Urn 2, W2, B), (Urn 2, W2, W1)\}

(b) The probability that two black balls are chosen is $\frac{1}{6}$.

(c) The probability that two balls of opposite colour are chosen is $\frac{2}{3}$.

Given:

Number of red balls = 8

Number of white balls = 5

Total number of balls = 8 + 5 = 13

Three balls are drawn at random from the bag.

To Find:

(a) The probability that all three balls drawn are white.

(b) The probability that all three balls drawn are red.

(c) The probability that one ball is red and two balls are white.

Solution:

When three balls are drawn at random from the bag, the total number of possible outcomes is the number of ways to choose 3 balls from 13, without regard to order. This can be calculated using combinations:

Total number of ways to draw 3 balls from 13 = $\binom{13}{3}$

$\binom{13}{3} = \frac{13!}{3!(13-3)!} = \frac{13!}{3!10!}$

$\binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 13 \times 2 \times 11 = 286$

Total number of possible outcomes in the sample space = 286.

(a) Probability that all three balls are white:

This event occurs if we choose 3 white balls from the 5 available white balls.

Number of ways to choose 3 white balls from 5 = $\binom{5}{3}$

$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$

Number of favorable outcomes for event (a) = 10.

The probability of this event is the number of favorable outcomes divided by the total number of possible outcomes.

P(All three balls are white) = $\frac{\text{Number of ways to choose 3 white}}{\text{Total number of ways to choose 3}}$

P(All three balls are white) = $\frac{10}{286}$

Simplifying the fraction by dividing both numerator and denominator by 2:

P(All three balls are white) = $\frac{\cancel{10}^{5}}{\cancel{286}_{143}} = \frac{5}{143}$

(b) Probability that all three balls are red:

This event occurs if we choose 3 red balls from the 8 available red balls.

Number of ways to choose 3 red balls from 8 = $\binom{8}{3}$

$\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Number of favorable outcomes for event (b) = 56.

P(All three balls are red) = $\frac{\text{Number of ways to choose 3 red}}{\text{Total number of ways to choose 3}}$

P(All three balls are red) = $\frac{56}{286}$

Simplifying the fraction by dividing both numerator and denominator by 2:

P(All three balls are red) = $\frac{\cancel{56}^{28}}{\cancel{286}_{143}} = \frac{28}{143}$

(c) Probability that one ball is red and two balls are white:

This event occurs if we choose 1 red ball from the 8 red balls AND 2 white balls from the 5 white balls.

Number of ways to choose 1 red ball from 8 = $\binom{8}{1} = 8$

Number of ways to choose 2 white balls from 5 = $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$

The number of ways to choose 1 red and 2 white balls is the product of the number of ways to make each selection:

Number of favorable outcomes for event (c) = $\binom{8}{1} \times \binom{5}{2} = 8 \times 10 = 80$

P(One red and two white) = $\frac{\text{Number of ways to choose 1 red and 2 white}}{\text{Total number of ways to choose 3}}$

P(One red and two white) = $\frac{80}{286}$

Simplifying the fraction by dividing both numerator and denominator by 2:

P(One red and two white) = $\frac{\cancel{80}^{40}}{\cancel{286}_{143}} = \frac{40}{143}$

Final Answers:

(a) P(All three balls are white) = $\frac{5}{143}$.

(b) P(All three balls are red) = $\frac{28}{143}$.

(c) P(One ball is red and two balls are white) = $\frac{40}{143}$.

Given:

The word is ASSASSINATION.

The letters are arranged at random in a row.

The letters in the word are A(3), S(4), I(2), N(2), T(1), O(1).

Total number of letters = $3 + 4 + 2 + 2 + 1 + 1 = 13$.

To Find:

(a) Probability that the four S’s come consecutively.

(b) Probability that the two I’s and two N’s come together.

(c) Probability that all A’s are not coming together.

(d) Probability that no two A’s are coming together.

Solution:

First, calculate the total number of distinct arrangements of the letters in the word ASSASSINATION.

The word has 13 letters with repetitions: 3 A's, 4 S's, 2 I's, 2 N's, 1 T, 1 O.

The total number of permutations is given by the formula $\frac{n!}{n_1! n_2! \dots n_k!}$, where $n$ is the total number of items and $n_i$ is the count of identical items of type $i$.

Total number of arrangements $= \frac{13!}{3! 4! 2! 2! 1! 1!} = \frac{13!}{3! 4! 2! 2!}$

Let $N$ denote the total number of arrangements.

$N = \frac{13!}{3! 4! 2! 2!} = \frac{6,227,020,800}{(6)(24)(2)(2)} = \frac{6,227,020,800}{576} = 10,810,800$

(a) Probability that the four S’s come consecutively:

Treat the four S's as a single block (SSSS). Now we are arranging 10 entities: the block (SSSS) and the remaining 9 letters (A, A, A, I, I, N, N, T, O).

The number of arrangements of these 10 entities, considering the repetitions of A (3 times), I (2 times), and N (2 times), is:

Number of favorable arrangements $= \frac{10!}{3! 2! 2! 1! 1!} = \frac{10!}{3! 2! 2!}$

Let $N_a$ be the number of favorable arrangements.

$N_a = \frac{3,628,800}{(6)(2)(2)} = \frac{3,628,800}{24} = 151,200$

The probability is the ratio of favorable arrangements to the total number of arrangements.

$P(\text{SSSS together}) = \frac{N_a}{N} = \frac{151200}{10810800}$

Simplifying the fraction:

$P(\text{SSSS together}) = \frac{\cancel{151200}^{1512}}{\cancel{10810800}_{108108}} = \frac{\cancel{1512}^{126}}{\cancel{108108}_{9009}} = \frac{\cancel{126}^{14}}{\cancel{9009}_{1001}} = \frac{\cancel{14}^{2}}{\cancel{1001}_{143}} = \frac{2}{143}$

(b) Probability that the two I’s and two N’s come together:

Interpret this as the two I's forming a unit (II) and the two N's forming a unit (NN). Now we are arranging 11 entities: the blocks (II), (NN), and the remaining 9 letters (A, A, A, S, S, S, S, T, O).

The number of arrangements of these 11 entities, considering the repetitions of A (3 times) and S (4 times), is:

Number of favorable arrangements $= \frac{11!}{3! 4! 1! 1! 1!} = \frac{11!}{3! 4!}$

Let $N_b$ be the number of favorable arrangements.

$N_b = \frac{39,916,800}{(6)(24)} = \frac{39,916,800}{144} = 277,200$

The probability is the ratio of favorable arrangements to the total number of arrangements.

$P(\text{II and NN together}) = \frac{N_b}{N} = \frac{277200}{10810800}$

Simplifying the fraction:

$P(\text{II and NN together}) = \frac{\cancel{277200}^{2772}}{\cancel{10810800}_{108108}} = \frac{\cancel{2772}^{231}}{\cancel{108108}_{9009}} = \frac{\cancel{231}^{1}}{\cancel{9009}_{39}} = \frac{1}{39}$

(c) Probability that all A’s are not coming together:

This is the complement of the event that all A's come together. Let $E'$ be the event that all A's are coming together.

$P(\text{All A's not together}) = 1 - P(E')$

To find $P(E')$, treat the three A's as a single block (AAA). Now we are arranging 11 entities: the block (AAA) and the remaining 10 letters (S, S, S, S, I, I, N, N, T, O).

The number of arrangements of these 11 entities, considering the repetitions of S (4 times), I (2 times), and N (2 times), is:

Number of arrangements with AAA together $= \frac{11!}{4! 2! 2! 1! 1!} = \frac{11!}{4! 2! 2!}$

Let $N_{c'}$ be the number of favorable arrangements for $E'$.

