Question

There are three boys and two girls. A committee of two boys is to be formed. Find the probability of the event that the committee contains only Boys.
[a] $\dfrac{3}{10}$
[b] $\dfrac{2}{5}$
[c] $\dfrac{1}{2}$
[d] $\dfrac{7}{10}$

Answer 1

Hint: Probability of event E = $\dfrac{n(E)}{n(S)}=\dfrac{\text{Favourable cases}}{\text{Total number of cases}}$ where S is called the sample space of the random experiment. Assume that E be the event that the committee formed consists of only boys. Find the number of ways in which a committee of only boys can be formed and hence find n(E). Find the number of ways in which a committee can be formed and hence find n(S). Hence find the probability of the event E.

Complete step-by-step answer:
Let E be the event: The committee formed is only of boys.
The number of ways in which a committee of only boys can be formed is equal to the number of ways of selecting 2 boys among a lot of 3 boys which can be done in $^{3}{{C}_{2}}=3$ ways.
Hence, we have n(E) = 3
The number of ways in which a committee can be formed is equal to the number of ways of selection of two persons among a lot of 5 persons, which can be done in $^{5}{{C}_{2}}=10$ ways
Hence, we have n(S) = 10
Hence, we have
$P\left( E \right)=\dfrac{n\left( E \right)}{n\left( S \right)}=\dfrac{3}{10}$
Hence the probability that the committee formed is only of boys is $\dfrac{3}{10}$
Hence option [a] is correct.

Note: [1] It is important to note that choosing of persons uniformly at random is important for the application of the above formula. If the choice is not random, then there is a bias factor in choosing, and the above formula is not applicable. In those cases, we use the conditional probability of an event.
[2] We can also calculate the number of favourable cases in E’
E’: At least one girl is present
So we have two cases
One girl is present. So we select 1 boy from 3 boys and 1 girl from 2 girls, which can be done in $^{3}{{C}_{1}}{{\times }^{2}}{{C}_{1}}=6$ ways
Two girls are present. So we select 2 girls from the lot of 2 girls, which can be done in $^{2}{{C}_{2}}=1$ way
Hence, we have
n(E’) = 6+1 = 7
Hence, we have
$P\left( E' \right)=\dfrac{7}{10}$
We know that $P\left( E' \right)=1-P\left( E \right)$
Hence, we have
$1-P\left( E \right)=\dfrac{7}{10}\Rightarrow P\left( E \right)=\dfrac{3}{10}$, which is the same as obtained above.
Hence option [a] is correct.