$\begin{array}{1 1}(A)\;\large\frac{19}{30}\\(B)\;\large\frac{29}{30}\\(C)\;\large\frac{9}{28}\\(D)\;\text{None of these}\end{array} $

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

Step 1:

Given 60 students

n(NCC)=30

n(NSS)=32

$n(NCC\cap NSS)=24$

$\therefore P(NCC)=\large\frac{30}{60}$

$P(NSS)=\large\frac{32}{60}$

$P(NCC \cap NSS)=\large\frac{24}{60}$

Step 2:

P(Student opted for NCC or NSS)

$\Rightarrow P(NCC \cup NSS)=P(NCC)+P(NSS)-P(NCC \cap NSS)$

$\Rightarrow \large\frac{30}{60}+\frac{32}{60}-\frac{24}{60}$

$\Rightarrow \large\frac{30+32-24}{60}$

$\Rightarrow \large\frac{38}{60}$

$\Rightarrow \large\frac{19}{30}$

$\therefore$ The students opted for NCC or NSS is $\large\frac{19}{30}$

Hence (A) is the correct answer.

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