Question

A committee of two persons is selected from two men and two women. What is the probability that the committee will have no man ?

Answer 1

Hint : In this question we will use the concept of probability of event ‘not A’ and combinations. In this we have to make some combinations for the statement that is given in the question and then by selecting the right one combination which is true for our question we will get our answer.

Complete step-by-step solution -
Given that , there’s a committee of 2 men and 2 woman
\[ \Rightarrow {\text{total number of persons in committee = }}4\]
Out of these 4 persons, two can be selected in $^4{C_2}{\text{ ways}}$,
$ \Rightarrow
  ^4{C_2}{\text{ = }}\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}{\text{ = }}\dfrac{{{\text{4}} \times {\text{3}} \times {\text{2!}}}}{{2! \times 2!}} \\
  ^4{C_2} = {\text{6 ways}} \\
 $
According to the question ,if there’s no men in the committee then there will be only two women in the committee.
So, out of two women ,two can be selected in $^2{C_2}{\text{ ways}}$
$
   \Rightarrow {{\text{ }}^2}{C_2}{\text{ = }}\dfrac{{2!}}{{2!\left( {2 - 2} \right)!}} \\
   \Rightarrow {{\text{ }}^2}{C_2} = {\text{1 way }} \\
$
Therefore ,the number of committees having no man = 1.
$\therefore $ Probability that the committee have no man,
\[ \Rightarrow {\text{ P}}\left( {{\text{no man}}} \right) = \dfrac{{^2{C_2}}}{{^4{C_2}}}{\text{ }} = \dfrac{1}{6}\]

Note : In this question first we have to make the combinations that are needed for this question, then we will identify in how many ways a committee with no man can be formed and then by putting the formula of probability ,i.e.
$ \Rightarrow {\text{ }}\dfrac{{{\text{favourable outcomes}}}}{{{\text{total number of outcomes}}}}$. And then we will get our answer.