If P(A) = 0.4, P(B) = 0.7 and P(A|B) = 0.6, then \(P(A\cup B)\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.68
From the definition of conditional probability, \(P(A|B) = \frac{P(A\cap B)}{P(B)}\), so
\(P(A\cap B) = P(A|B)\times P(B)\)
\(P(A\cap B) = 0.6\times 0.7 = 0.42\)
By the addition theorem:
\(P(A\cup B) = P(A)+P(B)-P(A\cap B)\)
\(P(A\cup B) = 0.4+0.7-0.42 = 0.68\)
Why the other options are incorrect
Option (1) 0.42 — Incorrect. This is \(P(A\cap B)\), not \(P(A\cup B)\).
Option (2) 0.82 — Incorrect. It uses \(P(A\cap B) = 0.4\times 0.7 = 0.28\), as if A and B were independent; they are not, since \(P(A|B) = 0.6 \ne P(A) = 0.4\).
Option (4) 0.28 — Incorrect. This is only the product P(A)P(B), which is not \(P(A\cup B)\).