Question

In a club with 6 male and 13 female​ members, a 4​-member committee will be randomly chosen. Find the probability that the committee contains at least 3 women.

Answer 1

$С(19,4) = \binom{19}{4} = \frac{19!}{4!(19-4)!}=\frac{(16)(17)(18)(19)}{(1)(2)(3)(4)} = 3660$

There are $\binom{6}{4}$ ways to choose 4 men out of 6 and $\binom{13}{0}$ ways to choose 0 women out of 13.

$\binom{6}{4} \binom{13}{0} = \frac{6!}{4!(6-4)!} \times \frac{13!}{0!(13-0)!}= 15$

There are $\binom{6}{3}$ ways to choose 3 men out of 6 and $\binom{13}{1}$ ways to choose 1 women out of 13.

$\binom{6}{3} \binom{13}{1} = \frac{6!}{3!(6-3)!} \times \frac{13!}{1!(13-1)!}=20 \times 13 = 260$

There are $\binom{6}{2}$ ways to choose 2 men out of 6 and $\binom{13}{2}$ ways to choose 2 women out of 13.

$\binom{6}{2} \binom{13}{2} = \frac{6!}{2!(6-2)!} \times \frac{13!}{2!(13-2)!}=15 \times 78 = 1170$

Find the probability that the committee contains at least 3 women.

$P(W≥3) = 1 - \frac{15+260+1170}{3660} = 1 -0.3948 =0.6052$