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(B) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 	imes 6 = 36$.Let $A$ be the event of getting a multiple of $3$ as th

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$\frac{1}{3}$

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(B) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 	imes 6 = 36$.Let $A$ be the event of getting a multiple of $3$ as the sum. The possible sums are $3, 6, 9, 12$.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 favorable outcomes $n(A) = 2 + 5 + 4 + 1 = 12$.Therefore,the required probability $P(A) = \frac{n(A)}{n(S)} = \frac{12}{36} = \frac{1}{3}$.

$2$ dice are thrown. Find the probability of getting a multiple of $3$ as the sum.

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