Let A and B be two events such that P(A) = 0.4, P(B) = 0.5 and \(P(A\cup B) = 0.7\). Consider the following statements:
(A) P(A|B) = 0.4
(B) A and B are independent.
(C) \(P(A\cap B') = 0.3\)
(D) \(P(B|A') = 0.5\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
By the addition theorem:
\(P(A\cap B) = P(A)+P(B)-P(A\cup B)\)
\(P(A\cap B) = 0.4+0.5-0.7 = 0.2\)
The parts outside the other event are
\(P(A\cap B') = 0.4-0.2 = 0.2\)
\(P(A'\cap B) = 0.5-0.2 = 0.3\)
Also \(P(A') = 0.6\) and \(P(A)P(B) = 0.4\times 0.5 = 0.2\).
So \(P(A\cap B) = 0.2 = P(A)P(B)\), \(P(A\cap B') = 0.2\), \(P(A|B) = 0.4\) and \(P(B|A') = 0.5\).
Checking the statements
(A) P(A|B) = 0.4 — Correct. \(\frac{P(A\cap B)}{P(B)} = \frac{0.2}{0.5} = 0.4\).
(B) A and B are independent. — Correct. \(P(A\cap B) = 0.2 = 0.4\times 0.5\), the product rule holds.
(C) \(P(A\cap B') = 0.3\) — Incorrect. \(P(A\cap B') = 0.4-0.2 = 0.2\).
(D) \(P(B|A') = 0.5\) — Correct. \(\frac{P(A'\cap B)}{P(A')} = \frac{0.3}{0.6} = 0.5\).