Let A and B be two events such that \(P(A|B) = P(A|B') = 0.3\) and P(B) = 0.4. Then \(P(A\cup B)\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.58
By the multiplication theorem, the part of A inside B and the part of A outside B are
\(P(A\cap B) = P(B)\,P(A|B) = 0.4\times 0.3 = 0.12\)
\(P(A\cap B') = P(B')\,P(A|B') = 0.6\times 0.3 = 0.18\)
The event \(A\cup B\) is made up of B and the part of A outside B, which do not overlap:
\(P(A\cup B) = P(B)+P(A\cap B')\)
\(P(A\cup B) = 0.4+0.18 = 0.58\)
Also \(P(A) = 0.12+0.18 = 0.3\), so \(P(A\cap B) = 0.12 = P(A)P(B)\): equal chances of A with and without B make A and B independent.
Why the other options are incorrect
Option (1) 0.70 — Incorrect. This is \(0.3+0.4\), which leaves out the overlap \(P(A\cap B) = 0.12\).
Option (2) 0.12 — Incorrect. This is \(P(A\cap B)\), the probability that both events happen.
Option (4) 0.88 — Incorrect. This is \(1-0.12\), the probability that A and B do not both happen.