A and B are independent events such that \(P(A\cup B) = 0.8\) and \(P(A\cap B') = 0.3\). Then \(P(A|A\cup B)\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.75
Let P(A) = a and P(B) = b. Since A and B are independent, so are A and B’, and A’ and B’:
\(P(A\cap B') = a(1-b) = 0.3\)
\(P(A'\cap B') = 1-P(A\cup B) = 0.2\)
\((1-a)(1-b) = 0.2\)
Adding the two equations:
\(a(1-b)+(1-a)(1-b) = 0.5\)
\(1-b = 0.5\)
\(b = 0.5,\quad a = \frac{0.3}{0.5} = 0.6\)
Since \(A\subset A\cup B\), the common part \(A\cap(A\cup B)\) is A itself. By the definition of conditional probability:
\( \displaystyle P(A|A\cup B) = \frac{P(A)}{P(A\cup B)} = \frac{0.6}{0.8} = 0.75 \)
Check: \(P(A\cup B) = 0.6+0.5-0.3 = 0.8\).
Why the other options are incorrect
Option (1) 0.6 — Incorrect. This is P(A); it still has to be divided by \(P(A\cup B)\).
Option (2) 0.5 — Incorrect. This is P(B).
Option (4) 0.375 — Incorrect. This is \(\frac{0.3}{0.8}\), which uses \(P(A\cap B')\) in place of the whole of P(A).