Two persons A and B throw three unbiased dice | vedclass

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(C) The total number of outcomes when throwing three dice is $6^3 = 216$. First,we find the number of ways to get a sum $S$ with three dice,denoted by

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$\frac{35}{216}$

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(C) The total number of outcomes when throwing three dice is $6^3 = 216$. First,we find the number of ways to get a sum $S$ with three dice,denoted by $n(S)$. The number of ways to get a sum $S = 13$ is $n(13) = 21$. The number of ways to get a sum $S > 13$ is $n(14) + n(15) + n(16) + n(17) + n(18)$. Using the formula for the number of ways to get a sum $S$ with $n$ dice,each having $f$ faces: $n(14) = 15, n(15) = 10, n(16) = 6, n(17) = 3, n(18) = 1$. Summing these: $15 + 10 + 6 + 3 + 1 = 35$. Thus,the probability that $B$ gets a sum higher than $13$ is $\frac{35}{216}$.

Two persons $A$ and $B$ throw three unbiased dice one after the other. If $A$ gets a sum of $13$,then the probability that $B$ gets a higher sum is:

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