Two dice are thrown together. The probability | vedclass

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(C) When two dice are thrown,the total number of outcomes is $6 	imes 6 = 36$. Let $E$ be the event that the sum of the numbers on the two dice is a

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(C) When two dice are thrown,the total number of outcomes is $6 	imes 6 = 36$. Let $E$ be the event that the sum of the numbers on the two dice is a multiple of $4$. The possible sums that are multiples of $4$ are $4, 8, 12$. The outcomes for sum $= 4$ are $(1, 3), (2, 2), (3, 1)$. The outcomes for sum $= 8$ are $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$. The outcomes for sum $= 12$ are $(6, 6)$. The total number of favorable outcomes is $3 + 5 + 1 = 9$. Therefore,the probability $P(E) = \frac{9}{36} = \frac{1}{4}$.

Two dice are thrown together. The probability that the sum of the two numbers will be a multiple of $4$ is

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