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(B) When two dice are thrown simultaneously,the total number of possible outcomes is $6 	imes 6 = 36$.Let $E$ be the event of getting a sum of $2$,$8

Vedclass · Accepted Answer

$\frac{7}{36}$

Vedclass · Answer

(B) When two dice are thrown simultaneously,the total number of possible outcomes is $6 	imes 6 = 36$.Let $E$ be the event of getting a sum of $2$,$8$,or $12$.$1$. Outcomes for sum $2$: $(1, 1)$ - Total $1$ outcome.$2$. Outcomes for sum $8$: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ - Total $5$ outcomes.$3$. Outcomes for sum $12$: $(6, 6)$ - Total $1$ outcome.Total favorable outcomes $= 1 + 5 + 1 = 7$.Therefore,the required probability $P(E) = \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{7}{36}$.

Two dice are thrown simultaneously. The probability of getting the sum $2$,$8$,or $12$ is

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