If A and B are two events such that \(P(A) = 0.6\), \(P(B) = 0.5\) and \(P(A \cup B) = 0.8\), then \(P(A' | B)\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.4
By the addition rule:
\( P(A \cap B) = P(A) + P(B) - P(A \cup B) \)
\( = 0.6 + 0.5 - 0.8 = 0.3 \)
The part of B outside A:
\( P(A' \cap B) = P(B) - P(A \cap B) \)
\( = 0.5 - 0.3 = 0.2 \)
By the definition of conditional probability:
\( \displaystyle P(A' | B) = \frac{P(A' \cap B)}{P(B)} \)
\( \displaystyle = \frac{0.2}{0.5} = 0.4 \)
Why the other options are incorrect
Option (1) 0.6 — Incorrect. This is \(P(A|B) = \frac{0.3}{0.5}\), the probability of A itself given B.
Option (2) 0.2 — Incorrect. This is \(P(A' \cap B)\), not yet divided by \(P(B) = 0.5\).
Option (4) 0.5 — Incorrect. This is \(P(B|A) = \frac{0.3}{0.6}\), which conditions on A instead of B.