Question

It is given that in a random experiment, events A and B are such that \( P(A) = \frac{1}{4} ,P(A|B) = \frac{1}{2} \) and \( P(B|A) = \frac{2}{3} \). Then \( P(B) \) is:

• A. \( \frac{1}{3} \)
• B. \( \frac{2}{3} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{1}{6} \)

Hint:

When dealing with conditional probabilities, always use Bayes' theorem to express relationships between events.

Answer 1

The Correct Option is A

We are given: \[ P(A|B) = \frac{1}{2}, \quad P(B|A) = \frac{2}{3}. \] We know that: \[ P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad P(B|A) = \frac{P(A \cap B)}{P(A)}. \] Let \( P(A \cap B) = p \), \( P(A) = p_1 \), and \( P(B) = p_2 \). Then: From \( P(A|B) = \frac{1}{2} \), we have: \[ \frac{p}{p_2} = \frac{1}{2} \quad \Rightarrow \quad p = \frac{p_2}{2}. \] From \( P(B|A) = \frac{2}{3} \), we have: \[ \frac{p}{p_1} = \frac{2}{3} \quad \Rightarrow \quad p = \frac{2p_1}{3}. \] Equating the two expressions for \( p \), we get: \[ \frac{p_2}{2} = \frac{2p_1}{3} \quad \Rightarrow \quad 3p_2 = 4p_1 \quad \Rightarrow \quad p_2 = \frac{4p_1}{3}. \] Now, we use Bayes' Theorem: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \quad \Rightarrow \quad \frac{p}{p_2} = \frac{1}{3}. \] So, the correct answer is \( P(B) = \frac{1}{3} \).