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(A) Given that $A$ and $B$ are independent events,$P(A \cap B) = P(A) \cdot P(B) = \frac{1}{6}$.The probability that neither occurs is $P(\bar{A} \cap

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

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(A) Given that $A$ and $B$ are independent events,$P(A \cap B) = P(A) \cdot P(B) = \frac{1}{6}$.The probability that neither occurs is $P(\bar{A} \cap \bar{B}) = P(\bar{A}) \cdot P(\bar{B}) = (1 - P(A))(1 - P(B)) = \frac{1}{3}$.Expanding this,we get $1 - (P(A) + P(B)) + P(A) \cdot P(B) = \frac{1}{3}$.Substituting $P(A) \cdot P(B) = \frac{1}{6}$,we have $1 - (P(A) + P(B)) + \frac{1}{6} = \frac{1}{3}$.$P(A) + P(B) = 1 + \frac{1}{6} - \frac{1}{3} = \frac{6 + 1 - 2}{6} = \frac{5}{6}$.Let $x = P(A)$ and $y = P(B)$. We have $x + y = \frac{5}{6}$ and $xy = \frac{1}{6}$.These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$,which is $t^2 - \frac{5}{6}t + \frac{1}{6} = 0$.Multiplying by $6$,we get $6t^2 - 5t + 1 = 0$.$(2t - 1)(3t - 1) = 0$,so $t = \frac{1}{2}$ or $t = \frac{1}{3}$.Thus,the probabilities are $\frac{1}{2}$ and $\frac{1}{3}$.

$A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Then the probabilities of the two events are respectively:

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