# There are three boys and two girls. A committee of two is to be formed. Find the probability of events that the committee contains one boy and one girl.

Hint: 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.