Two dice are thrown together. The probability | vedclass

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(C) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.We are looking for pairs $(x, y)$ such that $x + y$ is a multi

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(C) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.We are looking for pairs $(x, y)$ such that $x + y$ is a multiple of $4$. The possible sums are $4, 8, 12$.The favorable outcomes are:Sum $= 4$: $(1, 3), (2, 2), (3, 1)$Sum $= 8$: $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$Sum $= 12$: $(6, 6)$The total number of favorable outcomes is $3 + 5 + 1 = 9$.The probability is $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{9}{36} = \frac{1}{4}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$Probability$	$0.2$	$0.22$	$0.11$	$0.25$	$0.05$	$0.17$

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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