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

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(B) We are given: $P(A \cup B) = \frac{5}{6}$, $P(\bar{A}) = \frac{1}{4}$, and $P(B) = \frac{1}{3}$.First, calculate $P(A)$:$P(A) = 1 - P(\bar{A}) = 1

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independent events

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(B) We are given: $P(A \cup B) = \frac{5}{6}$, $P(\bar{A}) = \frac{1}{4}$, and $P(B) = \frac{1}{3}$.First, calculate $P(A)$:$P(A) = 1 - P(\bar{A}) = 1 - \frac{1}{4} = \frac{3}{4}$.Using the addition theorem of probability:$P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the values:$\frac{5}{6} = \frac{3}{4} + \frac{1}{3} - P(A \cap B)$.$\frac{5}{6} = \frac{9 + 4}{12} - P(A \cap B) = \frac{13}{12} - P(A \cap B)$.$P(A \cap B) = \frac{13}{12} - \frac{5}{6} = \frac{13 - 10}{12} = \frac{3}{12} = \frac{1}{4}$.Now, check for independence by calculating $P(A) \cdot P(B)$:$P(A) \cdot P(B) = \frac{3}{4} 	imes \frac{1}{3} = \frac{1}{4}$.Since $P(A \cap B) = P(A) \cdot P(B)$, the events $A$ and $B$ are independent.

If $A$ and $B$ are events such that $P(A \cup B) = \frac{5}{6}$, $P(\bar{A}) = \frac{1}{4}$, and $P(B) = \frac{1}{3}$, then $A$ and $B$ are

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