Let A and B be two events of a sample space S such that \(P(A)\ne 0\) and \(P(B)\ne 0\). Consider the following statements:
(A) \(P(A|B)+P(A'|B) = 1\)
(B) If \(A\subset B\), then \(P(B|A) = 1\).
(C) If A and B are mutually exclusive, then A and B are independent.
(D) \(P(S|B) = P(B)\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (B) only
By the definition of conditional probability, \(P(E|B) = \displaystyle\frac{P(E\cap B)}{P(B)}\) for any event E. Since \(A\cap B\) and \(A'\cap B\) together make up B,
\(P(A\cap B)+P(A'\cap B) = P(B)\)
If \(A\subset B\), then \(A\cap B = A\). Also \(S\cap B = B\). For mutually exclusive events \(P(A\cap B) = 0\), while \(P(A)P(B)\ne 0\).
So the two conditional probabilities given B add up to 1, \(P(B|A) = \frac{P(A)}{P(A)} = 1\), \(P(S|B) = \frac{P(B)}{P(B)} = 1\), and mutually exclusive events with non-zero probabilities fail the product rule.
Checking the statements
(A) \(P(A|B)+P(A'|B) = 1\) — Correct. The sum is \(\frac{P(A\cap B)+P(A'\cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1\).
(B) If \(A\subset B\), then \(P(B|A) = 1\). — Correct. \(P(B|A) = \frac{P(A\cap B)}{P(A)} = \frac{P(A)}{P(A)} = 1\).
(C) If A and B are mutually exclusive, then A and B are independent. — Incorrect. \(P(A\cap B) = 0\), but \(P(A)P(B)\) is not 0.
(D) \(P(S|B) = P(B)\) — Incorrect. \(P(S|B) = \frac{P(B)}{P(B)} = 1\).