Question

There are three boys and two girls. A committee of two is to be formed. Find the probability of the following events:
Event A: The committee contains at least one boy.
Event B: The committee contains one boy and one girl.

Answer 1

Hint: Here, the committee of two is to be formed from three boys and two girls. So we need to consider the boys and girls to be different from each other. Hence boys can be considered as $B_1, B_2$ and $B_3$ and girls can be considered as $G_1$ and $G_2$.

Complete step-by-step answer:
According to question,
A committee of two is formed from 3 boys and 2 girls. Let 3 boys be $B_1, B_2$ and $B_3$ and girls be $G_1$ and $G_2$.
Sample space =$\left\{ {B_1B_2, B_1B_3, B_2B_3, G_1G_2, B_1G_1, B_1G_2, B_2G_1, B_2G_2, B_3G_1, B_3G_2} \right\}$
$\therefore n(S) = 10$ (Total number of events in sample space)
Event A says the committee contains at least one boy.
$\therefore A = \left\{ {B_1B_2,B_1B_3,B_2B_3,B_1G_1,B_1G_2,B_2G_1,B_2G_2,B_3G_1,B_3G_2} \right\}$
$n(A) = 9$
Therefore, Probability of Event A, $P(A) = \dfrac{{n(A)}}{{n(S)}} = \dfrac{9}{{10}}$
Event B says the committee contains one boy and one girl.
$\therefore B = \left\{ {B_1G_1,B_1G_2,B_2G_1,B_2G_2,B_3G_1,B_3G_2} \right\}$
$n(B) = 6$
Therefore, Probability of Event B, $P(B) = \dfrac{{n(B)}}{{n(S)}} = \dfrac{6}{{10}}$

Note: For solving such questions, write all the sample space and then see for favorable events asked in the questions. Write separately the favorable events and then count the number of favorable events and total number events. We can solve this problem by concept of permutations and combinations as well.