AMC10 2014 A

AMC10 2014 A · Q17

AMC10 2014 A · Q17. It mainly tests Probability (basic), Casework.

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?

掷三个公平的六面骰子。两个骰子上的数值之和等于剩余一个骰子上的数值的概率是多少？

(A) $\frac{1}{6}$ $\frac{1}{6}$

(B) $\frac{13}{72}$ $\frac{13}{72}$

(D) $\frac{5}{24}$ $\frac{5}{24}$

(E) $\frac{2}{9}$ $\frac{2}{9}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Each roll of the three dice can be recorded as an ordered triple $(a, b, c)$ of the three values appearing on the dice. There are $6^3$ equally likely triples possible. For the sum of two of the values in the triple to equal the third value, the triple must be a permutation of one of the triples (1, 1, 2), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 2, 4), (2, 3, 5), (2, 4, 6), or (3, 3, 6). There are $3! = 6$ permutations of the values $(a, b, c)$ when $a, b,$ and $c$ are distinct, and 3 permutations of the values when two of the values are equal. Thus there are $6 \cdot 6 + 3 \cdot 3 = 45$ triples where the sum of two of the values equals the third. The requested probability is $\frac{45}{216} = \frac{5}{24}$.

三个骰子的每一次掷骰可以用有序三元组 $(a, b, c)$ 记录。有 $6^3$ 个等可能的的三元组。对于三元组中两个数值之和等于第三个数值的三元组，必须是以下三元组 (1, 1, 2)、(1, 2, 3)、(1, 3, 4)、(1, 4, 5)、(1, 5, 6)、(2, 2, 4)、(2, 3, 5)、(2, 4, 6) 或 (3, 3, 6) 的排列。当 $a, b, c$ 互异时，有 $3! = 6$ 个排列；当两个数值相等时，有 3 个排列。因此有利三元组数为 $6 \cdot 6 + 3 \cdot 3 = 45$。所求概率为 $\frac{45}{216} = \frac{5}{24}$。

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