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

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(D) Given,$P(A)=\frac{1}{2}$,$P(B)=\frac{1}{2}$ and $P(A \mid B)=\frac{1}{4}$.Using the conditional probability formula,$P(A \mid B) = \frac{P(A \cap

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$\frac{1}{8}$

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(D) Given,$P(A)=\frac{1}{2}$,$P(B)=\frac{1}{2}$ and $P(A \mid B)=\frac{1}{4}$.Using the conditional probability formula,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.Substituting the values,$\frac{1}{4} = \frac{P(A \cap B)}{1/2}$.Therefore,$P(A \cap B) = \frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$.By De Morgan's Law,$P(A^{\prime} \cap B^{\prime}) = P((A \cup B)^{\prime}) = 1 - P(A \cup B)$.Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{2} - \frac{1}{8} = 1 - \frac{1}{8} = \frac{7}{8}$.Thus,$P(A^{\prime} \cap B^{\prime}) = 1 - \frac{7}{8} = \frac{1}{8}$.

If $A$ and $B$ are two events such that $P(A)=\frac{1}{2}$,$P(B)=\frac{1}{2}$ and $P(A \mid B)=\frac{1}{4}$,then $P(A^{\prime} \cap B^{\prime})$ is

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