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(B) The total number of outcomes when two dice are rolled is $6 	imes 6 = 36$. Let the outcomes be $(x, y)$ where $x, y \in \{1, 2, 3, 5, 7, 11\}$. W

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$\frac{17}{36}$

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(B) The total number of outcomes when two dice are rolled is $6 	imes 6 = 36$. Let the outcomes be $(x, y)$ where $x, y \in \{1, 2, 3, 5, 7, 11\}$. We need the sum $x + y \leq 8$. The possible pairs $(x, y)$ are: If $x=1$,$y \in \{1, 2, 3, 5, 7\}$ ($5$ outcomes). If $x=2$,$y \in \{1, 2, 3, 5\}$ ($4$ outcomes). If $x=3$,$y \in \{1, 2, 3, 5\}$ ($4$ outcomes). If $x=5$,$y \in \{1, 2, 3\}$ ($3$ outcomes). If $x=7$,$y \in \{1\}$ ($1$ outcome). If $x=11$,no $y$ satisfies the condition. Total favorable outcomes $n(E) = 5 + 4 + 4 + 3 + 1 = 17$. Therefore,the probability $P(E) = \frac{n(E)}{n(S)} = \frac{17}{36}$.

Two dice are rolled. If both dice have six faces numbered $1, 2, 3, 5, 7,$ and $11,$ then the probability that the sum of the numbers on the top faces is less than or equal to $8$ is

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