College Algebra, Chapter 10, 10.3, Section 10.3, Problem 70

A mathematics department consists of ten men and eight women. Six mathematics faculty members are to be selected at random for the curriculum committee.

a.) What is the probability that two women and four men are selected?

There are 18 faculty members which consist of ten men and eight women that will be selected for six curriculum committee, so there are $C(18,6)$ ways to do this. Also, there are $C(8,2)$ and $C(10,4)$ ways to select two men out of 8 men and 4 women out of 10 women faculty members, respectively. Therefore the probability is

$\displaystyle \frac{10 C4}{18 C6} \times \frac{8 C2}{18C6} = \frac{5880}{344622096} = \frac{5}{293046} = 0.00001706$

b.) What is the probability that two or fewer women are selected?

In this case we need to find the probability that two, one or no women are selected. Thus, we have

$\displaystyle \frac{8C2}{18C6} + \frac{8C1}{18C6} + \frac{8C0}{18C6} = \frac{37}{18564} = 0.0019931$

c.) What is the probability that more than two women are selected?

In this case we can apply compliment of the probability that two or fewer women are selected. Thus, we have

$\displaystyle 1 - \frac{37}{18564} = 0.998$