If 2 P(A) = P(B) = frac{5}{13} and P(A mid B) | vedclass

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(A) Given that $2 P(A) = \frac{5}{13} \implies P(A) = \frac{5}{26}$.Given that $P(B) = \frac{5}{13}$.Given that $P(A \mid B) = \frac{2}{5}$.We know th

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$\frac{11}{26}$

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(A) Given that $2 P(A) = \frac{5}{13} \implies P(A) = \frac{5}{26}$.Given that $P(B) = \frac{5}{13}$.Given that $P(A \mid B) = \frac{2}{5}$.We know that $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.Therefore,$P(A \cap B) = P(A \mid B) 	imes P(B) = \frac{2}{5} 	imes \frac{5}{13} = \frac{2}{13}$.We use the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the values: $P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$.$P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26} = \frac{11}{26}$.

If $2 P(A) = P(B) = \frac{5}{13}$ and $P(A \mid B) = \frac{2}{5}$,then $P(A \cup B) =$ . . . . . . .

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