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(B) Given $P(\bar{A}) = \frac{2}{3}$,so $P(A) = 1 - \frac{2}{3} = \frac{1}{3}$.Given $P(A \cap \bar{B}) = \frac{1}{5}$. Since $A = (A \cap B) \cup (A

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$11$

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(B) Given $P(\bar{A}) = \frac{2}{3}$,so $P(A) = 1 - \frac{2}{3} = \frac{1}{3}$.Given $P(A \cap \bar{B}) = \frac{1}{5}$. Since $A = (A \cap B) \cup (A \cap \bar{B})$,we have $P(A) = P(A \cap B) + P(A \cap \bar{B})$.Thus,$P(A \cap B) = P(A) - P(A \cap \bar{B}) = \frac{1}{3} - \frac{1}{5} = \frac{5-3}{15} = \frac{2}{15}$.Now,$P(B) = P(A \cap B) + P(\bar{A} \cap B)$,so $P(\bar{A} \cap B) = P(B) - P(A \cap B) = \frac{4}{15} - \frac{2}{15} = \frac{2}{15}$.We need $P(B \mid (A \cup \bar{B})) = \frac{P(B \cap (A \cup \bar{B}))}{P(A \cup \bar{B})}$.$B \cap (A \cup \bar{B}) = (B \cap A) \cup (B \cap \bar{B}) = (A \cap B) \cup \emptyset = A \cap B$,so $P(B \cap (A \cup \bar{B})) = \frac{2}{15}$.$P(A \cup \bar{B}) = P(A) + P(\bar{B}) - P(A \cap \bar{B}) = \frac{1}{3} + (1 - \frac{4}{15}) - \frac{1}{5} = \frac{5}{15} + \frac{11}{15} - \frac{3}{15} = \frac{13}{15}$.So,$P(B \mid (A \cup \bar{B})) = \frac{2/15}{13/15} = \frac{2}{13}$.$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{3} + \frac{4}{15} - \frac{2}{15} = \frac{5+4-2}{15} = \frac{7}{15}$.Finally,$\sqrt{195[\frac{2}{13} + \frac{7}{15}]} = \sqrt{195[\frac{30+91}{195}]} = \sqrt{121} = 11$.

If $A$ and $B$ are two events of a random experiment such that $P(\bar{A})=\frac{2}{3}$,$P(B)=\frac{4}{15}$ and $P(A \cap \bar{B})=\frac{1}{5}$,then $\sqrt{195[P(B \mid(A \cup \bar{B}))+P(A \cup B)]} = $

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