# 21.(iii)    In a class of $\small 60$ students, $\small 30$ opted for NCC, $\small 32$ opted for NSS and $\small 24$ opted for both NCC and NSS. If one of these students is selected at random, find the probability that                (iii) The student has opted NSS but not NCC.

H Harsh Kankaria

Let A be the event that student opted for NCC and B be the event that the student opted for NSS.

Given,

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

Therefore, P(A) = $\inline \dpi{100} \frac{30}{60} = \frac{1}{2}$

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

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

(iii) Now,

Probability that the student has opted NSS but not NCC = P(B $\cap$ A' ) = P(B-A)

We know,

P(B-A) = P(B) -  P(A $\cap$ B)

$= \frac{8}{15}- \frac{2}{5} = \frac{8-6}{15}$

$= \frac{2}{15}$

Hence,the probability that the student has opted NSS but not NCC is $\frac{2}{15}$

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