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(B) When two dice are tossed,the total number of possible outcomes is $6 	imes 6 = 36$.The possible sums range from $2$ to $12$. The prime numbers in

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$\frac{5}{12}$

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(B) When two dice are tossed,the total number of possible outcomes is $6 	imes 6 = 36$.The possible sums range from $2$ to $12$. The prime numbers in this range are $\{2, 3, 5, 7, 11\}$.Now,we count the number of outcomes for each prime sum:- Sum $= 2$: $(1, 1)$ $ightarrow 1$ outcome- Sum $= 3$: $(1, 2), (2, 1)$ $ightarrow 2$ outcomes- Sum $= 5$: $(1, 4), (2, 3), (3, 2), (4, 1)$ $ightarrow 4$ outcomes- Sum $= 7$: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ $ightarrow 6$ outcomes- Sum $= 11$: $(5, 6), (6, 5)$ $ightarrow 2$ outcomesTotal favorable outcomes $= 1 + 2 + 4 + 6 + 2 = 15$.Therefore,the probability $= \frac{15}{36} = \frac{5}{12}$.

Two dice are tossed. The probability that the total score is a prime number is

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