Two unbiased dice are thrown once. Write all | vedclass

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(C) When two unbiased dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.The possible outcomes are:$(1,1), (1,2), (1,3), (1,4)

Vedclass · Accepted Answer

$\frac{1}{3}$

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(C) When two unbiased dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.The possible outcomes are:$(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 $H$ be the event that the sum of the numbers on the two dice is a factor of $18$. The factors of $18$ are $1, 2, 3, 6, 9, 18$. Since the minimum sum is $2$ and the maximum sum is $12$,the possible sums that are factors of $18$ are $2, 3, 6, 9$.Favorable outcomes for each sum:Sum $= 2: (1,1)$Sum $= 3: (1,2), (2,1)$Sum $= 6: (1,5), (2,4), (3,3), (4,2), (5,1)$Sum $= 9: (3,6), (4,5), (5,4), (6,3)$Total number of favorable outcomes $= 1 + 2 + 5 + 4 = 12$.Probability $P(H) = \frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{12}{36} = \frac{1}{3}$.

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 factor of $18$?

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