Question

In a school there are 3 sections A, B, and C. Section A contains 20 girls and 30 boys, section B contains 40 girls and 20 boys, and section C contains 10 girls and 30 boys. The probabilities of selecting section A, B, and C are 0.2, 0.3, and 0.5, respectively. If a student selected at random from the school is a girl, then the probability that she belongs to section A is

• A. $\frac{121}{200}$
• B. $\frac{16}{121}$
• C. $\frac{14}{81}$
• D. $\frac{16}{81}$

Hint:

Use Bayes’ theorem, $P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)}$, and compute $P(B)$ via the law of total probability for conditional probabilities across multiple groups.

Answer 1

The Correct Option is D

Let $G$ be the event that the student is a girl, and $A$, $B$, $C$ be the events that the student is from section A, B, or C, respectively. Given: - $P(A) = 0.2$, $P(B) = 0.3$, $P(C) = 0.5$. - Section A: 20 girls, 30 boys, total 50. $P(G \mid A) = \frac{20}{50} = \frac{2}{5}$. - Section B: 40 girls, 20 boys, total 60. $P(G \mid B) = \frac{40}{60} = \frac{2}{3}$. - Section C: 10 girls, 30 boys, total 40. $P(G \mid C) = \frac{10}{40} = \frac{1}{4}$. We need $P(A \mid G)$. Using Bayes’ theorem: \[ P(A \mid G) = \frac{P(G \mid A) P(A)}{P(G)} \] \[ P(G) = P(G \mid A) P(A) + P(G \mid B) P(B) + P(G \mid C) P(C) = \left( \frac{2}{5} \cdot 0.2 \right) + \left( \frac{2}{3} \cdot 0.3 \right) + \left( \frac{1}{4} \cdot 0.5 \right) \] \[ = 0.08 + 0.2 + 0.125 = 0.405 \] \[ P(A \mid G) = \frac{\frac{2}{5} \cdot 0.2}{0.405} = \frac{0.08}{0.405} = \frac{80}{405} = \frac{16}{81} \] Option (4) is correct. Options (1), (2), and (3) do not match the computed probability.