For two events A and B,if P(A) = P(A|B) = fra | vedclass

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(D) Given: $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{4}$,and $P(B|A) = \frac{1}{2}$.Step $1$: Check for independence.$P(A|B) = \frac{P(A \cap B)}{P(B)}

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All of the above

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(D) Given: $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{4}$,and $P(B|A) = \frac{1}{2}$.Step $1$: Check for independence.$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1}{4}$.Also,$P(A|B) = P(A)$ implies that $A$ and $B$ are independent events.Thus,option $A$ is true.Step $2$: Check $P(A'|B)$.Since $A$ and $B$ are independent,$A'$ and $B$ are also independent.Therefore,$P(A'|B) = P(A') = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.Thus,option $B$ is true.Step $3$: Check $P(B'|A')$.Since $A$ and $B$ are independent,$A'$ and $B'$ are also independent.Therefore,$P(B'|A') = P(B') = 1 - P(B)$.To find $P(B)$,use $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{1}{2}$.Since $A, B$ are independent,$P(A \cap B) = P(A)P(B)$.So,$\frac{P(A)P(B)}{P(A)} = P(B) = \frac{1}{2}$.Then $P(B') = 1 - \frac{1}{2} = \frac{1}{2}$.Thus,$P(B'|A') = \frac{1}{2}$,so option $C$ is true.Since $A, B,$ and $C$ are true,the correct answer is $D$.

For two events $A$ and $B$,if $P(A) = P(A|B) = \frac{1}{4}$ and $P(B|A) = \frac{1}{2}$,then which of the following is true?

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