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(B) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.We are looking for the probability that the sum of the numbers

Vedclass · Accepted Answer

$\frac{1}{12}$

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(B) When two dice are thrown,the total number of possible outcomes is $6 	imes 6 = 36$.We are looking for the probability that the sum of the numbers on the two dice is more than $10$.The possible sums greater than $10$ are $11$ and $12$.The outcomes that result in a sum of $11$ are $(5, 6)$ and $(6, 5)$.The outcome that results in a sum of $12$ is $(6, 6)$.Thus,the favorable outcomes are ${(5, 6), (6, 5), (6, 6)}$.The number of favorable outcomes is $3$.The probability is $\frac{	ext{Number of favorable outcomes}}{	ext{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12}$.

Two dice are thrown. The probability that the sum of numbers appearing is more than $10$ is:

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