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# 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 student students is selected at random,find the probability that the student opted for NCC or NSS

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

Toolbox:
• $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
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.