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(C) When a pair of dice is thrown,the total number of possible outcomes is $6 	imes 6 = 36$. We need to find the probability of getting a sum greater

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$ \frac{5}{12} $

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(C) When a pair of dice is thrown,the total number of possible outcomes is $6 	imes 6 = 36$. We need to find the probability of getting a sum greater than $7$. The outcomes with a sum greater than $7$ are: Sum $= 8$: $(2,6), (3,5), (4,4), (5,3), (6,2)$ ($5$ outcomes) Sum $= 9$: $(3,6), (4,5), (5,4), (6,3)$ ($4$ outcomes) Sum $= 10$: $(4,6), (5,5), (6,4)$ ($3$ outcomes) Sum $= 11$: $(5,6), (6,5)$ ($2$ outcomes) Sum $= 12$: $(6,6)$ ($1$ outcome) Total favorable outcomes $= 5 + 4 + 3 + 2 + 1 = 15$. Therefore,the required probability is $P = \frac{15}{36} = \frac{5}{12}$.

In a simultaneous throw of a pair of dice,the probability of getting a total more than $7$ is

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