Let A&B be the sets for NCC and NSS respectively.
Given n(A)=40,n(B)=30,n(A∩B)=20
∴n(A∪B)=n(A)+n(B)−n(A∩B)
=40+30−20=50
∴P(A∪B)=6050=65
∴P(A∪B)′=1−P(A∪B)
=1−65=61
In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is :
Held on 12 Jan 2019 · Verified 6 Jul 2026.
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