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

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(C) By the definition of conditional probability,$P\left( \frac{\overline{A}}{\overline{B}} \right) = \frac{P(\overline{A} \cap \overline{B})}{P(\over

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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