AMC10 2012 A

AMC10 2012 A · Q9

AMC10 2012 A · Q9. It mainly tests Basic counting (rules of product/sum), Probability (basic).

A pair of six-sided fair dice are labeled so that one die has only even numbers (two each of 2, 4, and 6), and the other die has only odd numbers (two each of 1, 3, and 5). The pair of dice is rolled. What is the probability that the sum of the numbers on the tops of the two dice is 7?

一对六面的公平骰子被标记，其中一个骰子只有偶数（各有两个2、4和6），另一个骰子只有奇数（各有两个1、3和5）。掷这对骰子，两个骰子上面数字之和为7的概率是多少？

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

(B) \frac{1}{5} \frac{1}{5}

(D) \frac{1}{3} \frac{1}{3}

(E) \frac{1}{2} \frac{1}{2}

Answer

Correct choice: (D)

正确答案：（D）

Solution

The sum could be 7 only if the even die showed 2 and the odd showed 5, the even showed 4 and the odd showed 3, or the even showed 6 and the odd showed 1. Each of these events can occur in $2\cdot2 = 4$ ways. Hence there are 12 ways for a 7 to occur. There are $6 \cdot 6 = 36$ possible outcomes, so the probability that a 7 occurs is $12/36 = 1/3$.

和为7只有在以下情况：偶数骰子显示2奇数显示5、偶数显示4奇数显示3，或偶数显示6奇数显示1。每种情况各有$2\cdot2 = 4$种方式。因此有12种方式得到7。总可能结果$6 \cdot 6 = 36$，概率为$12/36 = 1/3$。

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