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(C) Since $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \cdot P(B)$.Given $P(A \cap B) = \frac{1}{6}$,let $P(A) = x$ and $P(B) = y$.

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$\frac{1}{2}$ or $\frac{1}{3}$

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(C) Since $A$ and $B$ are independent events,we have $P(A \cap B) = P(A) \cdot P(B)$.Given $P(A \cap B) = \frac{1}{6}$,let $P(A) = x$ and $P(B) = y$. Thus,$xy = \frac{1}{6}$.The probability that neither occurs is $P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = \frac{1}{3}$.Therefore,$P(A \cup B) = 1 - \frac{1}{3} = \frac{2}{3}$.Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$,we get $x + y - \frac{1}{6} = \frac{2}{3}$,which implies $x + y = \frac{5}{6}$.Substituting $y = \frac{5}{6} - x$ into $xy = \frac{1}{6}$,we get $x(\frac{5}{6} - x) = \frac{1}{6}$.This simplifies to $6x^2 - 5x + 1 = 0$.Factoring the quadratic equation,we get $(2x - 1)(3x - 1) = 0$.Thus,$x = \frac{1}{2}$ or $x = \frac{1}{3}$.Hence,the probability of occurrence of $A$ is $\frac{1}{2}$ or $\frac{1}{3}$.

Let $A$ and $B$ be two independent events of a random experiment. If the probability that both $A$ and $B$ occur is $\frac{1}{6}$ and the probability that neither of them occur is $\frac{1}{3}$,then the probability of occurrence of $A$ is

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