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(A) We know that by the definition of conditional probability: $P(\frac{B}{A}) = \frac{P(A \cap B)}{P(A)}$ $\Rightarrow P(A) P(\frac{B}{A}) = P(A \cap

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) We know that by the definition of conditional probability: $P(\frac{B}{A}) = \frac{P(A \cap B)}{P(A)}$ $\Rightarrow P(A) P(\frac{B}{A}) = P(A \cap B)$ Now,using De Morgan's Law,we have: $P(\bar{A} \cup \bar{B}) = P(\overline{A \cap B})$ Using the property of complement events,$P(\bar{E}) = 1 - P(E)$: $P(\overline{A \cap B}) = 1 - P(A \cap B)$ Substituting the value of $P(A \cap B)$ from the first step: $P(\bar{A} \cup \bar{B}) = 1 - P(A) P(\frac{B}{A})$ Thus,option $A$ is the correct statement.

For two events $A$ and $B$,a true statement among the following is

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