Chapter 13.5, Problem 44E

To determine

To calculate: The probability that committee has at least three women.

Answer to Problem 44E

The probability that committee has at least three women is $\frac{59}{143}$ .

Explanation of Solution

Given information:

A subcommittee of 5 members is to be selected from a committee of 6 women and 7 men.

Formula used:

Probability of an event E is $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

If there are n objects taken r at a time then combination is defined as $C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$ .

Calculation:

Consider the provided information that subcommittee of 5 members is to be selected from a committee of 6 women and 7 men.

Since, order is not important use combination.

Recall that if there are n objects taken r at a time then combination is defined as $C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}$ .

Recall that probability of an event E is $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

The probability that committee has at least three women means the committee can have either 3 or 4 or 5 women to form a 5 member committee.

If there are 3 women then there are 2 men.

If there are 4 women then there is 1 man.

If there are 5 women then there are 0 men.

The events are mutually exclusive events.

The probability is,

  $\begin{array}{c}P\left(\text{at least 3 women}\right)=P\left(\text{3 women}\right)+P\left(\text{4 women}\right)+P\left(\text{5 women}\right)\\ =\frac{C\left(6,3\right)\cdot C\left(7,2\right)}{C\left(13,5\right)}+\frac{C\left(6,4\right)\cdot C\left(7,1\right)}{C\left(13,5\right)}+\frac{C\left(6,5\right)\cdot C\left(7,0\right)}{C\left(13,5\right)}\\ =\frac{420}{1287}+\frac{105}{1287}+\frac{6}{1287}\\ =\frac{59}{143}\end{array}$

Thus, the probability that committee has at least three women is $\frac{59}{143}$ .