Question

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

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 so as to have one man. If only one man is included the $4$ more women are needed for formation of committee of $5$. But the number of women are only $3$.

Hence, it is an impossible event.