Appeared in: 2024 (5 sets)
If A and B are events such that $P(A/B) = P(B/A) \neq 0$, then: (A) $A \subset B$, but $A \neq B$ (B) $A = B$ (C) $A \cap B = \phi$ (D) $P(A) = P(B)$
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Official Evaluation Steps
Step 1 (Formula / Governing Law):
The conditional probability of an event is defined by the formula:
$$P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) \neq 0$$ $$P(B|A) = \frac{P(B \cap A)}{P(A)}, \quad P(A) \neq 0$$Given that $P(A|B) = P(B|A) \neq 0$. (1/2 Mark)
Step 2 (Step-by-Step Working):
Equating the two expressions based on the given condition:
$$\frac{P(A \cap B)}{P(B)} = \frac{P(B \cap A)}{P(A)}$$Since $P(A \cap B) = P(B \cap A)$ (commutative property of intersection), and it is given that $P(A|B) \neq 0$, it implies $P(A \cap B) \neq 0$. We can divide both sides by $P(A \cap B)$:
$$\frac{1}{P(B)} = \frac{1}{P(A)}$$Taking the reciprocal of both sides:
$$P(A) = P(B)$$(1/4 Mark)
Step 3 (Final Answer):
Comparing the derived result with the given options, the correct option is (D).
Result: $P(A) = P(B)$
(1/4 Mark)