$N_{c'} = \frac{39,916,800}{(24)(2)(2)} = \frac{39,916,800}{96} = 415,800$

The probability of event $E'$ is:

$P(E') = \frac{N_{c'}}{N} = \frac{415800}{10810800}$

Simplifying the fraction:

$P(E') = \frac{\cancel{415800}^{4158}}{\cancel{10810800}_{108108}} = \frac{\cancel{4158}^{231}}{\cancel{108108}_{6006}} = \frac{\cancel{231}^{77}}{\cancel{6006}_{2002}} = \frac{\cancel{77}^{11}}{\cancel{2002}_{286}} = \frac{\cancel{11}^{1}}{\cancel{286}_{26}} = \frac{1}{26}$

Now calculate the probability that all A's are not coming together:

$P(\text{All A's not together}) = 1 - P(E') = 1 - \frac{1}{26} = \frac{26 - 1}{26} = \frac{25}{26}$

(d) Probability that no two A’s are coming together:

To ensure no two A's are adjacent, first arrange the other 10 letters (S, S, S, S, I, I, N, N, T, O).

Number of arrangements of the 10 non-A letters $= \frac{10!}{4! 2! 2! 1! 1!} = \frac{10!}{4! 2! 2!}$

$= \frac{3,628,800}{(24)(2)(2)} = \frac{3,628,800}{96} = 37,800$

When these 10 letters are arranged, they create 11 spaces where the three A's can be placed so that no two A's are adjacent.

_ L _ L _ L _ L _ L _ L _ L _ L _ L _ L _

Here, 'L' represents a non-A letter, and '_' represents a potential space for an A.

We need to choose 3 of these 11 spaces for the three identical A's. The number of ways to choose 3 spaces from 11 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Number of ways to choose spaces for A's $= \binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11!}{3!8!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 11 \times 5 \times 3 = 165$

The number of favorable arrangements (where no two A's are together) is the product of the number of ways to arrange the non-A letters and the number of ways to place the A's in the spaces.

Number of favorable arrangements $= (\text{Arrangements of non-A letters}) \times (\text{Ways to place A's})$

Number of favorable arrangements $= 37800 \times 165 = 6,237,000$

Let $N_d$ be the number of favorable arrangements.

$P(\text{No two A's together}) = \frac{N_d}{N} = \frac{6237000}{10810800}$

Simplifying the fraction:

$P(\text{No two A's together}) = \frac{\cancel{6237000}^{62370}}{\cancel{10810800}_{108108}} = \frac{\cancel{62370}^{3465}}{\cancel{108108}_{6006}} = \frac{\cancel{3465}^{1155}}{\cancel{6006}_{2002}} = \frac{\cancel{1155}^{165}}{\cancel{2002}_{286}} = \frac{\cancel{165}^{15}}{\cancel{286}_{26}} = \frac{15}{26}$

Given:

A standard deck of 52 cards.

One card is drawn at random.

To Find:

The probability of getting a king or a heart or a red card.

Solution:

Let S be the sample space of drawing a card from a deck of 52 cards.

Total number of outcomes in S is $|S| = 52$.

Let K be the event of getting a King.

Let H be the event of getting a Heart.

Let R be the event of getting a Red card.

We want to find the probability of the union of these three events, $P(K \cup H \cup R)$.

The number of outcomes in each event and their intersections are:

Number of Kings, $|K| = 4$

Number of Hearts, $|H| = 13$

Number of Red cards, $|R| = 26$ (13 Hearts + 13 Diamonds)

Number of Kings that are Hearts, $|K \cap H| = 1$ (King of Hearts)

Number of Kings that are Red, $|K \cap R| = 2$ (King of Hearts, King of Diamonds)

Number of Hearts that are Red, $|H \cap R| = 13$ (All Hearts are Red)

Number of cards that are King AND Heart AND Red, $|K \cap H \cap R| = 1$ (King of Hearts)

Using the Principle of Inclusion-Exclusion for three events:

$P(K \cup H \cup R) = P(K) + P(H) + P(R) - P(K \cap H) - P(K \cap R) - P(H \cap R) + P(K \cap H \cap R)$

The probabilities are:

$P(K) = \frac{|K|}{|S|} = \frac{4}{52}$

$P(H) = \frac{|H|}{|S|} = \frac{13}{52}$

$P(R) = \frac{|R|}{|S|} = \frac{26}{52}$

$P(K \cap H) = \frac{|K \cap H|}{|S|} = \frac{1}{52}$

$P(K \cap R) = \frac{|K \cap R|}{|S|} = \frac{2}{52}$

$P(H \cap R) = \frac{|H \cap R|}{|S|} = \frac{13}{52}$

$P(K \cap H \cap R) = \frac{|K \cap H \cap R|}{|S|} = \frac{1}{52}$

Substitute these values into the formula:

$P(K \cup H \cup R) = \frac{4}{52} + \frac{13}{52} + \frac{26}{52} - \frac{1}{52} - \frac{2}{52} - \frac{13}{52} + \frac{1}{52}$

$P(K \cup H \cup R) = \frac{4 + 13 + 26 - 1 - 2 - 13 + 1}{52}$

$P(K \cup H \cup R) = \frac{44 - 16}{52} = \frac{28}{52}$

Simplify the fraction:

$P(K \cup H \cup R) = \frac{\cancel{28}^{7}}{\cancel{52}_{13}} = \frac{7}{13}$

Alternate Solution:

Notice that all Heart cards are also Red cards. This means the set of Hearts (H) is a subset of the set of Red cards (R), i.e., $H \subseteq R$.

When finding the union of events K, H, and R, since H is included in R, the event "Heart or Red" ($H \cup R$) is simply the event "Red" (R).

Therefore, the event "King or Heart or Red" ($K \cup H \cup R$) is equivalent to the event "King or Red" ($K \cup R$).

We need to find $P(K \cup R)$. Using the formula for the union of two events:

$P(K \cup R) = P(K) + P(R) - P(K \cap R)$

We have:

$P(K) = \frac{4}{52}$

$P(R) = \frac{26}{52}$

$P(K \cap R)$ is the probability of getting a King and a Red card. There are 2 Red Kings (King of Hearts, King of Diamonds).

$P(K \cap R) = \frac{2}{52}$

Substitute these values into the formula:

$P(K \cup R) = \frac{4}{52} + \frac{26}{52} - \frac{2}{52}$

$P(K \cup R) = \frac{4 + 26 - 2}{52} = \frac{30 - 2}{52} = \frac{28}{52}$

Simplify the fraction:

$P(K \cup R) = \frac{\cancel{28}^{7}}{\cancel{52}_{13}} = \frac{7}{13}$

The probability of getting a king or a heart or a red card is $\frac{7}{13}$.

Given:

Sample space $S = \{e_1, e_2, e_3, e_4, e_5, e_6, e_7, e_8, e_9\}$.

Probabilities of elementary outcomes:

$P(e_1) = 0.08$, $P(e_2) = 0.08$

$P(e_3) = 0.10$, $P(e_4) = 0.10$, $P(e_5) = 0.10$

$P(e_6) = 0.20$, $P(e_7) = 0.20$

$P(e_8) = 0.07$, $P(e_9) = 0.07$

Events $A = \{e_1, e_5, e_8\}$ and $B = \{e_2, e_5, e_8, e_9\}$.

To Find:

(a) $P(A)$, $P(B)$, and $P(A \cap B)$.

(b) $P(A \cup B)$ using the addition law.

(c) Composition of $A \cup B$ and $P(A \cup B)$ by adding probabilities.

(d) $P(\overline{B})$ from $P(B)$ and directly from elementary outcomes of $\overline{B}$.

Solution:

The probability of an event is the sum of the probabilities of the elementary outcomes comprising the event.

(a) Calculate P(A), P(B), and P(A $\cap$ B):

$A = \{e_1, e_5, e_8\}$

$P(A) = P(e_1) + P(e_5) + P(e_8)$

$P(A) = 0.08 + 0.10 + 0.07 = 0.25$

$B = \{e_2, e_5, e_8, e_9\}$

$P(B) = P(e_2) + P(e_5) + P(e_8) + P(e_9)$

$P(B) = 0.08 + 0.10 + 0.07 + 0.07 = 0.32$

The intersection of A and B, $A \cap B$, consists of the elementary outcomes common to both A and B.

