Let A and B be two events such that P(overlin | vedclass

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(A) $P(\overline{A \cup B}) = \frac{1}{6} \implies P(A \cup B) = 1 - \frac{1}{6} = \frac{5}{6}$.Given $P(A \cap B) = \frac{1}{4}$ and $P(\bar{A}) = \f

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independent but not equally likely.

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(A) $P(\overline{A \cup B}) = \frac{1}{6} \implies P(A \cup B) = 1 - \frac{1}{6} = \frac{5}{6}$.Given $P(A \cap B) = \frac{1}{4}$ and $P(\bar{A}) = \frac{1}{4}$,we have $P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.Using the addition theorem,$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the values: $\frac{5}{6} = \frac{3}{4} + P(B) - \frac{1}{4}$.$\frac{5}{6} = \frac{1}{2} + P(B) \implies P(B) = \frac{5}{6} - \frac{1}{2} = \frac{5-3}{6} = \frac{2}{6} = \frac{1}{3}$.Now,check for independence: $P(A) \cdot P(B) = \frac{3}{4} 	imes \frac{1}{3} = \frac{1}{4} = P(A \cap B)$.Since $P(A \cap B) = P(A) \cdot P(B)$,the events are independent.Since $P(A) = \frac{3}{4}$ and $P(B) = \frac{1}{3}$,$P(A) 
eq P(B)$,so they are not equally likely.

Let $A$ and $B$ be two events such that $P(\overline{A \cup B}) = \frac{1}{6}$,$P(A \cap B) = \frac{1}{4}$ and $P(\bar{A}) = \frac{1}{4}$,where $\bar{A}$ stands for the complement of the event $A$. Then the events $A$ and $B$ are

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