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 us define the events as, A= student opted for NCC, B= student opted for NSS

We have, n(A)= 30, n(B)= 32, $n(A\cap B)=24$ and n(S)= 60

$\therefore P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}=\frac{30}{60},P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{32}{60}$ and $P(A\cap B)=\frac{n(A\cap B)}{n\left(S\right)}=\frac{24}{60}$

(i) Probability that the student opted for NCC or NSS $P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$
$=\frac{30}{60}+\frac{32}{60}-\frac{24}{60}=\frac{38}{60}=\frac{19}{30}$

(ii) Probability that the student has opted neither NCC nor NSS, $P(A\cup B{)}^{\prime }=1-P(A\cup B)$
$=1-\frac{19}{30}=\frac{11}{30}$

(iii) Probability that the student has opted NSS but not NCC, $P(B\cap A^{\prime })=p\left(B\right)-P(A\cap B)$
$=\frac{32}{60}-\frac{24}{60}=\frac{8}{60}=\frac{2}{15}$