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(B) Let $A$ be the event that the sum of the scores is a multiple of $4$.The possible outcomes for $A$ are: $\{(1,3), (2,2), (3,1), (2,6), (3,5), (4,4

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$\frac{1}{9}$

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(B) Let $A$ be the event that the sum of the scores is a multiple of $4$.The possible outcomes for $A$ are: $\{(1,3), (2,2), (3,1), (2,6), (3,5), (4,4), (5,3), (6,2), (6,6)\}$.Thus,the number of outcomes in $A$ is $n(A) = 9$.Let $B$ be the event that the score $4$ appears at least once.We are interested in $B \cap A$,which is the set of outcomes where the sum is a multiple of $4$ $AND$ $4$ appears at least once.Looking at set $A$,the outcomes containing $4$ are: $\{(4,4)\}$.Thus,$B \cap A = \{(4,4)\}$ and $n(B \cap A) = 1$.The conditional probability $P(B|A)$ is given by $\frac{n(B \cap A)}{n(A)}$.$P(B|A) = \frac{1}{9}$.

$A$ die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of $4$. Then the conditional probability that the score $4$ has appeared at least once is

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