$A \cap B = \{e_1, e_5, e_8\} \cap \{e_2, e_5, e_8, e_9\} = \{e_5, e_8\}$

$P(A \cap B) = P(e_5) + P(e_8)$

$P(A \cap B) = 0.10 + 0.07 = 0.17$

(b) Using the addition law of probability, calculate P(A $\cup$ B):

The addition law of probability states that for any two events A and B:

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

Using the values calculated in part (a):

$P(A \cup B) = 0.25 + 0.32 - 0.17$

$P(A \cup B) = 0.57 - 0.17 = 0.40$

(c) List the composition of the event A $\cup$ B, and calculate P(A $\cup$ B) by adding the probabilities of the elementary outcomes:

The union of A and B, $A \cup B$, consists of the elementary outcomes that are in A or in B or in both.

$A \cup B = \{e_1, e_5, e_8\} \cup \{e_2, e_5, e_8, e_9\} = \{e_1, e_2, e_5, e_8, e_9\}$

Now, calculate $P(A \cup B)$ by summing the probabilities of these elementary outcomes:

$P(A \cup B) = P(e_1) + P(e_2) + P(e_5) + P(e_8) + P(e_9)$

$P(A \cup B) = 0.08 + 0.08 + 0.10 + 0.07 + 0.07$

$P(A \cup B) = 0.16 + 0.10 + 0.14 = 0.40$

This result matches the result obtained using the addition law in part (b).

(d) Calculate P($\overline{B}$) from P(B), also calculate P($\overline{B}$) directly from the elementary outcomes of $\overline{B}$:

Calculate $P(\overline{B})$ from $P(B)$ using the complement rule: $P(\overline{B}) = 1 - P(B)$.

From part (a), $P(B) = 0.32$.

$P(\overline{B}) = 1 - 0.32 = 0.68$

Now, calculate $P(\overline{B})$ directly from the elementary outcomes of $\overline{B}$. The complement of B, $\overline{B}$, consists of all elementary outcomes in the sample space S that are not in B.

$S = \{e_1, e_2, e_3, e_4, e_5, e_6, e_7, e_8, e_9\}$

$B = \{e_2, e_5, e_8, e_9\}$

$\overline{B} = \{e_1, e_3, e_4, e_6, e_7\}$

Calculate $P(\overline{B})$ by summing the probabilities of these elementary outcomes:

$P(\overline{B}) = P(e_1) + P(e_3) + P(e_4) + P(e_6) + P(e_7)$

$P(\overline{B}) = 0.08 + 0.10 + 0.10 + 0.20 + 0.20$

$P(\overline{B}) = 0.08 + 0.20 + 0.40 = 0.68$

Both methods give the same result for $P(\overline{B})$.

Solution (a):

Given:

A single toss of a fair die.

Event: An odd number appears.

To Find:

The probability of getting an odd number.

Solution:

The sample space for a single toss of a fair die is $S = \{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes is $|S| = 6$.

Let E be the event that an odd number appears. The odd numbers in the sample space are 1, 3, and 5.

The favorable outcomes for event E are $\{1, 3, 5\}$.

The number of favorable outcomes is $|E| = 3$.

The probability of event E is given by:

$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$p = \frac{|E|}{|S|} = \frac{3}{6}$

Simplify the fraction:

$p = \frac{\cancel{3}^{1}}{\cancel{6}_{2}} = \frac{1}{2}$

Solution (b):

Given:

Two tosses of a fair coin.

Event: At least one head appears.

To Find:

The probability of getting at least one head.

Solution:

The sample space for two tosses of a fair coin is $S = \{HH, HT, TH, TT\}$.

The total number of possible outcomes is $|S| = 4$.

Let E be the event that at least one head appears. The outcomes in E are those that contain at least one H.

The favorable outcomes for event E are $\{HH, HT, TH\}$.

The number of favorable outcomes is $|E| = 3$.

The probability of event E is given by:

$p = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$p = \frac{|E|}{|S|} = \frac{3}{4}$

Solution (c):

Given:

Drawing a single card from a well-shuffled ordinary deck of 52 cards.

Event: Getting a king or a 9 of hearts or a 3 of spades.

To Find:

The probability of getting a king or a 9 of hearts or a 3 of spades.

Solution:

The total number of possible outcomes when drawing a single card from a deck of 52 is 52.

Total number of possible outcomes = 52.

Let K be the event of getting a King.

Let H9 be the event of getting the 9 of hearts.

Let S3 be the event of getting the 3 of spades.

We are looking for the probability of the event $K \cup H9 \cup S3$.

Number of Kings in a deck = 4.

Number of 9 of hearts = 1.

Number of 3 of spades = 1.

These three events (getting a King, getting the 9 of hearts, getting the 3 of spades) are mutually exclusive, as no card can be both a King, a 9 of hearts, and a 3 of spades simultaneously.

The number of favorable outcomes is the sum of the number of outcomes in each event:

Number of favorable outcomes $= |K| + |H9| + |S3| = 4 + 1 + 1 = 6$.

The probability is given by:

$p = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$p = \frac{6}{52}$

Simplify the fraction:

$p = \frac{\cancel{6}^{3}}{\cancel{52}_{26}} = \frac{3}{26}$

Solution (d):

Given:

A single toss of a pair of fair dice.

Event: The sum of the numbers is 6.

To Find:

The probability that the sum of the numbers is 6.

Solution:

When a pair of fair dice is tossed, the total number of possible outcomes is $6 \times 6 = 36$.

Total number of possible outcomes = 36.

Let E be the event that the sum of the numbers is 6. The pairs of numbers that sum to 6 are:

$(1, 5)$ (Die 1 shows 1, Die 2 shows 5)

$(2, 4)$

$(3, 3)$

$(4, 2)$

$(5, 1)$

The favorable outcomes for event E are $\{(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)\}$.

The number of favorable outcomes is $|E| = 5$.

The probability of event E is given by:

$p = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$p = \frac{|E|}{|S|} = \frac{5}{36}$

Given:

A non-leap year.

The event is having 53 Tuesdays or 53 Wednesdays.

To Find:

The probability of the given event.

Solution:

A non-leap year has 365 days.

We know that $365 = 52 \times 7 + 1$.

This means a non-leap year has 52 full weeks and 1 extra day.

The 52 full weeks contain exactly 52 Tuesdays and 52 Wednesdays.

The occurrence of a 53rd Tuesday or a 53rd Wednesday depends entirely on the nature of the single extra day.

The possible outcomes for the extra day are the 7 days of the week: {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}.

Total number of possible outcomes for the extra day = 7.

Let A be the event that the extra day is a Tuesday. If the extra day is Tuesday, the year will have 53 Tuesdays.

The number of favorable outcomes for event A = 1 (Tuesday).

$P(\text{53 Tuesdays}) = P(\text{extra day is Tuesday}) = \frac{1}{7}$.

Let B be the event that the extra day is a Wednesday. If the extra day is Wednesday, the year will have 53 Wednesdays.

The number of favorable outcomes for event B = 1 (Wednesday).

$P(\text{53 Wednesdays}) = P(\text{extra day is Wednesday}) = \frac{1}{7}$.

We are looking for the probability of having 53 Tuesdays OR 53 Wednesdays. This is the probability of the union of events A and B, i.e., $P(A \cup B)$.

Events A and B are mutually exclusive because the extra day cannot be both Tuesday and Wednesday simultaneously.

For mutually exclusive events, $P(A \cup B) = P(A) + P(B)$.

$P(\text{53 Tuesdays or 53 Wednesdays}) = P(\text{53 Tuesdays}) + P(\text{53 Wednesdays})$

$P(A \cup B) = \frac{1}{7} + \frac{1}{7} = \frac{2}{7}$.

