Let A and B be two events such that P(A) = fr | vedclass

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(A) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the given values: $\frac{3}{4} = \frac{3}{8} + \frac{5}{8} - P(A \cap B)$.$\fr

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$\frac{1}{5}$

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(A) We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the given values: $\frac{3}{4} = \frac{3}{8} + \frac{5}{8} - P(A \cap B)$.$\frac{3}{4} = 1 - P(A \cap B) \Rightarrow P(A \cap B) = 1 - \frac{3}{4} = \frac{1}{4}$.Now, calculate $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{5/8} = \frac{1}{4} 	imes \frac{8}{5} = \frac{2}{5}$.Since $P(A'|B) = 1 - P(A|B)$, we have $P(A'|B) = 1 - \frac{2}{5} = \frac{3}{5}$.Finally, $P(A'|B) - P(A|B) = \frac{3}{5} - \frac{2}{5} = \frac{1}{5}$.

Let $A$ and $B$ be two events such that $P(A) = \frac{3}{8}$, $P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$. Then $P(A'|B) - P(A|B) =$ . . . . . .

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