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

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(A) We know that for any two events $A$ and $B$,the addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Rearranging

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$\frac{7}{12}$

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(A) We know that for any two events $A$ and $B$,the addition theorem of probability states that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Rearranging this,we get $P(A \cup B) + P(A \cap B) = P(A) + P(B)$.Given that $P(A \cup B) + P(A \cap B) = \frac{7}{8}$ and $P(A) = 2P(B)$,which implies $P(B) = \frac{P(A)}{2}$.Substituting these into the equation: $\frac{7}{8} = P(A) + \frac{P(A)}{2}$.$\frac{7}{8} = \frac{3P(A)}{2}$.$P(A) = \frac{7}{8} 	imes \frac{2}{3} = \frac{14}{24} = \frac{7}{12}$.

If $A$ and $B$ are two events such that $P(A \cup B) + P(A \cap B) = \frac{7}{8}$ and $P(A) = 2P(B)$,then $P(A) = $

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