The probability of having 53 Tuesdays or 53 Wednesdays in a non-leap year is $\frac{2}{7}$.

This corresponds to option (B).

Given:

Three numbers are chosen from the integers 1 to 20.

To Find:

The probability that the three chosen numbers are not consecutive.

Solution:

Let S be the sample space of choosing 3 numbers from the 20 integers {1, 2, ..., 20}. The total number of ways to choose 3 numbers from 20 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Total number of outcomes = $\binom{20}{3}$

$\binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = \frac{6840}{6} = 1140$

Total number of possible outcomes = 1140.

Let A be the event that the three chosen numbers are consecutive.

Consecutive sets of 3 numbers from 1 to 20 are of the form $(n, n+1, n+2)$.

The possible values for the first number $n$ are from 1 up to the point where $n+2 \leq 20$.

$n+2 \leq 20 \implies n \leq 18$.

So, the possible consecutive sets are:

(1, 2, 3)

(2, 3, 4)

...

(18, 19, 20)

The number of such consecutive sets is equal to the number of possible values for $n$, which is 18.

Number of favorable outcomes for event A = 18.

The probability of event A (getting 3 consecutive numbers) is:

$P(A) = \frac{\text{Number of consecutive sets}}{\text{Total number of sets}} = \frac{18}{1140}$

We are asked to find the probability that the three chosen numbers are not consecutive. This is the complement of event A, denoted as $A'$.

$P(A') = P(\text{not consecutive}) = 1 - P(A)$

$P(A') = 1 - \frac{18}{1140}$

$P(A') = \frac{1140 - 18}{1140} = \frac{1122}{1140}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor. Both are divisible by 6.

$1122 \div 6 = 187$

$1140 \div 6 = 190$

$P(A') = \frac{\cancel{1122}^{187}}{\cancel{1140}_{190}} = \frac{187}{190}$

The probability that the three chosen numbers are not consecutive is $\frac{187}{190}$.

Comparing this result with the given options, we find that it matches option (B).

Given:

A pack of 52 playing cards.

2 cards are accidentally dropped (chosen) from the pack.

To Find:

The probability that the two missing cards are of different colours.

Solution:

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

When 2 cards are dropped, we are essentially choosing 2 cards from the deck without replacement and without considering the order.

The total number of ways to choose 2 cards from 52 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Total number of ways to choose 2 cards from 52 = $\binom{52}{2}$

$\binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326$

Total number of possible outcomes = 1326.

We want to find the number of ways to choose 2 cards such that they are of different colours. This means one card is red and the other card is black.

Number of ways to choose 1 red card from 26 red cards = $\binom{26}{1} = 26$.

Number of ways to choose 1 black card from 26 black cards = $\binom{26}{1} = 26$.

The number of ways to choose 1 red card and 1 black card is the product of the number of ways to make each choice.

Number of favorable outcomes = $\binom{26}{1} \times \binom{26}{1} = 26 \times 26 = 676$

The probability that the two missing cards are of different colours is the ratio of the number of favorable outcomes to the total number of possible outcomes.

Probability (different colours) = $\frac{\text{Number of ways to choose 1 red and 1 black}}{\text{Total number of ways to choose 2 cards}}$

Probability = $\frac{676}{1326}$

Simplify the fraction by dividing both numerator and denominator by their greatest common divisor. Both are divisible by 26.

$676 \div 26 = 26$

$1326 \div 26 = 51$

Probability = $\frac{\cancel{676}^{26}}{\cancel{1326}_{51}} = \frac{26}{51}$

The probability that the missing cards are of different colours is $\frac{26}{51}$.

This corresponds to option (C).

Given:

Seven persons are to be seated in a row.

To Find:

The probability that two particular persons sit next to each other.

Solution:

Let the seven persons be $P_1, P_2, P_3, P_4, P_5, P_6, P_7$. Let the two particular persons be $P_1$ and $P_2$.

Total Number of Arrangements:

The total number of ways to arrange 7 distinct persons in a row is $7!$.

Total number of arrangements $= 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$

Favorable Arrangements (Two particular persons sit together):

Consider the two particular persons ($P_1$ and $P_2$) as a single unit. Now we are arranging this unit and the remaining 5 persons, making a total of 6 entities to arrange.

The number of ways to arrange these 6 entities is $6!$.

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

Within the unit of the two particular persons, they can be arranged in $2!$ ways ($P_1P_2$ or $P_2P_1$).

$2! = 2 \times 1 = 2$

The total number of arrangements where the two particular persons sit next to each other is the product of the number of ways to arrange the 6 entities and the number of ways to arrange the persons within the unit.

Number of favorable arrangements $= 6! \times 2! = 720 \times 2 = 1440$

Probability:

The probability that the two particular persons sit next to each other is the ratio of the number of favorable arrangements to the total number of arrangements.

Probability $= \frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}}$

Probability $= \frac{6! \times 2!}{7!}$

We can write $7!$ as $7 \times 6!$.

Probability $= \frac{6! \times 2}{7 \times 6!}$

Cancel out $6!$ from the numerator and denominator.

Probability $= \frac{2}{7}$

The probability that two particular persons sit next to each other when seven persons are seated in a row is $\frac{2}{7}$.

This corresponds to option (C).

Given:

The digits available are 0, 2, 3, 5.

Four-digit numbers are formed using these digits without repetition.

To Find:

The probability that a randomly formed four-digit number is divisible by 5.

Solution:

First, we need to find the total number of four-digit numbers that can be formed using the digits 0, 2, 3, 5 without repetition. A four-digit number cannot have 0 in the thousands place.

Number of choices for the thousands place = 3 (can be 2, 3, or 5).

Number of choices for the hundreds place = 3 (any of the remaining 3 digits).

Number of choices for the tens place = 2 (any of the remaining 2 digits).

Number of choices for the units place = 1 (the last remaining digit).

Total number of four-digit numbers = $3 \times 3 \times 2 \times 1 = 18$.

Total number of possible outcomes = 18.

Next, we need to find the number of these four-digit numbers that are divisible by 5. A number is divisible by 5 if its units digit is 0 or 5.

Case 1: The units digit is 0.

Units place: 1 choice (0).

Thousands place: 3 choices (2, 3, 5, as 0 is used).

Hundreds place: 2 choices (remaining 2 digits).

Tens place: 1 choice (remaining 1 digit).

Number of four-digit numbers ending in 0 = $1 \times 3 \times 2 \times 1 = 6$.

Case 2: The units digit is 5.

Units place: 1 choice (5).

Thousands place: 2 choices (2 or 3, as 0 cannot be in the thousands place and 5 is used).

Hundreds place: 2 choices (0 and the remaining digit from {2, 3}).

Tens place: 1 choice (the last remaining digit).

Number of four-digit numbers ending in 5 = $1 \times 2 \times 2 \times 1 = 4$.

Total number of favorable outcomes (numbers divisible by 5) = (Numbers ending in 0) + (Numbers ending in 5) = $6 + 4 = 10$.

The probability that a randomly formed number is divisible by 5 is the ratio of the number of favorable outcomes to the total number of possible outcomes.

Probability = $\frac{\text{Number of numbers divisible by 5}}{\text{Total number of 4-digit numbers}}$

Probability = $\frac{10}{18}$

Simplify the fraction:

Probability = $\frac{\cancel{10}^{5}}{\cancel{18}_{9}} = \frac{5}{9}$

The probability that the number is divisible by 5 is $\frac{5}{9}$.

This corresponds to option (D).

Given:

A and B are mutually exclusive events.

To Find:

The correct relationship between $P(A)$ and $P(\overline{B})$.

Solution:

Since A and B are mutually exclusive events, their intersection is the empty set.

$A \cap B = \emptyset$

This means that if event A occurs, event B cannot occur. If event B does not occur, then its complement $\overline{B}$ must occur.

Consider an outcome $\omega$ in the sample space.

If $\omega \in A$, then since A and B are mutually exclusive, $\omega \notin B$.

