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(A) When two balanced dice are rolled,the total number of possible outcomes is $6 	imes 6 = 36$.We need to find the probability that the sum of the n

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$5$

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(A) When two balanced dice are rolled,the total number of possible outcomes is $6 	imes 6 = 36$.We need to find the probability that the sum of the numbers on the two dice is greater than $8$.The possible sums greater than $8$ are $9, 10, 11,$ and $12$.The outcomes resulting in these sums are: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 $= 4 + 3 + 2 + 1 = 10$.The probability is $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{10}{36} = \frac{5}{18}$.

Two balanced dice are rolled. Find the probability that the sum of numbers on two dice is greater than $8$. (in $/18$)

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