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

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(C) Given that $P(A \cup B) = \frac{5}{6}$,$P(A \cap B) = \frac{1}{3}$,and $P(\bar{B}) = \frac{1}{3}$.Since $P(B) = 1 - P(\bar{B})$,we have $P(B) = 1

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$\frac{1}{2}$

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(C) Given that $P(A \cup B) = \frac{5}{6}$,$P(A \cap B) = \frac{1}{3}$,and $P(\bar{B}) = \frac{1}{3}$.Since $P(B) = 1 - P(\bar{B})$,we have $P(B) = 1 - \frac{1}{3} = \frac{2}{3}$.Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the values: $\frac{5}{6} = P(A) + \frac{2}{3} - \frac{1}{3}$.$\frac{5}{6} = P(A) + \frac{1}{3}$.$P(A) = \frac{5}{6} - \frac{1}{3} = \frac{5-2}{6} = \frac{3}{6} = \frac{1}{2}$.

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

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