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(D) Given $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{4}$,and $P(B|A) = \frac{1}{2}$.From $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{1}{2}$,we get $P(A \c

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All of these

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(D) Given $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{4}$,and $P(B|A) = \frac{1}{2}$.From $P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{1}{2}$,we get $P(A \cap B) = \frac{1}{2} 	imes P(A) = \frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}$.From $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1}{4}$,we get $P(B) = \frac{P(A \cap B)}{1/4} = \frac{1/8}{1/4} = \frac{1}{2}$.Since $P(A \cap B) = \frac{1}{8}$ and $P(A) 	imes P(B) = \frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$,we have $P(A \cap B) = P(A)P(B)$,so $A$ and $B$ are independent.$P(A'|B) = 1 - P(A|B) = 1 - \frac{1}{4} = \frac{3}{4}$.Since $A$ and $B$ are independent,$A'$ and $B'$ are also independent. Thus,$P(B'|A') = P(B') = 1 - P(B) = 1 - \frac{1}{2} = \frac{1}{2}$.Therefore,all options are correct.

For two events $A$ and $B$,if $P(A) = P(A|B) = \frac{1}{4}$ and $P(B|A) = \frac{1}{2}$,then:

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