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

Committee of two is to be formed from $3$ boys and $2$ girls.
Let $3$ boys be $B_1,B_2,B_3$ and $2$ girls be $G_1,G_2$.
$S=\{B_1{B}_2,B_1{B}_3,B_2{B}_3,G_1{G}_2,B_1{G}_1,B_1{G}_2,B_2{G}_1,B_2{G}_2,B_3{G}_1,B_3{G}_2\}$
$\therefore n\left(S\right)=10$
$A$ is the event that the committee contains atleast one boy.
$\therefore A=\{B_1{B}_2,B_1{B}_3,B_2{B}_3,B_1{G}_1,B_1{G}_2,B_2{G}_1,B_2{G}_2,B_3{G}_1,B_3{G}_2\}$
$n\left(A\right)=9$
$\therefore P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{9}{10}$
and $B$ is the event that the committee contains one boy and one girl.
$\therefore B=\{B_1{G}_1,B_1{G}_2,B_2{G}_1,B_2{G}_2,B_3{G}_1,B_3{G}_2\}$
$n\left(B\right)=6$
$\therefore P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{6}{10}=\frac{3}{5}$