If A and B are two events such that P(A) = 0.3, P(B) = 0.4 and \(P(A\cup B) = 0.5\), then \(P(A|B')\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{6}\)
By the addition theorem:
\(P(A\cap B) = P(A)+P(B)-P(A\cup B)\)
\(P(A\cap B) = 0.3+0.4-0.5 = 0.2\)
The part of A outside B is
\(P(A\cap B') = P(A)-P(A\cap B)\)
\(P(A\cap B') = 0.3-0.2 = 0.1\)
Also \(P(B') = 1-0.4 = 0.6\). By the definition of conditional probability:
\( \displaystyle P(A|B') = \frac{P(A\cap B')}{P(B')} = \frac{0.1}{0.6} = \frac{1}{6} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{3}\) — Incorrect. This is \(\frac{0.1}{0.3} = P(B'|A)\), which divides by P(A) instead of P(B').
Option (2) \(\displaystyle\frac{1}{2}\) — Incorrect. This is \(\frac{0.2}{0.4} = P(A|B)\), the condition B instead of B'.
Option (4) \(\displaystyle\frac{5}{6}\) — Incorrect. This is \(P(A'|B') = \frac{0.5}{0.6}\), the probability that A does not happen, given B'.