Question

There are $threeboys$ and $twogirls$. A committee of $two$is to be formed. Find the probability of the event that the committee contains $atleastonegirl$

Answer 1

Step-1: Given Information

Number of boys = $3$

Number of girls = $2$

Total number of people = $3+2=5$

Step-2: Find the number of outcomes such that the committee of two be formed from three boys and girls

The number of outcomes such that the committee of two be formed from three boys and girls. This is given by the formula ${}^{n}c_r$(that is for selecting $r$ objects out of $n$ objects).

Here, we have$n=5$(since total $5$ people are there)

$r=2$(selecting a committee of $2$people).

So, ${}^{n}c_r$$={}^{5}c_2$

${}^{5}c_2=\frac{5!}{3!2!}$

${}^{5}c_2=\frac{5\times 4\times 3\times 2\times 1}{3\times 2\times 1\times 2\times 1}$

${}^{5}c_2=10$

Total number of outcomes = $10$

Step-3: Find the number of outcomes such that the committee has at least one girl

Let the girls are $G_1$, $G_2$and boys are $B_1$, $B_2$, and $B_3$.

Let $A$be the event that the committee contains at least one girl.

So, $A=(G1,B1),(G1,B2),(G1,B3),(G2,B1),(G2,B2),(G2,B3),(G1,G2)$

$n\left(A\right)=7$

So the number of favorable outcomes, $n\left(A\right)=7$

The probability of the event that the committee contains $atleastonegirl$ = $=\frac{Numberoffavourableoutcomes}{Totalnumberofoutcomes}=\frac{7}{10}$

Hence, the probability of the event that the committee contains $atleastonegirl$is $\frac{7}{10}$