VerifiedHint: Here we will use the concept of combination. First find the selection of 2 people from the committee of 5 which gives us the total number of cases. Later find the selection of a boy and a girl from the committee which gives the favorable case.
Complete step by step answer:
There are three boys and two girls and we have to select two person from the sample so, total number of ways of selecting them is
${}^5{C_2} = \dfrac{{5!}}{{3! \times 2!}} = 10$
Number of ways for selecting the committee of 1 girl & 1 boy =$ {}^2{C_1} \times {}^3{C_1} $
$ \Rightarrow 2 \times 3 = 6 $
So, required probability is = $\dfrac{favorable\;ways}{total\;ways}$
$\Rightarrow \dfrac{6}{{10}} $
$ \Rightarrow \left( {\dfrac{3}{5}} \right)$
So, this is our required probability.
NOTE: - In these types of questions always find the first total number of ways, then favorable ways, then divide favorable ways by total ways, we will get the required probability.
