Committee of 4 From 8 Boys, 4 Girls — P(Exactly 2 Girls | At Least 1 Girl)

NCERT Exemplar, Class 12 Mathematics — Probability.

Problem

A committee of 4 students is selected at random from a group consisting of 8 boys and 4 girls. Given that there is at least one girl in the committee, find the probability that there are exactly 2 girls in the committee.

Key insight. “Exactly 2 girls” is automatically a subset of “at least 1 girl” — every committee with exactly 2 girls already satisfies the given condition. So $P(B\cap A)$ in the conditional probability formula is simply $P(B)$ itself, and the only real work is computing $P(A)$, the probability of the given condition.

Watch this explained step by step →  ·  More on the Shiwam’s Classes channel

Approach

Let $A$ be the event “at least one girl in the committee” and $B$ be “exactly two girls.” Since $B$ is entirely contained within $A$, the required probability is $P(B|A) = \dfrac{P(B)}{P(A)}$. Compute $P(A)$ as $1-P(\text{no girls})$, and $P(B)$ directly using combinations, then divide.

Solution

Step 1 — Define the events

Let $A$ = “at least one girl in the committee” and $B$ = “exactly two girls in the committee.” Since every committee with exactly 2 girls automatically has at least one girl, $B \subseteq A$, so $B\cap A = B$.

Step 2 — Find the total number of ways to choose the committee

$$\text{Total ways} = {}^{12}C_4 = \frac{12\times11\times10\times9}{4\times3\times2\times1} = 495$$

Step 3 — Find P(A) using the complement

The complement of “at least one girl” is “no girls at all” (all 4 members are boys):

$$P(\text{no girls}) = \frac{{}^4C_0 \times {}^8C_4}{{}^{12}C_4} = \frac{1\times70}{495} = \frac{70}{495}$$

$$P(A) = 1-\frac{70}{495} = \frac{425}{495}$$

Step 4 — Find P(B)

$$P(B) = \frac{{}^4C_2\times{}^8C_2}{{}^{12}C_4} = \frac{6\times28}{495} = \frac{168}{495}$$

Step 5 — Apply the conditional probability formula

$$P(B|A) = \frac{P(B\cap A)}{P(A)} = \frac{P(B)}{P(A)} = \frac{\frac{168}{495}}{\frac{425}{495}} = \frac{168}{425}$$

Answer

Common mistakes

• Computing $P(B\cap A)$ as if $B$ and $A$ were unrelated events, by trying to count committees with “at least one girl AND exactly two girls” as a separate, more complex calculation. Since $B$ is entirely contained in $A$, $B\cap A$ is simply $B$ — no extra counting needed.
• Using $1-P(\text{no girls})$ as the denominator but forgetting to cancel the common ${}^{12}C_4$ in the final division — carrying it through unnecessarily just adds arithmetic without changing the answer.

Practise next

• A committee of 3 is selected from 6 boys and 3 girls. Given at least one girl is selected, find the probability of exactly 2 girls, using the same conditional probability approach.