Question

A committee of five persons is selected from $4$ men and $3$ women. What is the probability that the committee will have two women

Answer 1

It is given that there are $4$ men and $3$ women which means the total number of person are $7$.

The number of ways of selecting five person is:

$n\left(S\right){=}^7{C}_5=\frac{7!}{(7-5)!\times 5!}=\frac{7!}{2!\times 5!}=\frac{7\times 6\times 5!}{2\times 1\times 5!}=7\times 3=21$

Let $A$ denote the event in which a committee of $5$ is selected as to have two women, therefore,

$n\left(A\right){=}^4{C}_3{\times }^3{C}_2=\frac{4!}{(4-3)!\times 3!}\times \frac{3!}{(3-2)!\times 2!}=\frac{4!}{1!\times 3!}\times \frac{3!}{1!\times 2!}=\frac{4\times 3!}{1\times 3!}\times \frac{3\times 2!}{1\times 2!}=4\times 3=12$

Thus, probability that the committee will have two women is:

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{12}{21}=\frac{4}{7}$

Hence, probability that the committee will have two women is $\frac{4}{7}$.