Question

In a class of $60$ students, $40$ opted for $NCC$, $30$ opted for $NSS$ and $20$ opted for both $NCC$ and $NSS$. If one of these students is selected at random, then the probability that the student selected has opted neither for $NCC$ nor for $NSS$ is :

• A. $\frac{2}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{5}{6}$

Answer 1

The correct option is B $\frac{1}{6}$

Since there are a total of $60$ students in the class, therefore,

$n\left(S\right)=60$

Let $A$ and $B$ be the event that a student opted for NCC and NSS respectively.

Given:-

$n\left(A\right)=40$

$n\left(B\right)=30$

$n(A\cap B)=20$

Therefore,

$P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{40}{60}=\frac{2}{3}$

$P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{30}{60}=\frac{1}{2}$

$P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{20}{60}=\frac{1}{3}$

Now, as we know that,

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$\therefore P(A\cup B)=\frac{2}{3}+\frac{1}{2}-\frac{1}{3}$

$\Rightarrow P(A\cup B)=\frac{4+3-2}{6}=\frac{5}{30}$

Now,

$P(A^{\prime }\cap B^{\prime })=P{(A\cup B)}^{\prime }=1-P(A\cup B)$

$\therefore P(A^{\prime }\cup B^{\prime })=1-\frac{5}{6}=\frac{6-5}{6}=\frac{1}{6}$

Thus the probability that the student selected has opted neither for NCC nor for NSS is $\frac{1}{6}$.

Hence the correct answer is $\left(B\right)\frac{1}{6}$.