Answer to Question #87493 in Statistics and Probability for Eric

Question

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In a club with

7 male and

11 female​ members, a

3​-member committee will be randomly chosen. Find the probability that the committee contains at least

2 women.

Answer 1

We have a committee that will have 3 people and at least 2 of them will be women. There are 7 men and 11 women that can be on the committee. How many ways can we select 3 people?

$$C(18, 3)=\begin{pmatrix} 18 \\ 3 \end{pmatrix}={18! \over 3!(18-3)!}={18(17)(16) \over 1(2)(3)}=816$$

There are $\begin{pmatrix} 7 \\ 3 \end{pmatrix}$ ways to choose three men out of seven and $\begin{pmatrix} 11 \\ 0 \end{pmatrix}$ ways to choose zero women out of eleven.

$$\begin{pmatrix} 7 \\ 3 \end{pmatrix}\begin{pmatrix} 11 \\ 0 \end{pmatrix}={7! \over 3!(7-3)!}*{11! \over 0!(11-0)!}=35$$

There are$\begin{pmatrix} 7 \\ 2 \end{pmatrix}$ ways to choose two men out of seven and$\begin{pmatrix} 11 \\ 1 \end{pmatrix}$ ways to choose one woman out of eleven.

$$\begin{pmatrix} 7 \\ 2 \end{pmatrix}\begin{pmatrix} 11 \\ 1 \end{pmatrix}={7! \over 2!(7-2)!}*{11! \over 1!(11-1)!}=231$$

Find the probability that the committee contains at least 2 women.

$$P(W \geq 2)=1-{35+231\over816}={275 \over 408}\approx0.674$$