Question

Let \( A, B \) be two events of a sample space such that \( P(A) = \frac{1}{4} \), \( P(B|A) = \frac{1}{2} \), \( P(A|B) = \frac{1}{4} \). If \( \overline{B} \) is the complement of \( B \), then \( P(A | \overline{B}) \) is _____

• A. \( \dfrac{3}{4} \)
• B. \( \dfrac{1}{4} \)
• C. \( \dfrac{1}{3} \)
• D. \( \dfrac{2}{3} \)

Hint:

When dealing with conditional probabilities involving complements, use \( P(A|\overline{B}) = \frac{P(A) - P(A \cap B)}{1 - P(B)} \) to simplify.

Answer 1

The Correct Option is B

We are given: \[ P(A) = \frac{1}{4},\quad P(B|A) = \frac{1}{2},\quad P(A|B) = \frac{1}{4} \] From the definition of conditional probability: \[ P(A \cap B) = P(B) \cdot P(A|B) = P(B) \cdot \frac{1}{4} \] Also: \[ P(A \cap B) = P(A) \cdot P(B|A) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \] So equating the two expressions for \( P(A \cap B) \): \[ \frac{1}{8} = P(B) \cdot \frac{1}{4} \Rightarrow P(B) = \frac{1}{2} \] Now we calculate \( P(A \cap \overline{B}) \): \[ P(A \cap \overline{B}) = P(A) - P(A \cap B) = \frac{1}{4} - \frac{1}{8} = \frac{1}{8} \] And: \[ P(\overline{B}) = 1 - P(B) = 1 - \frac{1}{2} = \frac{1}{2} \] Therefore: \[ P(A|\overline{B}) = \frac{P(A \cap \overline{B})}{P(\overline{B})} = \frac{\frac{1}{8}}{\frac{1}{2}} = \frac{1}{4} \]