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(D) Given that $A$ and $B$ are mutually exclusive events,so $A \cap B = \phi$,which implies $P(A \cap B) = 0$.By the definition of conditional probabi

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$\frac{P(A)}{1-P(B)}$

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(D) Given that $A$ and $B$ are mutually exclusive events,so $A \cap B = \phi$,which implies $P(A \cap B) = 0$.By the definition of conditional probability,$P(A \mid \bar{B}) = \frac{P(A \cap \bar{B})}{P(\bar{B})}$.Since $A = (A \cap B) \cup (A \cap \bar{B})$ and these are disjoint,$P(A) = P(A \cap B) + P(A \cap \bar{B})$.Thus,$P(A \cap \bar{B}) = P(A) - P(A \cap B) = P(A) - 0 = P(A)$.Also,$P(\bar{B}) = 1 - P(B)$.Therefore,$P(A \mid \bar{B}) = \frac{P(A)}{1 - P(B)}$.

If $A$ and $B$ are mutually exclusive events with $P(B) \neq 1$,then $P(A \mid \bar{B})$ is equal to (Here $\bar{B}$ is the complement of the event $B$)

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