A and B are independent events such that P(A) = 2P(B) and \(P(A\cap B) = 0.18\). Then \(P(A\cup B)\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.72
Let P(B) = p, so that P(A) = 2p. For independent events the chance of both is the product:
\(P(A\cap B) = 2p\times p = 2p^{2}\)
\(2p^{2} = 0.18\)
\(p^{2} = 0.09\)
A probability cannot be negative, so
\(P(B) = 0.3,\quad P(A) = 0.6\)
By the addition theorem:
\(P(A\cup B) = P(A)+P(B)-P(A\cap B)\)
\(P(A\cup B) = 0.6+0.3-0.18 = 0.72\)
Why the other options are incorrect
Option (1) 0.9 — Incorrect. This is 0.6 + 0.3; the case when both occur is counted twice.
Option (2) 0.54 — Incorrect. This is \(0.6\times 0.7+0.4\times 0.3\), the chance that exactly one of A and B occurs.
Option (4) 0.28 — Incorrect. This is \(0.4\times 0.7\), the chance that neither A nor B occurs.