In an experiment of throwing three fair dice, | vedclass

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(A) The total number of outcomes when throwing three dice is $n = 6 	imes 6 	imes 6 = 216$. The event $A$ of getting a sum of $16$ consists of the o

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$\frac{1}{36}$

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(A) The total number of outcomes when throwing three dice is $n = 6 	imes 6 	imes 6 = 216$. The event $A$ of getting a sum of $16$ consists of the outcomes: $A = \{(6, 6, 4), (6, 4, 6), (4, 6, 6), (5, 5, 6), (5, 6, 5), (6, 5, 5)\}$. The number of favorable outcomes is $r = 6$. Therefore,the probability $P(A) = \frac{r}{n} = \frac{6}{216} = \frac{1}{36}$.

In an experiment of throwing three fair dice,the probability of getting a sum of $16$ is ..........

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