Two unbiased dice are thrown once. Write all | vedclass

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(D) When two unbiased dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$. The sample space $S$ is given by:$S = \{(1,1), (1,2)

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$\frac{5}{9}$

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(D) When two unbiased dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$. The sample space $S$ is given by:$S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\}$Let $E$ be the event that the sum of the numbers on the two dice is a multiple of $3$ or $4$. The possible sums range from $2$ to $12$. Multiples of $3$ are $3, 6, 9, 12$. Multiples of $4$ are $4, 8, 12$.Thus,we need the sum to be in the set $\{3, 4, 6, 8, 9, 12\}$.Outcomes with sum $3$: $(1,2), (2,1)$Outcomes with sum $4$: $(1,3), (2,2), (3,1)$Outcomes with sum $6$: $(1,5), (2,4), (3,3), (4,2), (5,1)$Outcomes with sum $8$: $(2,6), (3,5), (4,4), (5,3), (6,2)$Outcomes with sum $9$: $(3,6), (4,5), (5,4), (6,3)$Outcomes with sum $12$: $(6,6)$Total favorable outcomes = $2 + 3 + 5 + 5 + 4 + 1 = 20$.Probability $P(E) = \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{20}{36} = \frac{5}{9}$.

Two unbiased dice are thrown once. Write all the possible outcomes of this experiment. What is the probability that the sum of numbers on two dice is a multiple of $3$ or $4$?

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