 QUESTION

# 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.

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.