In a single throw of two dice,the probability | vedclass

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

Vedclass · Accepted Answer

$\frac{5}{12}$

Vedclass · Answer

(C) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.We need to find the probability of getting a sum greater than $7$,which means the sum can be $8, 9, 10, 11,$ or $12$.The favorable 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 $= 5 + 4 + 3 + 2 + 1 = 15$.Required probability $= \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{15}{36} = \frac{5}{12}$.

In a single throw of two dice,the probability of getting a sum of more than $7$ is

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