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:

Let the universal set be the whole class of 60 students. We denote by $$n(\text{NCC})$$ the number of students who opted for NCC, by $$n(\text{NSS})$$ the number who opted for NSS, and by $$n(\text{NCC} \cap \text{NSS})$$ the number who opted for both NCC and NSS.

According to the question, we have $$n(\text{NCC}) = 40,$$ $$n(\text{NSS}) = 30,$$ $$n(\text{NCC} \cap \text{NSS}) = 20.$$

First, we find the total number of students who opted for at least one of the two activities. For two sets, the principle of inclusion-exclusion states

$$n(\text{NCC} \cup \text{NSS}) = n(\text{NCC}) + n(\text{NSS}) - n(\text{NCC} \cap \text{NSS}).$$

Substituting the given numbers, we get $$n(\text{NCC} \cup \text{NSS}) = 40 + 30 - 20 = 50.$$

Now, the students who opted for neither NCC nor NSS are outside this union. So their count is obtained by subtracting the size of the union from the total class strength:

$$n(\text{neither}) = 60 - n(\text{NCC} \cup \text{NSS}) = 60 - 50 = 10.$$

The experiment is to choose one student at random from the class. The probability that the selected student belongs to the “neither” category equals the ratio of favourable outcomes to total outcomes, namely

$$\text{Probability} = \frac{n(\text{neither})}{\text{total students}} = \frac{10}{60} = \frac{1}{6}.$$

Hence, the correct answer is Option A.