Question

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, then match the probability of events of Column I with their respective values in Column II and choose the correct option from the codes given below.

Column I		Column II
A	The student opted for NCC or NSS.	1	$\frac{2}{15}$
B	The student has opted neither NCC nor NSS	2	$\frac{11}{30}$
C	The student has opted NSS but not NCC	3	$ \frac{19}{30}$

• A. A - (1), B -(2), C -(3)
• B. A - (2), B -(3), C -(1)
• C. A - (2), B -(1), C -(3)
• D. A - (3), B -(2), C -(1)

Answer 1

Answer: (D)

Let $A$ and $B$ denote the students in NCC and NSS respectively.
Here, $ n(A)=30, n(B)=32$
$n(A \cap B)=24$
$(\because 24$ students opt for both NCC and NSS i.e., they are common in both)
$\therefore P(A)=\frac{30}{60} $
$ P(B)=\frac{32}{60} $ and $ P(A \cap B)=\frac{24}{60}$
A. $P$ (student opted for NCC or NSS)
$P(A \cup B) =P(A)+P(B)-P(A \cap B) $
$[\because P(A \cup B)=P(A \circ B)]$
$ =\frac{30}{60}+\frac{32}{60}-\frac{24}{60}$
$ =\frac{30+32-24}{60}=\frac{62-24}{60}=\frac{38}{60}=\frac{19}{30}$
B. $P$ (student has opted neither NCC nor NSS)
$=1-P$ (student opted for NCC or NSS)
$=1-\frac{19}{30}=\frac{30-19}{30}=\frac{11}{30}$
C. $P$ (the student has opted NSS but not NCC)
$ =P(B)-P(A \cap B)$
$ =\frac{32}{60}-\frac{24}{60}=\frac{8}{60}=\frac{2}{15}$