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(C) The total number of outcomes when throwing two dice is $n(S) = 6 	imes 6 = 36$.Let $E$ be the event that the sum of the numbers on the two dice i

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$\frac{35}{36}$

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(C) The total number of outcomes when throwing two dice is $n(S) = 6 	imes 6 = 36$.Let $E$ be the event that the sum of the numbers on the two dice is $12$. The only outcome for this event is $(6, 6)$,so $n(E) = 1$.The probability of getting a sum of $12$ is $P(E) = \frac{n(E)}{n(S)} = \frac{1}{36}$.The event of getting a sum less than $12$ is the complement of the event of getting a sum of $12$,denoted as $\overline{E}$.Therefore,the required probability is $P(\overline{E}) = 1 - P(E) = 1 - \frac{1}{36} = \frac{35}{36}$.

What is the probability of getting a total of less than $12$ in the throw of two dice?

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