In a class of 80 students, 48 opted for NCC, 48 opted for NSS
and 32 opted to both NCC and NSS. If one of these students
is selected at random. Find the probability that
(i) The students opted for NCC or NSS
(ii)The students has opted neither NCC nor NSS
(iii) The students has opted NSS but not NCC
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1
Answer:
(i) We know that P(A∪B)=P(A)+P(B)−P(A∩B)
∴P(A∪B)=
2
1
+
15
8
−
5
2
=
30
15+16−12
=
30
19
Thus the probability that the selected student has opted for NCC or NSS is
30
19
(ii) P(not A and not B)
= P(A' and B')
= P(A
′
∩B
′
)
= P(A∪B)
′
[(A
′
∩B
′
)=(A∪B)
′
(by De Morgan's law)]
=1−P(A∪B)
=1−P(AorB)
=1−
30
19
=
30
11
Thus the probability that the selected students has neither opted for NCC nor NSS is
30
11
(iii) Number of students who have opted for NSS but not NCC
=n(B−A)=n(B)−n(A∩B)
=32−24=8
So, the probability that the student has opted for NSS but not NCC is
60
8
=
15
2
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