Question

掷三枚公平的六面骰子。恰有两枚骰子的点数之和等于剩下一枚骰子的点数的概率是多少？

Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?

$\dfrac16$

$\dfrac{13}{72}$

$\dfrac7{36}$

$\dfrac5{24}$

$\dfrac29$

答案：D

知识点：骰子（概率）分类讨论

难度评级：1540

解答：

若一枚骰子的点数是另外两枚之和，它必然是三枚中最大的。

先选哪一枚是这个和，有 $3$ 种方式。

最大骰子的点数不能为 $1$ ，因为两个正整数之和不可能是 $1$ 。

它取其余每个点数的概率都是 $\dfrac{1}{6}$ 。

和为 $2$ 时有 $1$ 种有序数对，和为 $3$ 时有 $2$ 种，和为 $4$ 时有 $3$ 种，和为 $5$ 时有 $4$ 种，和为 $6$ 时有 $5$ 种。

另外两枚骰子共有 $6^2 = 36$ 种有序结果，所以所求概率为 $3 \cdot \dfrac{1}{6} \cdot \dfrac{1 + 2 + 3 + 4 + 5}{36} = \dfrac{5}{24}.$

所以正确答案是 D。

Note that if one die is the sum of the other two dice, then it is strictly greater than the other two dice.

There are $3$ ways to choose which of the dice is the sum of the other two, which makes it the greatest.

This die cannot be $1,$ since there is no way to sum two positive integers to get $1.$

There is a $\dfrac{1}{6}$ chance that this die is any of the other numbers.

There is $1$ way to get a sum of $2,$ $2$ ways for $3,$ $3$ for $4,$ $4$ for $5,$ and $5$ for $6.$

We take these numbers of ways out of a total of $6^2 = 36$ possibilities. The desired probability is then $3 \cdot \dfrac{1}{6} \cdot \dfrac{1 + 2 + 3 + 4 + 5}{36} = \dfrac{5}{24}.$

Thus, D is the correct answer.

2014 AMC 10A 第 17 题

答案：D

难度评级：1540

解答：

其他年份的第 17 题