Question

In a class of $60$ students, $30$ opted for NCC, $32$ opted for NSS and $24$ opted for both NCC and NSS. If one of these students is selected at random, find the probability that
(i) The student opted for NCC or NSS
(ii) The student has opted neither NCC nor NSS
(iii) The student has opted NSS but not NCC

Answer 1

Let $A$ be the event in which the selected student has opted for NCC and $B$ be the event in which the selected student has opted for NSS
Total number of students $=60$
Given, number of students who have opted for NCC i.e. $n\left(A\right)=30$
$\therefore P\left(A\right)=\frac{30}{60}=\frac{1}{2}$

Also, given number of students who have opted for NSS i.e. $n\left(B\right)=32$
$\therefore P\left(B\right)=\frac{32}{60}=\frac{8}{15}$

Also, given number of students who have opted for both NCC and NSS i.e. $n(A\cap B)=24$
$\therefore P(A\cap B)=\frac{24}{60}=\frac{2}{5}$

(i) We know that $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$\therefore P(A\cup B)=\frac{1}{2}+\frac{8}{15}-\frac{2}{5}$ = $\frac{15+16-12}{30}=\frac{19}{30}$
Thus the probability that the selected student has opted for NCC or NSS is $\frac{19}{30}$

(ii) P(not A and not B)
= P(A' and B')
= $P(A^{\prime }\cap B^{\prime })$
= $P{(A\cup B)}^{\prime }$ [$(A^{\prime }\cap B^{\prime })={(A\cup B)}^{\prime }$ (by De Morgan's law)]
$=1-P(A\cup B)$
$=1-P\left(AorB\right)$
$=1-\frac{19}{30}$
$=\frac{11}{30}$
Thus the probability that the selected students has neither opted for NCC nor NSS is $\frac{11}{30}$

(iii) Number of students who have opted for NSS but not NCC
$=n(B-A)=n\left(B\right)-n(A\cap B)$

$=32-24=8$

So, the probability that the student has opted for NSS but not NCC is $\frac{8}{60}=\frac{2}{15}$