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(C) When $2$ dice are rolled,the total number of outcomes is $6 	imes 6 = 36$.Let $A$ be the event that the sum is divisible by $3$. The possible sum

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$\frac{5}{9}$

Vedclass · Answer

(C) When $2$ dice are rolled,the total number of outcomes is $6 	imes 6 = 36$.Let $A$ be the event that the sum is divisible by $3$. The possible sums are $3, 6, 9, 12$.The outcomes for $A$ are: $(1,2), (2,1), (1,5), (5,1), (2,4), (4,2), (3,3), (3,6), (6,3), (4,5), (5,4), (6,6)$.Number of outcomes in $A$,$n(A) = 12$.Let $B$ be the event that the sum is divisible by $4$. The possible sums are $4, 8, 12$.The outcomes for $B$ are: $(1,3), (3,1), (2,2), (2,6), (6,2), (3,5), (5,3), (4,4), (6,6)$.Number of outcomes in $B$,$n(B) = 9$.The intersection $A \cap B$ is the event that the sum is divisible by both $3$ and $4$ (i.e.,divisible by $12$).The only outcome with a sum of $12$ is $(6,6)$.So,$n(A \cap B) = 1$.The probability $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{12}{36} + \frac{9}{36} - \frac{1}{36} = \frac{20}{36} = \frac{5}{9}$.

$2$ dice are rolled together. What is the probability that the sum of the $2$ numbers is divisible by $3$ or by $4$?

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