Two dice are thrown together. If the numbers | vedclass

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(C) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.The condition given is that the numbers appearing on the two d

Vedclass · Accepted Answer

$\frac{2}{15}$

Vedclass · Answer

(C) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.The condition given is that the numbers appearing on the two dice are different. The cases where the numbers are the same are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)$,which are $6$ outcomes.Therefore,the number of outcomes where the numbers are different is $36 - 6 = 30$.Now,we need the probability that the sum is $6$,given that the numbers are different. The pairs that result in a sum of $6$ are $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$.Since the numbers must be different,we exclude $(3, 3)$. The remaining favorable outcomes are $(1, 5), (2, 4), (4, 2), (5, 1)$,which are $4$ outcomes.The required probability is $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes with different numbers}} = \frac{4}{30} = \frac{2}{15}$.

Two dice are thrown together. If the numbers appearing on the two dice are different,then what is the probability that the sum is $6$?

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