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(B) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 	imes 6 = 36$.Let $A$ be the event of getting a sum of at least $10$

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$1/6$

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(B) When $2$ dice are thrown,the total number of possible outcomes is $n(S) = 6 	imes 6 = 36$.Let $A$ be the event of getting a sum of at least $10$ (i.e.,sum is $10, 11,$ or $12$).The favorable outcomes are:For sum $= 10$: $(4, 6), (5, 5), (6, 4)$For sum $= 11$: $(5, 6), (6, 5)$For sum $= 12$: $(6, 6)$Thus,the set of favorable outcomes is $A = \{(4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)\}$.The number of favorable outcomes is $n(A) = 6$.The probability is $P(A) = \frac{n(A)}{n(S)} = \frac{6}{36} = \frac{1}{6}$.

$2$ dice are thrown. Find the probability of getting a sum of at least $10$.

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