Question

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

(i) no man?

(ii) one man?

(iii) two men?

Answer 1

The total number of persons = 2 + 2 = 4.
Out of these four person. two can be selected in ${}^4{C}_2$ ways.

(i) No men in the committee of two means there will be two women in the committee.

Out of two women, two can be selected in ${}^2{C}_2$ = 1 way.

Therefore. P (no man) $=\frac{{}^2{C}_2}{{}^4{C}_2}=\frac{1\times 2\times 1}{4\times 3}=\frac{1}{6}$

(ii) One man in the committee means that there is one woman.
One man out of 2 can be selected in ${}^2{C}_1$ ways and one woman out of 2 can be selected in ${}^2{C}_1$ ways.

Together they can be selected in ${}^2{C}_1{\times }^2{C}_1$ ways.

Therefore, P (One man) $\frac{{}^2{C}_1{\times }^2{C}_1}{{}^4{C}_2}=\frac{2\times 2}{2\times 3}=\frac{2}{3}$

(iii) Two men can be selected in ${}^2{C}_2$ way.

Hence, P (Two men) $=\frac{{}^2{C}_2}{{}^4{C}_2}=\frac{1}{{}^4{C}_2}=\frac{1}{6}$