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

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(A) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.The outcomes that result in a sum of $7$ are: $\{(6, 1), (5, 2

Vedclass · Accepted Answer

$\frac{5}{18}$

Vedclass · Answer

(A) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.The outcomes that result in a sum of $7$ are: $\{(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)\}$. There are $6$ such outcomes.The outcomes that result in a sum of $9$ are: $\{(6, 3), (5, 4), (4, 5), (3, 6)\}$. There are $4$ such outcomes.Since these events are mutually exclusive,the total number of favorable outcomes is $6 + 4 = 10$.The required probability is $\frac{10}{36} = \frac{5}{18}$.

In a single throw of two dice,the probability of obtaining a total of $7$ or $9$ is:

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