Question

In a club with 9 male and 12 female​ members, a ​6 -member committee will be randomly chosen. Find the probability that the committee contains at least 5 women.

Answer 1

We divide this up into several cases; either there are five women, or six women on the committee.

Let's first consider the situation with 5 women.

 We may consider the construction of a committee as a two-step process: select the women serving on it, and then the men. There are 12 women, so there are $\binom{12}{5}$ ways to select five of them to serve on the committee;. likewise, there are $\binom{9}{1}$ ways to select one man to serve. Since the formation of a committee results from the independent performance of these two steps, we may form any of $\binom{12}{5}*\binom{9}{1}$ different committees.

To calculate this:

$$\binom{12}{5}*\binom{9}{1}\implies \frac{12!}{5!(12-5)!}*\frac{9!}{1!(9-1)!}=792*9=7128$$

The second and case can be shown, via very simple modifications of the above argument, to contribute $\binom{12}{6}*\binom{9}{0}$ possible committees. Thus, the total number of possible committees is

$$\binom{12}{5}*\binom{9}{1}+\binom{12}{6}*\binom{9}{0}= 7128+924*1=8052$$