Chapter 4.4, Problem 38E

Choosing officers: A committee consists of ten women and eight men. Three committee members will be chosen as officers.

1. a. How many different choices are possible?
2. b. How many different choices are possible if all the officers are to be women?
3. c. How many different choices are possible if all the officers are to be men?
4. d. What is the probability that all the officers are women?
5. e. What is the probability that at least one officer is a man?

To determine

Find the number of different choices.

Answer to Problem 38E

There are 816 different choices are possible.

Explanation of Solution

Calculation:

It is given that the committee consists 10 women and 8 men. Three of the committee members are selected as officers.

Here, order does not matter. Hence, combination is applied.

Combination Rule:

The number of combinations of r objects from n distinct objects is ${}_n{C}_r$ and is given by the formula:

${}_n{C}_r=\frac{n!}{r!\left(n-r\right)!}$

The committee consists 18$\left(=10+8\right)$, three of the members are selected as officers.

Substitute 18 for “n” and 3 for “r” in combination rule.

$\begin{array}{c}N\left(\text{Choices}\right){=}_{18}{C}_3\\ =\frac{18!}{3!\left(18-3\right)!}\\ =\frac{18!}{3!\times 15!}\\ =\frac{18\times 16\times 17\times 15!}{3\times 2\times 1\times 15!}\\ =\frac{18\times 16\times 17}{3\times 2\times 1}\\ =816\end{array}$

Thus, there are 816 different choices are possible.

To determine

Find the number of different choices if all the officers are to be women.

Answer to Problem 38E

There are 120 different choices are possible if all the officers are to be women.

Explanation of Solution

Calculation:

All the officers are women.

The committee consists of 10 women.

Here, order does not matter. Hence, combination is applied.

Substitute 10 for “n” and 3 for “r” in combination rule.

$\begin{array}{c}N\left(\text{Choices with all officers are women}\right){=}_{10}{C}_3\\ =\frac{10!}{3!\left(10-3\right)!}\\ =\frac{10!}{3!\times 7!}\\ =\frac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}\\ =\frac{10\times 9\times 8}{3\times 2\times 1}\\ =120\end{array}$

Thus, there are 120 different choices are possible if all the officers are to be women.

To determine

Find the number of different choices if all the officers are to be men.

Answer to Problem 38E

There are 56 different choices are possible if all the officers are to be men.

Explanation of Solution

Calculation:

All the officers are women.

The committee consists of 8 men.

Here, order does not matter. Hence, combination is applied.

Substitute 8 for “n” and 3 for “r” in combination rule.

$\begin{array}{c}N\left(\text{Choices with all officers are men}\right){=}_8{C}_3\\ =\frac{8!}{3!\left(8-3\right)!}\\ =\frac{8!}{3!\times 5!}\\ =\frac{8\times 7\times 6\times 5!}{3\times 2\times 1\times 5!}\\ =\frac{8\times 7\times 6}{3\times 2\times 1}\\ =56\end{array}$

Thus, there are 56 different choices are possible if all the officers are to be men.

To determine

Find the probability that all the officers are women.

Answer to Problem 38E

The probability that all the officers are women is 0.1471.

Explanation of Solution

Calculation:

The probability of an event can be obtained as shown below:

$P\left(A\right)=\frac{\text{number of outcomes in }A}{\text{ number of outcomes in the sample space}}$

Event A denotes that all the officers are women.

From part (a), it is clear that there are 816 different choices are possible.

From part (b), it is clear that there are 120 different choices are possible if all the officers are to be women.

Substitute 120 for “number of outcomes in A” and 816 for “number of outcomes in the sample space”

The required probability is obtained as follows:

$\begin{array}{c}P\left(A\right)=\frac{120}{816}\\ =0.14705\\ =0.1471\end{array}$

Thus, the probability that all the officers are women is 0.1471.

To determine

Find the probability that at least one of the officers is a man.

Answer to Problem 38E

The probability that at least one of the officers is a man is 0.8529.

Explanation of Solution

Calculation:

Event A denotes that all the officers are women.

Required probability can be obtained as follows:

$\begin{array}{c}P\left(\text{at least one officer is a man}\right)=1-P\left(\text{none}\text{of the officer is a man}\right)\\ =1-P\left(\text{all of the officer is a women}\right)\\ 1-P\left(A\right)\end{array}$

From part (d), it is clear that probability that all the officers are women is 0.1471. Substitute this value in the above formula.

Therefore,

$\begin{array}{c}P\left(\text{at least one officer is a man}\right)=1-0.1471\\ =0.8529\end{array}$

Thus, the probability that at least one of the officers is a man is 0.8529.