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.
(a) $\dfrac{1}{5}$
(b) $\dfrac{2}{5}$
(c) $\dfrac{3}{5}$
(d) $\dfrac{4}{5}$

Answer 1

Hint: Calculate the number of ways to choose 1 boy out of 3 boys. Calculate the number of ways to choose 1 girl out of 2 girls. Calculate the number of ways to choose 2 people out of 5 people. Use the fact that the probability of any event is the ratio of the number of favourable outcomes to the total number of possible outcomes.

Complete step-by-step answer:

We have 3 boys and 2 girls. We have to calculate the probability of choosing 2 members of a committee such that it contains 1 girl and 1 boy.

We know that we can choose ‘r’ objects out of ‘n’ objects ${}^{n}{{C}_{r}}$ ways.

Substituting $n=3,r=1$ in the above formula, the number of ways to choose 1 boy out of 3 boys $={}^{3}{{C}_{1}}$ ways.

Substituting $n=2,r=1$ in the above formula, the number of ways to choose 1 girl out of 2 girls$={}^{2}{{C}_{1}}$ ways.

So, the total number of ways to choose 1 boy out of three boys and 1 girl out of 2 girls

$={}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}}$ ways.

Substituting $n=5,r=2$ in the above formula, the number of ways to choose 2 people out of

5 $={}^{5}{{C}_{2}}$ ways.

We know that the probability of any event is the ratio of the number of favourable outcomes to the total number of possible outcomes.

Thus, the probability of choosing 1 boy and 1 girl in a committee of 5

$=\dfrac{{}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}}}{{}^{5}{{C}_{1}}}$.

Simplifying the above expression, probability of choosing 1 boy and 1 girl in a committee of 5

$=\dfrac{\dfrac{3!}{1!2!}\times \dfrac{2!}{1!1!}}{\dfrac{5!}{2!3!}}=\dfrac{3\times

2}{\dfrac{5\times 4}{2}}=\dfrac{6}{10}=\dfrac{3}{5}$ .

Hence, the probability of choosing 1 boy and 1 girl in a committee of 5 is $\dfrac{3}{5}$, which is option (b).

Note: One must remember that the probability of any event lies in the range $\left[ 0,1 \right]$, where having 0 probability denotes that event can’t occur, and having probability equal to 1 denotes that the event will surely happen. We must know the formula${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.