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

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(B) Given $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{5}$,and $P(A \cup B) = \frac{1}{2}$.First,we find $P(A \cap B)$ using the formula $P(A \cup B) = P(A)

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$\frac{5}{8}$

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(B) Given $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{5}$,and $P(A \cup B) = \frac{1}{2}$.First,we find $P(A \cap B)$ using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:$P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{2} = \frac{10 + 6 - 15}{30} = \frac{1}{30}$.Now,we calculate $P(A \cap B')$ and $P(B \cap A')$:$P(A \cap B') = P(A) - P(A \cap B) = \frac{1}{3} - \frac{1}{30} = \frac{10 - 1}{30} = \frac{9}{30} = \frac{3}{10}$.$P(B \cap A') = P(B) - P(A \cap B) = \frac{1}{5} - \frac{1}{30} = \frac{6 - 1}{30} = \frac{5}{30} = \frac{1}{6}$.Also,$P(B') = 1 - P(B) = 1 - \frac{1}{5} = \frac{4}{5}$ and $P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$.Therefore,$P(A \mid B') + P(B \mid A') = \frac{P(A \cap B')}{P(B')} + \frac{P(B \cap A')}{P(A')} = \frac{9/30}{4/5} + \frac{5/30}{2/3} = \frac{9}{30} 	imes \frac{5}{4} + \frac{5}{30} 	imes \frac{3}{2} = \frac{3}{10} 	imes \frac{5}{4} + \frac{1}{6} 	imes \frac{3}{2} = \frac{3}{8} + \frac{1}{4} = \frac{3 + 2}{8} = \frac{5}{8}$.

If $A$ and $B$ are two events such that $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{5}$,and $P(A \cup B) = \frac{1}{2}$,then $P(A \mid B') + P(B \mid A')$ is equal to

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