Answer in Statistics and Probability for Eric #87493

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}n 18 \n 3nend{pmatrix}={18! over 3!(18-3)!}={18(17)(16) over 1(2)(3)}=816”

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

“begin{pmatrix}n 7 \n 3nend{pmatrix}begin{pmatrix}n 11 \n 0nend{pmatrix}={7! over 3!(7-3)!}*{11! over 0!(11-0)!}=35”

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

“begin{pmatrix}n 7 \n 2nend{pmatrix}begin{pmatrix}n 11 \n 1nend{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+231over816}={275 over 408}approx0.674”