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(C) By the definition of conditional probability,we have:$P\left( \frac{\bar{A}}{\bar{B}} \right) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})}$Using D

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

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(C) By the definition of conditional probability,we have:$P\left( \frac{\bar{A}}{\bar{B}} \right) = \frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})}$Using De Morgan's Law,we know that $\bar{A} \cap \bar{B} = \overline{A \cup B}$.Therefore,$P(\bar{A} \cap \bar{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B)$.Substituting this into the formula,we get:$P\left( \frac{\bar{A}}{\bar{B}} \right) = \frac{1 - P(A \cup B)}{P(\bar{B})}$.

If $A$ and $B$ are two events such that $P(A) \neq 0$ and $P(B) \neq 1$,then $P\left( \frac{\bar{A}}{\bar{B}} \right) = .....$

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