A and B are mutually exclusive events such that P(A) = 0.2 and P(B) = 0.3. Then \(P(A|A\cup B)\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.4
Since A and B are mutually exclusive, \(P(A\cap B) = 0\), so
\(P(A\cup B) = 0.2+0.3 = 0.5\)
The event A lies inside \(A\cup B\), so \(A\cap(A\cup B) = A\). By the definition of conditional probability:
\( \displaystyle P(A|A\cup B) = \frac{P(A)}{P(A\cup B)} \)
\( \displaystyle = \frac{0.2}{0.5} = 0.4 \)
Why the other options are incorrect
Option (1) 0.6 — Incorrect. This is \(\frac{0.3}{0.5} = P(B|A\cup B)\), the chance that it was B.
Option (2) 0.5 — Incorrect. This is \(P(A\cup B)\), the probability of the given event.
Option (3) 0.2 — Incorrect. This is P(A) itself; the condition “A or B has happened” raises the chance of A.