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(A) Let $X$ be the event that the sum of the numbers on the two dice is greater than $9$.Let $Y$ be the event that the number on die $B$ is $5$.The sa

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$\frac{1}{3}$

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(A) Let $X$ be the event that the sum of the numbers on the two dice is greater than $9$.Let $Y$ be the event that the number on die $B$ is $5$.The sample space for event $Y$ is $\{(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\}$. Thus, $n(Y) = 6$.The event $X \cap Y$ represents the outcomes where the number on die $B$ is $5$ $AND$ the sum is greater than $9$.The possible outcomes for $X \cap Y$ are $\{(5,5), (6,5)\}$. Thus, $n(X \cap Y) = 2$.The conditional probability is given by $P(X|Y) = \frac{n(X \cap Y)}{n(Y)}$.$P(X|Y) = \frac{2}{6} = \frac{1}{3}$.

Two dice $A$ and $B$ are rolled. If it is known that the number on $B$ is $5$, then the probability that the sum of the numbers on the two dice will be greater than $9$ is

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