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(B) Given that $A$ and $B$ are independent events,$P(\overline{A} | B) = P(\overline{A}) = 1 - P(A)$.Since $P(\overline{A} | B) = \frac{2}{5}$,we have

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$\frac{9}{100}$

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(B) Given that $A$ and $B$ are independent events,$P(\overline{A} | B) = P(\overline{A}) = 1 - P(A)$.Since $P(\overline{A} | B) = \frac{2}{5}$,we have $1 - P(A) = \frac{2}{5}$,which implies $P(A) = 1 - \frac{2}{5} = \frac{3}{5}$.Given $P(A) + P(B) = \frac{3}{4}$,substituting $P(A) = \frac{3}{5}$ gives $\frac{3}{5} + P(B) = \frac{3}{4}$.Thus,$P(B) = \frac{3}{4} - \frac{3}{5} = \frac{15 - 12}{20} = \frac{3}{20}$.Since $A$ and $B$ are independent,$P(A \cap B) = P(A) 	imes P(B) = \frac{3}{5} 	imes \frac{3}{20} = \frac{9}{100}$.

Let $A$ and $B$ be two independent events such that $P(A) + P(B) = \frac{3}{4}$ and $P(\overline{A} | B) = \frac{2}{5}$. Then,$P(A \cap B)$ is -

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