If $\omega \notin B$, then $\omega \in \overline{B}$.

Thus, any outcome that is in A is also in $\overline{B}$. This implies that event A is a subset of event $\overline{B}$.

$A \subseteq \overline{B}$

For any two events $E_1$ and $E_2$ in the same sample space, if $E_1 \subseteq E_2$, then the probability of $E_1$ is less than or equal to the probability of $E_2$.

Since $A \subseteq \overline{B}$, we have:

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

The correct relationship is $P(A) \leq P(\overline{B})$.

This corresponds to option (A).

Given:

For two events A and B, the probability of their union is equal to the probability of their intersection: $P(A \cup B) = P(A \cap B)$.

To Determine:

Which of the given statements must be true based on the given condition.

Solution:

We use the Addition Law of Probability for any two events A and B:

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

Given that $P(A \cup B) = P(A \cap B)$, we can substitute this into the addition law equation:

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

Add $P(A \cap B)$ to both sides of the equation:

$2 P(A \cap B) = P(A) + P(B)$

We know that the probability of the intersection of two events is always less than or equal to the probability of each individual event:

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

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

From $2 P(A \cap B) = P(A) + P(B)$ and $P(A \cap B) \leq P(A)$, we get:

$P(A) + P(B) \leq 2 P(A)$

$P(B) \leq P(A)$

Similarly, from $2 P(A \cap B) = P(A) + P(B)$ and $P(A \cap B) \leq P(B)$, we get:

$P(A) + P(B) \leq 2 P(B)$

$P(A) \leq P(B)$

Since both $P(B) \leq P(A)$ and $P(A) \leq P(B)$ must be true, we conclude that:

$P(A) = P(B)$

Substitute $P(B) = P(A)$ back into the equation $2 P(A \cap B) = P(A) + P(B)$:

$2 P(A \cap B) = P(A) + P(A)$

$2 P(A \cap B) = 2 P(A)$

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

Since $P(A) = P(B)$ and $P(A) = P(A \cap B)$, we also have $P(B) = P(A \cap B)$.

The given condition $P(A \cup B) = P(A \cap B)$ is equivalent to $P(A) = P(B) = P(A \cap B)$.

This implies that the probability of event A is equal to the probability of event B, and all of this probability is contained within their intersection. This means the events A and B are probabilistically equivalent (they differ only by sets of probability zero), i.e., $P(A \Delta B) = 0$.

Now let's examine the given options. We need to determine which relationship between $P(A)$ and $P(\overline{B})$ must be true given $P(A) = P(B)$.

The probability of the complement of B is $P(\overline{B}) = 1 - P(B)$. Since $P(B) = P(A)$, we have $P(\overline{B}) = 1 - P(A)$.

Let $p = P(A)$. Then $P(B) = p$ and $P(\overline{B}) = 1 - p$. The condition $P(A) = P(B) = P(A \cap B)$ is satisfied if $A$ and $B$ are the same event with probability $p$, where $p$ can be any value such that $0 \leq p \leq 1$.

Now let's check the options by comparing $p$ and $1-p$:

(A) $P(A) \leq P(\overline{B}) \implies p \leq 1 - p \implies 2p \leq 1 \implies p \leq 0.5$. This is true if $P(A) \leq 0.5$, but not necessarily true for all $P(A)$ (e.g., if $P(A)=0.6$).

(B) $P(A) \geq P(\overline{B}) \implies p \geq 1 - p \implies 2p \geq 1 \implies p \geq 0.5$. This is true if $P(A) \geq 0.5$, but not necessarily true for all $P(A)$ (e.g., if $P(A)=0.4$).

(C) $P(A) < P(\overline{B}) \implies p < 1 - p \implies 2p < 1 \implies p < 0.5$. This is true if $P(A) < 0.5$, but not necessarily true for all $P(A)$ (e.g., if $P(A)=0.5$ or $P(A)=0.6$).

Since the condition $P(A \cup B) = P(A \cap B)$ implies $P(A) = P(B)$, but does not restrict the value of $P(A)$ to be either $\leq 0.5$ or $\geq 0.5$ or $< 0.5$, none of the inequalities in options (A), (B), or (C) must necessarily be true.

Therefore, the correct answer is (D).

The final answer is $\boxed{D}$.

Given:

6 boys and 6 girls are to sit in a row.

Total number of people = 6 boys + 6 girls = 12.

To Find:

The probability that all the girls sit together.

Solution:

Let's assume all boys are distinct and all girls are distinct for counting permutations.

Total Number of Arrangements:

The total number of ways to arrange 12 distinct people in a row is given by the factorial of the total number of people.

Total number of arrangements $= 12!$

Favorable Arrangements (All girls sit together):

To count the arrangements where all 6 girls sit together, we treat the group of 6 girls as a single unit or block.

Now, we are arranging the 6 boys and this single unit of 6 girls. This gives us a total of $6 + 1 = 7$ entities to arrange.

The number of ways to arrange these 7 distinct entities is $7!$.

Within the unit of 6 girls, the girls themselves can be arranged in $6!$ ways.

The total number of favorable arrangements where all girls sit together is the product of the number of ways to arrange the 7 entities and the number of ways to arrange the girls within their unit.

Number of favorable arrangements $= (\text{Number of ways to arrange 7 entities}) \times (\text{Number of ways to arrange 6 girls within unit})$

Number of favorable arrangements $= 7! \times 6!$

Probability:

The probability that all the girls sit together is the ratio of the number of favorable arrangements to the total number of arrangements.

Probability $= \frac{\text{Number of favorable arrangements}}{\text{Total number of arrangements}}$

Probability $= \frac{7! \times 6!}{12!}$

We can simplify this expression:

Recall that $12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7!$

Probability $= \frac{7! \times 6!}{12 \times 11 \times 10 \times 9 \times 8 \times 7!}$

Cancel out $7!$ from the numerator and the denominator:

Probability $= \frac{6!}{12 \times 11 \times 10 \times 9 \times 8}$

Substitute the value of $6! = 720$:

Probability $= \frac{720}{12 \times 11 \times 10 \times 9 \times 8}$

Calculate the product in the denominator: $12 \times 11 \times 10 \times 9 \times 8 = 95,040$

Probability $= \frac{720}{95040}$

Simplify the fraction:

Probability $= \frac{\cancel{720}^{72}}{\cancel{95040}_{9504}}$

Now simplify $\frac{72}{9504}$ by dividing both numerator and denominator by 72:

$72 \div 72 = 1$

$9504 \div 72 = 132$

Probability $= \frac{1}{132}$

The probability that all the girls sit together is $\frac{1}{132}$.

This corresponds to option (C).

Given:

A single letter is selected at random from the word ‘PROBABILITY’.

To Find:

The probability that the selected letter is a vowel.

Solution:

First, list the letters in the word ‘PROBABILITY’ and count the total number of letters.

The letters are P, R, O, B, A, B, I, L, I, T, Y.

Total number of letters in the word = 11.

Now, identify the vowels in the word. The vowels are A, E, I, O, U.

In the word ‘PROBABILITY’, the vowels are O, A, I, I.

Count the number of vowel letters in the word.

Number of vowel letters = 4 (O, A, I, I).

The probability of selecting a vowel is the ratio of the number of vowel letters to the total number of letters.

Probability (vowel) $= \frac{\text{Number of vowel letters}}{\text{Total number of letters}}$

Probability (vowel) $= \frac{4}{11}$

The probability that the selected letter is a vowel is $\frac{4}{11}$.

This corresponds to option (B).

Given:

Let A be the event that A fails in the examination, so $P(A) = 0.2$.

Let B be the event that B fails in the examination, so $P(B) = 0.3$.

To Find:

The probability that either A or B fails, which is $P(A \cup B)$.

Solution:

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

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

We are given $P(A) = 0.2$ and $P(B) = 0.3$. Substitute these values into the formula:

$P(A \cup B) = 0.2 + 0.3 - P(A \cap B)$

$P(A \cup B) = 0.5 - P(A \cap B)$

The term $P(A \cap B)$ represents the probability that both A and B fail. We know that the probability of any event must be non-negative.

