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(A) When a pair of dice is thrown,the total number of outcomes is $6 	imes 6 = 36$. The sums that are multiples of $3$ are $3, 6, 9,$ and $12$. The f

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$1: 2$

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(A) When a pair of dice is thrown,the total number of outcomes is $6 	imes 6 = 36$. The sums that are multiples of $3$ are $3, 6, 9,$ and $12$. The favourable outcomes are: Sum $3$: $(1, 2), (2, 1)$ ($2$ outcomes) Sum $6$: $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$ ($5$ outcomes) Sum $9$: $(3, 6), (4, 5), (5, 4), (6, 3)$ ($4$ outcomes) Sum $12$: $(6, 6)$ ($1$ outcome) Total number of favourable outcomes $= 2 + 5 + 4 + 1 = 12$. Number of unfavourable outcomes $= 36 - 12 = 24$. The odds in favour are defined as $\frac{	ext{Number of favourable outcomes}}{	ext{Number of unfavourable outcomes}} = \frac{12}{24} = \frac{1}{2}$ or $1: 2$.

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

The odds in favour of getting a sum that is a multiple of $3$,when a pair of dice is thrown,is:

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