Question

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

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 at least two men, therefore,

$n\left(A\right)={(}^4{C}_2{\times }^3{C}_3)+{(}^4{C}_3{\times }^3{C}_2)+{(}^4{C}_4{\times }^3{C}_1)$

$=(\frac{4!}{(4-2)!\times 2!}\times \frac{3!}{(3-3)!\times 3!})+(\frac{4!}{(4-3)!\times 3!}\times \frac{3!}{(3-2)!\times 2!})+(\frac{4!}{(4-4)!\times 4!}\times \frac{3!}{(3-1)!\times 1!})$

$=(\frac{4!}{2!\times 2!}\times \frac{3!}{0!\times 3!})+(\frac{4!}{1!\times 3!}\times \frac{3!}{1!\times 2!})+(\frac{4!}{0!\times 4!}\times \frac{3!}{2!\times 1!})$
$=(\frac{4\times 3\times 2!}{2\times 1\times 2!}\times \frac{3!}{1\times 3!})+(\frac{4\times 3!}{1\times 3!}\times \frac{3\times 2!}{1\times 2!})+(\frac{4!}{1\times 4!}\times \frac{3\times 2!}{1\times 2!})$

$=(6\times 1)+(4\times 3)+(1\times 3)=6+12+3=21$

Thus, probability that the committee will have at least two men is:

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{21}{21}=1$

Hence, probability that the committee will have at least two men is $1$.