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

Substitute this inequality into the expression for $P(A \cup B)$:

$P(A \cup B) = 0.5 - P(A \cap B) \leq 0.5 - 0$

$P(A \cup B) \leq 0.5$

The maximum value of $P(A \cup B)$ is 0.5, which occurs when $P(A \cap B) = 0$ (i.e., if the events of A and B failing are mutually exclusive).

The minimum value of $P(A \cap B)$ cannot exceed the probabilities of A failing or B failing individually. That is, $P(A \cap B) \leq P(A) = 0.2$ and $P(A \cap B) \leq P(B) = 0.3$. So, $P(A \cap B) \leq \min(0.2, 0.3) = 0.2$.

Using the maximum possible value for $P(A \cap B)$: $P(A \cup B) = 0.5 - P(A \cap B) \geq 0.5 - 0.2 = 0.3$.

So, the probability $P(A \cup B)$ is always between 0.3 and 0.5 (inclusive): $0.3 \leq P(A \cup B) \leq 0.5$.

Based on the inequality $P(A \cup B) \leq 0.5$, we can evaluate the given options:

(A) $P(A \cup B) > 0.5$: This is not always true. For example, if $P(A \cap B) > 0$, then $P(A \cup B) < 0.5$.

(B) $P(A \cup B) = 0.5$: This is only true in the specific case where $P(A \cap B) = 0$ (mutually exclusive failure events), but not necessarily for all cases.

(C) $P(A \cup B) \leq 0.5$: This is always true because $P(A \cap B)$ cannot be negative.

(D) $P(A \cup B) = 0$: This is only true if $P(A) = 0$ and $P(B) = 0$, which is not the case here.

The probability that either A or B fails is always less than or equal to 0.5.

This corresponds to option (C).

Given:

The probability that at least one of the events A and B occurs is 0.6. This means the probability of the union of A and B is 0.6.

$P(A \cup B) = 0.6$

A and B occur simultaneously with probability 0.2. This means the probability of the intersection of A and B is 0.2.

$P(A \cap B) = 0.2$

To Find:

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

Solution:

We know the relationship between the probability of an event and the probability of its complement:

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

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

We want to find $P(\overline{A}) + P(\overline{B}) = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.

To find $P(A) + P(B)$, we can use the Addition Law of Probability:

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

Substitute the given values:

$0.6 = P(A) + P(B) - 0.2$

Add 0.2 to both sides of the equation to find the sum $P(A) + P(B)$:

$0.6 + 0.2 = P(A) + P(B)$

$P(A) + P(B) = 0.8$

Now substitute this sum into the expression for $P(\overline{A}) + P(\overline{B})$:

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

$P(\overline{A}) + P(\overline{B}) = 2 - 0.8$

$P(\overline{A}) + P(\overline{B}) = 1.2$

The value of $P(\overline{A}) + P(\overline{B})$ is 1.2.

This corresponds to option (C).

Given:

Two events M and N.

To Find:

The probability that at least one of them occurs.

Solution:

The phrase "at least one of the events M and N occurs" corresponds to the union of the two events, denoted as $M \cup N$.

The probability of the union of any two events M and N is given by the Addition Law of Probability.

The Addition Law states that:

$P(M \cup N) = P(M) + P(N) - P(M \cap N)$

where $P(M \cap N)$ is the probability that both events M and N occur simultaneously (the intersection of M and N).

Comparing this formula with the given options, we find that it matches option (B).

The probability that at least one of the events M and N occurs is $P(M) + P(N) - P(M \cap N)$.

This corresponds to option (B).

Given:

Let G be the event that the person sees the giraffes.

Let B be the event that the person sees the bears.

We are given:

$P(G) = 0.72$

$P(B) = 0.84$

$P(G \cap B) = 0.52$

To Determine:

Whether these probabilities are consistent with the rules of probability.

Solution:

For any two events G and B, the probability of their union (seeing either giraffes or bears or both) is given by the Addition Law of Probability:

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

Substitute the given probabilities into the formula:

$P(G \cup B) = 0.72 + 0.84 - 0.52$

$P(G \cup B) = 1.56 - 0.52$

$P(G \cup B) = 1.04$

According to the axioms of probability, the probability of any event must be between 0 and 1 (inclusive).

That is, for any event E, $0 \leq P(E) \leq 1$.

In this case, $P(G \cup B) = 1.04$, which is greater than 1.

Since the probability of the union of the two events exceeds 1, the given probabilities are inconsistent and cannot represent probabilities of real-world events.

Therefore, the statement is False.

Given:

Let P be the event that the student passes the examination.

$P(P) = 0.73$

Let C be the event that the student gets a compartment.

$P(C) = 0.13$

The probability that the student will either pass or get compartment is $P(P \cup C) = 0.96$.

To Determine:

Whether the given probabilities are consistent.

Solution:

In the context of a single examination result, the event of passing and the event of getting a compartment are mutually exclusive. A student cannot simultaneously pass the examination and get a compartment in the same subject(s) or examination attempt.

For mutually exclusive events P and C, the probability of their union is given by the Addition Law of Probability:

$P(P \cup C) = P(P) + P(C)$

Substitute the given probabilities for $P(P)$ and $P(C)$ into this formula:

$P(P \cup C) = 0.73 + 0.13$

$P(P \cup C) = 0.86$

The calculated probability of the student either passing or getting a compartment is 0.86.

However, the problem states that the probability of the student either passing or getting compartment is 0.96.

$0.86 \neq 0.96$

Since the given probability of the union (0.96) does not match the probability calculated using the Addition Law for mutually exclusive events (0.86), the given probabilities are inconsistent with the rules of probability for mutually exclusive events.

Therefore, the statement is False.

Given:

The probabilities of the number of mistakes made by a typist are given as:

$P(\text{0 mistakes}) = 0.12$

$P(\text{1 mistake}) = 0.25$

$P(\text{2 mistakes}) = 0.36$

$P(\text{3 mistakes}) = 0.14$

$P(\text{4 mistakes}) = 0.08$

$P(\text{5 or more mistakes}) = 0.11$

To Determine:

Whether these probabilities are consistent.

Solution:

The given events (making 0, 1, 2, 3, 4, or 5 or more mistakes) form a set of mutually exclusive and exhaustive outcomes for the number of mistakes made by the typist. This means that the sum of the probabilities of these events must be equal to 1.

Sum of probabilities $= P(\text{0}) + P(\text{1}) + P(\text{2}) + P(\text{3}) + P(\text{4}) + P(\text{5 or more})$

Sum of probabilities $= 0.12 + 0.25 + 0.36 + 0.14 + 0.08 + 0.11$

Let's add these values:

$0.12 + 0.25 = 0.37$

$0.37 + 0.36 = 0.73$

$0.73 + 0.14 = 0.87$

$0.87 + 0.08 = 0.95$

$0.95 + 0.11 = 1.06$

The sum of the given probabilities is 1.06.

According to the axioms of probability, the sum of the probabilities of all elementary outcomes in a sample space must be exactly 1.

Since the sum of the probabilities in this case is 1.06, which is greater than 1, the given probabilities are inconsistent.

Therefore, the statement is False.

Given:

Let A be the event that candidate A is selected.

Let B be the event that candidate B is selected.

We are given:

$P(A) = 0.5$

$P(A \cap B) \leq 0.3$

We want to determine if it is possible that $P(B) = 0.7$.

To Determine:

Whether the statement "it is possible that the probability of B getting selected is 0.7" is True or False, given the other conditions.

Solution:

Let's assume that $P(B) = 0.7$. We need to check if this assumption is consistent with the given conditions $P(A) = 0.5$ and $P(A \cap B) \leq 0.3$, along with the fundamental axioms of probability ($0 \leq P(E) \leq 1$ for any event E, and $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$).

