If A and B are two events such that P(A|B) = 0.5, P(B|A) = 0.4 and \(P(A\cap B) = 0.12\), then \(P(A\cup B)\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.42
By the multiplication theorem, \(P(A\cap B) = P(A)\,P(B|A)\) and also \(P(A\cap B) = P(B)\,P(A|B)\). So
\( \displaystyle P(A) = \frac{P(A\cap B)}{P(B|A)} = \frac{0.12}{0.4} = 0.3 \)
\( \displaystyle P(B) = \frac{P(A\cap B)}{P(A|B)} = \frac{0.12}{0.5} = 0.24 \)
By the addition theorem:
\(P(A\cup B) = P(A)+P(B)-P(A\cap B)\)
\(P(A\cup B) = 0.3+0.24-0.12 = 0.42\)
Why the other options are incorrect
Option (1) 0.54 — Incorrect. This is P(A) + P(B); the common part 0.12 has not been subtracted.
Option (2) 0.66 — Incorrect. This adds the common part 0.12 instead of subtracting it.
Option (3) 0.30 — Incorrect. This is P(A) only.