If two fair dice are rolled,then the probabil | vedclass

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(C) When two fair dice are rolled,the total number of possible outcomes is $6 	imes 6 = 36$. We want the sum of the numbers on the upper faces to be

Vedclass · Accepted Answer

$\frac{5}{18}$

Vedclass · Answer

(C) When two fair dice are rolled,the total number of possible outcomes is $6 	imes 6 = 36$. We want the sum of the numbers on the upper faces to be at least $9$,which means the sum can be $9, 10, 11,$ or $12$. The favorable outcomes are: Sum $= 9: (3, 6), (4, 5), (5, 4), (6, 3)$ Sum $= 10: (4, 6), (5, 5), (6, 4)$ Sum $= 11: (5, 6), (6, 5)$ Sum $= 12: (6, 6)$ The total number of favorable outcomes is $4 + 3 + 2 + 1 = 10$. Therefore,the required probability is $\frac{10}{36} = \frac{5}{18}$.

If two fair dice are rolled,then the probability that the sum of the numbers on the upper faces is at least $9$,is

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