If A and B are events such that P(A | B) = P( | vedclass

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(C) Given that $P(A | B) = P(B | A)$.Using the definition of conditional probability,we have $P(A | B) = \frac{P(A \cap B)}{P(B)}$ and $P(B | A) = \fr

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$P(A) = P(B)$

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(C) Given that $P(A | B) = P(B | A)$.Using the definition of conditional probability,we have $P(A | B) = \frac{P(A \cap B)}{P(B)}$ and $P(B | A) = \frac{P(A \cap B)}{P(A)}$.Substituting these into the given equation:$\frac{P(A \cap B)}{P(B)} = \frac{P(A \cap B)}{P(A)}$.Case $1$: If $P(A \cap B) 
eq 0$,we can divide both sides by $P(A \cap B)$,which gives $\frac{1}{P(B)} = \frac{1}{P(A)}$,implying $P(A) = P(B)$.Case $2$: If $P(A \cap B) = 0$,then $0 = 0$,which is always true,but in the context of standard probability problems,the implication $P(A) = P(B)$ is the intended relationship.Thus,the correct answer is $C$.

If $A$ and $B$ are events such that $P(A | B) = P(B | A)$,then

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