Question

If $ P(\bar{A}) = 0.3,\ P(B) = 0.4,\ P(A \cap \bar{B}) = 0.5 $, then find $ P(B / (A \cup \bar{B})) $

• A. 0.25
• B. 0.6
• C. 0.45
• D. 0.55

Hint:

Use conditional probability: \( P(B | E) = \frac{P(B \cap E)}{P(E)} \), and compute with given complements.

Answer 1

The Correct Option is A

We know: \[ P(A) = 1 - P(\bar{A}) = 0.7,\quad P(\bar{B}) = 0.6 \Rightarrow P(A \cup \bar{B}) = 1 - P(\bar{A} \cap B) \] But: \[ P(A \cap \bar{B}) = 0.5 \Rightarrow P(A \cup \bar{B}) = P(A) + P(\bar{B}) - 0.5 = 0.7 + 0.6 - 0.5 = 0.8 \] Now: \[ P(B \cap (A \cup \bar{B})) = P(B) - P(B \cap \bar{A}) = 0.4 - (P(B) - P(A \cap B)) \Rightarrow P(A \cup \bar{B}) = 0.8,\ P(B \cap (A \cup \bar{B})) = 0.2 \] \[ \Rightarrow P(B / (A \cup \bar{B})) = \frac{0.2}{0.8} = 0.25 \]