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

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(A) Given: $P(A \cup B) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{4}$,and $P(\overline{A}) = \frac{2}{3}$.Since $P(A) = 1 - P(\overline{A})$,we have $P(A

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$5/12$

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(A) Given: $P(A \cup B) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{4}$,and $P(\overline{A}) = \frac{2}{3}$.Since $P(A) = 1 - P(\overline{A})$,we have $P(A) = 1 - \frac{2}{3} = \frac{1}{3}$.Using the addition theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the values: $\frac{3}{4} = \frac{1}{3} + P(B) - \frac{1}{4}$.$P(B) = \frac{3}{4} + \frac{1}{4} - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3}$.We need to find $P(\overline{A} \cap B)$.Note that $P(\overline{A} \cap B) = P(B) - P(A \cap B)$.$P(\overline{A} \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{8 - 3}{12} = \frac{5}{12}$.

If $A$ and $B$ are events such that $P(A \cup B) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{4}$,and $P(\overline{A}) = \frac{2}{3}$,then what is the value of $P(\overline{A} \cap B)$?

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