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(B) Let $F$ be the event that the sum of the numbers on the $3$ dice is $15$. The possible outcomes for the sum $15$ are permutations of $(6, 6, 3)$,$

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$\frac{3}{10}$

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(B) Let $F$ be the event that the sum of the numbers on the $3$ dice is $15$. The possible outcomes for the sum $15$ are permutations of $(6, 6, 3)$,$(6, 5, 4)$,and $(5, 5, 5)$.Number of ways to get $(6, 6, 3)$ is $\frac{3!}{2!} = 3$.Number of ways to get $(6, 5, 4)$ is $3! = 6$.Number of ways to get $(5, 5, 5)$ is $1$.Total number of favorable outcomes for event $F$ is $3 + 6 + 1 = 10$.Let $E$ be the event that the number $5$ does not appear on any of the $3$ dice.We need to find the conditional probability $P(E|F) = \frac{n(E \cap F)}{n(F)}$.$E \cap F$ represents the outcomes where the sum is $15$ and the number $5$ does not appear.From the combinations above,only the set $(6, 6, 3)$ does not contain the number $5$.The number of ways to get $(6, 6, 3)$ is $3$.Thus,$n(E \cap F) = 3$.Therefore,$P(E|F) = \frac{3}{10}$.

When $3$ dice are thrown at a time,the sum of the numbers appeared on $3$ dice was found to be $15$. Then the probability that the number $5$ does not appear on any one of the dice is

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