Evaluate P(A cup B),if 2 P(A) = P(B) = frac{5 | vedclass

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(A) Given that $2 P(A) = P(B) = \frac{5}{13}$.This implies $P(A) = \frac{5}{26}$ and $P(B) = \frac{5}{13}$.We are given $P(A|B) = \frac{2}{5}$.Using t

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$\frac{11}{26}$

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(A) Given that $2 P(A) = P(B) = \frac{5}{13}$.This implies $P(A) = \frac{5}{26}$ and $P(B) = \frac{5}{13}$.We are given $P(A|B) = \frac{2}{5}$.Using the formula for conditional probability,$P(A|B) = \frac{P(A \cap B)}{P(B)}$.Therefore,$P(A \cap B) = P(A|B) 	imes P(B) = \frac{2}{5} 	imes \frac{5}{13} = \frac{2}{13}$.Using the addition theorem of probability,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the values,$P(A \cup B) = \frac{5}{26} + \frac{5}{13} - \frac{2}{13}$.$P(A \cup B) = \frac{5}{26} + \frac{10}{26} - \frac{4}{26} = \frac{5 + 10 - 4}{26} = \frac{11}{26}$.

Evaluate $P(A \cup B)$,if $2 P(A) = P(B) = \frac{5}{13}$ and $P(A|B) = \frac{2}{5}$.

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