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(A) We need to find the conditional probability of getting a sum of $15$ given that the first throw is $4$.Let $A$ be the event that the sum of the th

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$\frac{1}{18}$

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(A) We need to find the conditional probability of getting a sum of $15$ given that the first throw is $4$.Let $A$ be the event that the sum of the three throws is $15$,and $B$ be the event that the first throw is $4$.We need to calculate $P(A|B) = \frac{n(A \cap B)}{n(B)}$.First,consider $n(B)$: If the first throw is fixed as $4$,the remaining two throws can each result in any of $6$ outcomes. Thus,$n(B) = 6 	imes 6 = 36$.Next,consider $n(A \cap B)$: We need the sum of the three numbers $(4, x, y)$ to be $15$,where $1 \le x, y \le 6$. This implies $4 + x + y = 15$,or $x + y = 11$.The possible pairs $(x, y)$ that satisfy $x + y = 11$ are $(5, 6)$ and $(6, 5)$.Thus,there are $2$ favorable outcomes: $(4, 5, 6)$ and $(4, 6, 5)$. So,$n(A \cap B) = 2$.The conditional probability is $P(A|B) = \frac{2}{36} = \frac{1}{18}$.

$A$ die is thrown three times and the sum of the numbers obtained is $15$. What is the probability that the number $4$ appears in the first throw?

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