For any two events A and B, the probability of their intersection is always less than or equal to the probability of each individual event:

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

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

If we assume $P(A) = 0.5$ and $P(B) = 0.7$, these inequalities become:

$P(A \cap B) \leq 0.5$

$P(A \cap B) \leq 0.7$

Combining these, we must have $P(A \cap B) \leq \min(0.5, 0.7) = 0.5$.

We are also given the condition $P(A \cap B) \leq 0.3$.

For the assumption $P(B) = 0.7$ to be possible, there must exist a value for $P(A \cap B)$ that satisfies both $P(A \cap B) \leq 0.5$ and $P(A \cap B) \leq 0.3$. The combined upper bound is $P(A \cap B) \leq \min(0.5, 0.3) = 0.3$. So, any value of $P(A \cap B)$ between 0 and 0.3 (inclusive) would satisfy these conditions from the individual probabilities.

Now let's consider the Addition Law of Probability:

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

Using the assumed $P(A) = 0.5$ and $P(B) = 0.7$:

$P(A \cup B) = 0.5 + 0.7 - P(A \cap B) = 1.2 - P(A \cap B)$

We know that the probability of the union of any two events must be between 0 and 1 (inclusive):

$0 \leq P(A \cup B) \leq 1$

Substituting the expression for $P(A \cup B)$:

$0 \leq 1.2 - P(A \cap B) \leq 1$

From the right side of the inequality: $1.2 - P(A \cap B) \leq 1 \implies 1.2 - 1 \leq P(A \cap B) \implies 0.2 \leq P(A \cap B)$.

From the left side of the inequality: $0 \leq 1.2 - P(A \cap B) \implies P(A \cap B) \leq 1.2$. This is always true since $P(A \cap B) \leq P(A) = 0.5$ (and $0.5 \leq 1.2$).

So, if $P(A) = 0.5$ and $P(B) = 0.7$, the probability of the intersection must satisfy $P(A \cap B) \geq 0.2$.

Combining the constraints on $P(A \cap B)$: We are given $P(A \cap B) \leq 0.3$, and we derived $P(A \cap B) \geq 0.2$ from the assumption $P(B) = 0.7$ and $P(A) = 0.5$.

Thus, for $P(B) = 0.7$ to be possible, there must exist a value of $P(A \cap B)$ such that $0.2 \leq P(A \cap B) \leq 0.3$.

Since the interval $[0.2, 0.3]$ is a valid range for a probability (all values are between 0 and 1), and it satisfies both the condition from $P(A), P(B)$ and the separately given condition $P(A \cap B) \leq 0.3$, it is possible for $P(B)$ to be 0.7 under the given circumstances.

For instance, if $P(A)=0.5$, $P(B)=0.7$, and $P(A \cap B)=0.25$, then all conditions are met:

$P(A)=0.5$ (given)

$P(B)=0.7$ (assumed possible)

$P(A \cap B)=0.25 \leq 0.3$ (given condition satisfied)

$P(A \cap B)=0.25 \leq \min(P(A), P(B)) = \min(0.5, 0.7) = 0.5$ (fundamental rule satisfied)

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.7 - 0.25 = 1.2 - 0.25 = 0.95$. Since $0 \leq 0.95 \leq 1$, the probability of the union is valid.

Since we found a possible scenario where all conditions are met, the statement is True.

Therefore, the statement is True.

Given Statement:

The probability of intersection of two events A and B is always less than or equal to those favourable to the event A.

To Determine:

Whether the given statement is True or False.

Solution:

Let A and B be any two events in a sample space S.

The intersection of events A and B is the event $A \cap B$, which consists of all outcomes that are common to both event A and event B.

"Those favourable to the event A" refers to the outcomes that constitute event A itself. The probability of event A is $P(A)$.

The statement is claiming that $P(A \cap B) \leq P(A)$.

By definition, the set of outcomes in the intersection $A \cap B$ is a subset of the set of outcomes in A.

$A \cap B \subseteq A$

A fundamental property of probability states that if an event $E_1$ is a subset of another event $E_2$ (i.e., $E_1 \subseteq E_2$), then the probability of $E_1$ is less than or equal to the probability of $E_2$.

Since $A \cap B \subseteq A$, it follows that $P(A \cap B) \leq P(A)$.

This property holds true for any two events A and B.

Therefore, the statement "The probability of intersection of two events A and B is always less than or equal to those favourable to the event A" is True.

Given:

Let A and B be two events.

We are given the following probabilities:

$P(A) = 0.7$

$P(B) = 0.3$

$P(A \cap B) = 0.4$

To Determine:

Whether the given probabilities are consistent with the rules of probability.

Solution:

For any two events A and B, the probability of their intersection ($A \cap B$) must be less than or equal to the probability of each individual event ($A$ and $B$).

That is, we must have:

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

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

Let's check the first condition with the given values:

$0.4 \leq 0.7$

This condition is satisfied.

Now, let's check the second condition:

$0.4 \leq 0.3$

This condition is NOT satisfied. The probability of both events occurring (0.4) cannot be greater than the probability of event B occurring alone (0.3), because the outcomes where both A and B occur are a subset of the outcomes where B occurs.

Since one of the fundamental rules of probability is violated, the given probabilities are inconsistent.

Therefore, the statement is False.

Given Statement:

The sum of probabilities of two students getting distinction in their final examinations is 1.2.

To Determine:

Whether the given statement is True or False.

Solution:

Let $S_1$ be the event that the first student gets distinction in their final examinations.

Let $S_2$ be the event that the second student gets distinction in their final examinations.

The statement claims that the sum of the probabilities of these two events is 1.2, i.e., $P(S_1) + P(S_2) = 1.2$.

According to the axioms of probability, the probability of any event must be a number between 0 and 1, inclusive.

$0 \leq P(S_1) \leq 1$

$0 \leq P(S_2) \leq 1$

The statement asks if it is possible for the sum of two valid probabilities to be 1.2. Let's see if we can find values for $P(S_1)$ and $P(S_2)$ that satisfy both the probability axioms and the given sum.

For example, let $P(S_1) = 0.6$ and $P(S_2) = 0.6$. Both these probabilities are valid since $0 \leq 0.6 \leq 1$.

Their sum is $P(S_1) + P(S_2) = 0.6 + 0.6 = 1.2$.

Another example: Let $P(S_1) = 0.5$ and $P(S_2) = 0.7$. Both are valid probabilities, and their sum is $0.5 + 0.7 = 1.2$.

Since it is possible to have two distinct events whose individual probabilities sum up to 1.2 (as long as each individual probability is less than or equal to 1), the statement that the sum *is* 1.2 in a specific scenario is possible within the rules of probability.

Note that this is different from the probability of the union of the two events, $P(S_1 \cup S_2)$, which represents the probability that at least one of the students gets distinction. The probability of the union must always be less than or equal to 1, i.e., $P(S_1 \cup S_2) \leq 1$. However, the statement is about the sum of the individual probabilities, not the probability of the union.

Since we can find valid probabilities $P(S_1)$ and $P(S_2)$ whose sum is 1.2, the statement that the sum of probabilities is 1.2 is possible.

Therefore, the statement is True.

Given:

Let W be the event that the home team wins.

$P(W) = 0.77$

Let T be the event that the home team ties.

$P(T) = 0.08$

To Find:

The probability that the home team will lose the game.

Solution:

In a football game, there are three possible mutually exclusive and exhaustive outcomes for the home team: Win, Tie, or Lose.

Let L be the event that the home team loses.

The sum of the probabilities of all mutually exclusive and exhaustive events in a sample space must equal 1.

$P(W) + P(T) + P(L) = 1$

Substitute the given probabilities into the equation:

$0.77 + 0.08 + P(L) = 1$

Add the known probabilities:

$0.85 + P(L) = 1$

Subtract 0.85 from 1 to find $P(L)$:

$P(L) = 1 - 0.85$

$P(L) = 0.15$

The probability that the home team will lose the game is 0.15.

