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(B) Let $P(A) = x$ and $P(B) = y$. Since $A$ and $B$ are independent,$P(A \cap B) = P(A)P(B) = xy = \frac{1}{6}$.Also,$P(A^c \cap B^c) = P(A^c)P(B^c)

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$\frac{1}{3}$

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(B) Let $P(A) = x$ and $P(B) = y$. Since $A$ and $B$ are independent,$P(A \cap B) = P(A)P(B) = xy = \frac{1}{6}$.Also,$P(A^c \cap B^c) = P(A^c)P(B^c) = (1-x)(1-y) = \frac{1}{3}$.Expanding the second equation: $1 - x - y + xy = \frac{1}{3}$.Substituting $xy = \frac{1}{6}$: $1 - (x+y) + \frac{1}{6} = \frac{1}{3}$.$x+y = 1 + \frac{1}{6} - \frac{1}{3} = 1 + \frac{1-2}{6} = 1 - \frac{1}{6} = \frac{5}{6}$.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 $6$: $6t^2 - 5t + 1 = 0$.$(2t-1)(3t-1) = 0$.So,$t = \frac{1}{2}$ or $t = \frac{1}{3}$.Since $P(A) > P(B)$,we have $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{3}$.Thus,the probability of the occurrence of $B$ is $\frac{1}{3}$.

$A$ and $B$ are two independent events of a random experiment and $P(A) > P(B)$. If 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 probability of the occurrence of $B$ is

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