NCERT Class 11 Mathematics — Probability, Exercise 11.2, Question 21.
Problem
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, find the probability that:
(a) the student opted for NCC or NSS, (b) the student has opted for neither NCC nor NSS, (c) the student has opted for NSS but not NCC.
Key insight. With “both” given directly, this is a straight application of the addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ — the other two parts fall out of that one number by the complement rule and by subtracting the overlap.
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Approach
Let $A$ be the event “the student opted for NCC” and $B$ be “the student opted for NSS.” The counts given translate directly into probabilities out of 60, and the addition rule for probability handles the overlap between NCC and NSS in one step. The remaining two parts follow from that first result — neither is the complement of “NCC or NSS,” and “NSS but not NCC” is what’s left of NSS once the overlap is removed.
Solution
Step 1 — Convert counts to probabilities
Out of 60 students:
$$P(A) = \frac{30}{60} = \frac12, \qquad P(B) = \frac{32}{60} = \frac{8}{15}, \qquad P(A\cap B) = \frac{24}{60} = \frac25$$
Step 2 — (a) P(NCC or NSS)
By the addition rule,
$$P(A\cup B) = P(A)+P(B)-P(A\cap B)$$
Writing every term over the common denominator 30:
$$P(A\cup B) = \frac{15}{30}+\frac{16}{30}-\frac{12}{30} = \frac{19}{30}$$
This is the probability because “opted for NCC or NSS” covers every student in NCC, every student in NSS, or both — exactly the union of the two events.
Step 3 — (b) P(neither NCC nor NSS)
“Neither” is precisely the complement of “NCC or NSS” — a student who is in neither event is, by definition, not in the union:
$$P(\text{neither}) = 1-P(A\cup B) = 1-\frac{19}{30} = \frac{11}{30}$$
Step 4 — (c) P(NSS but not NCC)
A student who opted for NSS but not NCC is in $B$ but outside $A\cap B$, so this probability is what remains of $P(B)$ once the overlap with NCC is removed:
$$P(B \text{ but not } A) = P(B)-P(A\cap B) = \frac{32}{60}-\frac{24}{60} = \frac{8}{60} = \frac{2}{15}$$
Answer
$$P(\text{NCC or NSS})=\frac{19}{30},\qquad P(\text{neither})=\frac{11}{30},\qquad P(\text{NSS but not NCC})=\frac{2}{15}$$
Common mistakes
- Adding P(NCC) and P(NSS) without subtracting the overlap. Doing so double-counts the 24 students who are in both, inflating $P(A\cup B)$ past its true value of $\frac{19}{30}$.
- Computing “neither” from raw counts instead of using the complement of the union. It’s easy to try to count the “neither” students directly and make an arithmetic slip; going via $1-P(A\cup B)$ is both faster and safer once part (a) is already correct.
- Confusing “NSS but not NCC” with $P(B)$ alone. The phrase specifically excludes the overlap, so $P(B)$ on its own overstates the answer by exactly the 24-student overlap.
Practise next
- In a class of 50 students, 20 play cricket, 25 play football and 10 play both. Find the probability that a randomly selected student plays cricket or football, plays neither, and plays cricket but not football.
Show answer
$\tfrac{7}{10}$, $\tfrac{3}{10}$ and $\tfrac15$. By inclusion–exclusion, $|C\cup F|=20+25-10=35$.
So $P(C\cup F)=\tfrac{35}{50}=\tfrac{7}{10}$, and neither is the complement, $\tfrac{15}{50}=\tfrac3{10}$.
Cricket but not football is $20-10=10$ students, so $\tfrac{10}{50}=\tfrac15$ — subtracting the overlap from the cricket total, not from the class.
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