The completed statement is: The probability that the home team will win an upcoming football game is 0.77, the probability that it will tie the game is 0.08, and the probability that it will lose the game is 0.15.

Given:

The sample space consists of four elementary outcomes: $e_1, e_2, e_3, e_4$.

The probabilities of three of these outcomes are given:

$P(e_1) = 0.1$

$P(e_2) = 0.5$

$P(e_3) = 0.1$

To Find:

The probability of the elementary outcome $e_4$, $P(e_4)$.

Solution:

A fundamental property of probabilities for elementary outcomes is that the sum of the probabilities of all distinct elementary outcomes in a sample space must equal 1.

For the given sample space $\{e_1, e_2, e_3, e_4\}$, the sum of probabilities is:

$P(e_1) + P(e_2) + P(e_3) + P(e_4) = 1$

Substitute the given probabilities into this equation:

$0.1 + 0.5 + 0.1 + P(e_4) = 1$

Sum the known probabilities:

$0.7 + P(e_4) = 1$

Solve for $P(e_4)$ by subtracting 0.7 from 1:

$P(e_4) = 1 - 0.7$

$P(e_4) = 0.3$

The probability of $e_4$ is 0.3.

The completed statement is: If e₁, e₂, e₃, e₄ are the four elementary outcomes in a sample space and P(e₁) = .1, P(e₂) = .5, P (e₃) = .1, then the probability of e₄ is 0.3.

Given:

Sample space $S = \{1, 2, 3, 4, 5, 6\}$.

Event $E = \{1, 3, 5\}$.

To Find:

The complement of event E, denoted as $\overline{E}$ or $E'$.

Solution:

The complement of an event E with respect to the sample space S is the set of all outcomes in S that are not in E.

$\overline{E} = \{x \in S \mid x \notin E\}$

The elements in S are {1, 2, 3, 4, 5, 6}.

The elements in E are {1, 3, 5}.

To find the elements in $\overline{E}$, we remove the elements of E from S:

$S \setminus E = \{1, 2, 3, 4, 5, 6\} \setminus \{1, 3, 5\} = \{2, 4, 6\}$

So, the complement of E is $\overline{E} = \{2, 4, 6\}$.

The completed statement is: Let S = {1, 2, 3, 4, 5, 6} and E = {1, 3, 5}, then $\overline{E}$ is {2, 4, 6}.

Given:

A and B are two events.

$P(A) = 0.3$

$P(B) = 0.2$

$P(A \cap B) = 0.1$

To Find:

The value of $P(A \cap \overline{B})$.

Solution:

The event $A \cap \overline{B}$ represents the outcomes that are in event A but not in event B. This is sometimes referred to as the event "A occurs, but B does not occur".

The event A can be partitioned into two disjoint events: the part that is also in B ($A \cap B$) and the part that is not in B ($A \cap \overline{B}$).

$A = (A \cap B) \cup (A \cap \overline{B})$

Since the events $(A \cap B)$ and $(A \cap \overline{B})$ are mutually exclusive (they have no outcomes in common), the probability of their union is the sum of their probabilities:

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

We are given $P(A)$ and $P(A \cap B)$, and we want to find $P(A \cap \overline{B})$. Rearrange the formula to solve for $P(A \cap \overline{B})$:

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

Substitute the given values:

$P(A \cap \overline{B}) = 0.3 - 0.1$

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

The value of $P(A \cap \overline{B})$ is 0.2.

The completed statement is: If A and B are two events associated with a random experiment such that P ( A) = 0.3, P (B) = 0.2 and P (A ∩ B) = 0.1, then the value of P (A ∩ $\overline{B}$) is 0.2.

Given:

$P(A) = 0.5$

$P(B) = 0.3$

A and B are mutually exclusive events.

To Find:

The probability of neither A nor B.

Solution:

The event "neither A nor B" means that event A does not occur and event B does not occur. This corresponds to the intersection of the complements of A and B, which is $\overline{A} \cap \overline{B}$.

According to De Morgan's Law, the complement of the union of two events is equal to the intersection of their complements:

$\overline{A} \cap \overline{B} = (A \cup B)'$

So, $P(\text{neither A nor B}) = P(\overline{A} \cap \overline{B}) = P((A \cup B)')$.

Using the complement rule, the probability of the complement of an event is 1 minus the probability of the event:

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

We need to find $P(A \cup B)$. Since A and B are mutually exclusive events, the probability of their union is the sum of their individual probabilities:

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

Substitute the given probabilities:

$P(A \cup B) = 0.5 + 0.3$

$P(A \cup B) = 0.8$

Now, substitute this value back into the expression for the probability of neither A nor B:

$P(\text{neither A nor B}) = 1 - P(A \cup B)$

$P(\text{neither A nor B}) = 1 - 0.8$

$P(\text{neither A nor B}) = 0.2$

The probability of neither A nor B is 0.2.

The completed statement is: The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then the probability of neither A nor B is 0.2.

Solution:

Let's match each probability in Column C₁ with the most appropriate description in Column C₂ based on the interpretation of probability values.

(a) 0.95: This is a high probability, close to 1. It signifies that the event is very likely to occur.

This matches description (iv) Very likely to happen.

(b) 0.02: This is a very low positive probability, close to 0. It signifies that the event has a very small chance of occurring.

This matches description (v) Very little chance of happening.

(c) – 0.3: Probabilities must always be non-negative. A negative value is not a valid probability.

This matches description (i) An incorrect assignment.

(d) 0.5: This probability represents an equal chance of the event happening or not happening (50% chance).

This matches description (iii) As much chance of happening as not.

(e) 0: This probability signifies that the event is impossible; it has no chance of occurring.

This matches description (ii) No chance of happening.

The matches are:

(a) $\to$ (iv)

(b) $\to$ (v)

(c) $\to$ (i)

(d) $\to$ (iii)

(e) $\to$ (ii)

Solution:

Let's match the descriptions in the left column with the corresponding properties in the right column.

(a) If $E_1$ and $E_2$ are the two mutually exclusive events:

Mutually exclusive events are events that cannot occur at the same time. This means their intersection is the empty set ($\phi$).

This corresponds to property (iv) $E_1 \cap E_2 = \phi$.

(b) If $E_1$ and $E_2$ are the mutually exclusive and exhaustive events:

Mutually exclusive means $E_1 \cap E_2 = \phi$.

Exhaustive events are events whose union covers the entire sample space S, i.e., $E_1 \cup E_2 = S$.

This corresponds to property (iii) $E_1 \cap E_2 = \phi$, $E_1 \cup E_2 = S$.

(c) If $E_1$ and $E_2$ have common outcomes, then:

Having common outcomes means their intersection is not empty, $E_1 \cap E_2 \neq \phi$. The property listed in option (ii), $(E_1 – E_2) \cup (E_1 \cap E_2) = E_1$, is a general set identity that is always true for any two sets $E_1$ and $E_2$, regardless of whether they have common outcomes or not. The set $E_1$ can be partitioned into the part not in $E_2$ ($E_1 - E_2$ or $E_1 \setminus E_2$) and the part in $E_2$ ($E_1 \cap E_2$). The union of these disjoint parts is $E_1$. Given the options, this is the only set identity presented.

This corresponds to property (ii) $(E_1 – E_2) \cup (E_1 \cap E_2) = E_1$.

(d) If $E_1$ and $E_2$ are two events such that $E_1 \subset E_2$:

If $E_1$ is a subset of $E_2$, then every outcome in $E_1$ is also an outcome in $E_2$. The intersection of $E_1$ and $E_2$ contains all outcomes that are in both $E_1$ and $E_2$. Since all outcomes of $E_1$ are in $E_2$, the intersection is simply $E_1$.

This corresponds to property (i) $E_1 \cap E_2 = E_1$.

The final matches are:

(a) $\to$ (iv)

(b) $\to$ (iii)

(c) $\to$ (ii)

(d) $\to$ (i)