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 The student has opted neither $NCC$ nor $NSS$.

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

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

$= P(A ^{\prime}$ and  $B ^{\prime})$

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

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

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

$=1- P ( A$ or $B )$

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

$=\frac{11}{30}$

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