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(B) Let $S$ be the set of outcomes where the sum of three dice is $8$. The possible combinations (unordered) are:$(1, 1, 6) \rightarrow \frac{3!}{2!}

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$\frac{3}{7}$

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(B) Let $S$ be the set of outcomes where the sum of three dice is $8$. The possible combinations (unordered) are:$(1, 1, 6) \rightarrow \frac{3!}{2!} = 3$ permutations$(1, 2, 5) \rightarrow 3! = 6$ permutations$(1, 3, 4) \rightarrow 3! = 6$ permutations$(2, 2, 4) \rightarrow \frac{3!}{2!} = 3$ permutations$(2, 3, 3) \rightarrow \frac{3!}{2!} = 3$ permutationsTotal number of outcomes $n(S) = 3 + 6 + 6 + 3 + 3 = 21$.Let $E$ be the event that at least one die shows a $4$. The favorable outcomes are from $(1, 3, 4)$ and $(2, 2, 4)$.Number of favorable outcomes $n(E) = 6 + 3 = 9$.The probability $P(E) = \frac{n(E)}{n(S)} = \frac{9}{21} = \frac{3}{7}$.

Three dice are thrown. Given that the sum of the numbers on them is $8$,what is the probability that at least one of them shows a $4$?

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