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(D) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 	imes 6 = 36$.Let $A$ be the event of getting a sum more than $7$. Th

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$\frac{5}{12}$

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(D) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 	imes 6 = 36$.Let $A$ be the event of getting a sum more than $7$. The possible outcomes 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 $n(A) = 5 + 4 + 3 + 2 + 1 = 15$.Therefore,the required probability $P(A) = \frac{n(A)}{n(S)} = \frac{15}{36} = \frac{5}{12}$.

$2$ dice are thrown. Find the probability of getting a sum more than $7$.

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