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 $6 	imes 6 = 36$.Let $E_1$ be the event of getting a sum of $7$. The favorable

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$\frac{5}{18}$

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(A) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.Let $E_1$ be the event of getting a sum of $7$. The favorable outcomes are: $(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)$. The number of outcomes is $6$.Let $E_2$ be the event of getting a sum of $9$. The favorable outcomes are: $(6, 3), (5, 4), (4, 5), (3, 6)$. The number of outcomes is $4$.Since these events are mutually exclusive,the total number of favorable outcomes is $6 + 4 = 10$.The required probability is $P = \frac{	ext{Total favorable outcomes}}{	ext{Total possible outcomes}} = \frac{10}{36} = \frac{5}{18}$.

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

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