In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS.

Solution:

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

Number of students who have opted for NCC = 30

$\therefore P(A)=\frac{30}{60}=\frac{1}{2}$

Number of students who have opted for NSS = 32

$\therefore P(B)=\frac{32}{60}=\frac{8}{15}$

Number of students who have opted for both NCC and NSS = 24

$\therefore \mathrm{P}(\mathrm{A}$ and $\mathrm{B})=\frac{24}{60}=\frac{2}{5}$

(i) We know that P(A or B) = P(A) + P(B) – P(A and B)

$\therefore P(A$ or $B)=\frac{1}{2}+\frac{8}{15}-\frac{2}{5}=\frac{15+16-12}{30}=\frac{19}{30}$

Thus, the probability that the selected student has opted for NCC or NSS is $\frac{19}{30}$.

(ii)

$P$ (not $A$ and not $B$ )

$=\mathrm{P}\left(\mathrm{A}^{\prime}\right.$ and $\left.\mathrm{B}^{\prime}\right)$

$=\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)$

$=\mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime} \quad\left[\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=(\mathrm{A} \cup \mathrm{B})^{\prime}(\right.$ by De Morgan's law $\left.)\right]$

$=1-P(A \cup B)$

$=1-\mathrm{P}(\mathrm{A}$ or $\mathrm{B})$

$=1-\frac{19}{30}$

$=\frac{11}{30}$

Thus, the probability that the selected students has neither opted for NCC nor NSS is $\frac{11}{30}$.

(iii) The given information can be represented by a Venn diagram as

It is clear that

Number of students who have opted for NSS but not NCC

= n(B – A) = n(B) – n(A ∩ B) = 32 – 24 = 8

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