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

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(A) We are given $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{4}$,and $P(A \cap B) = \frac{1}{5}$.First,calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B) =

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$\frac{37}{40}$

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(A) We are given $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{4}$,and $P(A \cap B) = \frac{1}{5}$.First,calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{3} + \frac{1}{4} - \frac{1}{5} = \frac{20 + 15 - 12}{60} = \frac{23}{60}$.We need to find $P\left( \frac{\overline{B}}{\overline{A}} ight)$. By the definition of conditional probability,$P\left( \frac{\overline{B}}{\overline{A}} ight) = \frac{P(\overline{B} \cap \overline{A})}{P(\overline{A})}$.Using De Morgan's Law,$\overline{B} \cap \overline{A} = \overline{A \cup B}$,so $P(\overline{B} \cap \overline{A}) = 1 - P(A \cup B) = 1 - \frac{23}{60} = \frac{37}{60}$.Also,$P(\overline{A}) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$.Therefore,$P\left( \frac{\overline{B}}{\overline{A}} ight) = \frac{37/60}{2/3} = \frac{37}{60} 	imes \frac{3}{2} = \frac{37}{40}$.

If $A$ and $B$ are two events such that $P(A) = \frac{1}{3}$,$P(B) = \frac{1}{4}$ and $P(A \cap B) = \frac{1}{5}$,then $P\left( \frac{\overline{B}}{\overline{A}} \right) = $

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