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:

Since there are a total of 60 students in the class, therefore, n(S) = 60
Let A and B be the event that a student opted for NCC and NSS respectively.
Given:-
n(A) = 40
n(B) = 30
n(A∩B) = 20
Therefore,
P(A) = n(A)/n(S) = 40/60 = 2/3
P(B) = n(B)/n(S) = 30/60 = 1/2
P(A∩B) = n(A∩B)/n(S) = 20/60 = 1/3
Now, as we know that,
P(A∪B) = P(A) + P(B) − P(A∩B)
∴P(A∪B) = 2/3 + 1/2 − 1/3
⇒P(A∪B) = (4 + 3)/6 = 7/6
Now, P(A'∩B') = P(A∪B)' = 1 − P(A∪B)
∴P(A'∩B') = 1 − 7/6 = −1/6
This is incorrect. Let's recalculate P(A∪B):
P(A∪B) = P(A) + P(B) - P(A∩B) = 40/60 + 30/60 - 20/60 = 50/60 = 5/6
Now, P(A'∩B') = P(A∪B)' = 1 - P(A∪B) = 1 - 5/6 = 1/6
Thus the probability that the student selected has opted neither for NCC nor for NSS is 1/6.
Hence the correct answer is (B) 1/6. Note that the options given do not include 1/6. There may be an error in the question or the provided options.