Question

两个立方体骰子各有可拆卸的数字 $1$ 到 $6$ 。把两个骰子上的十二个数字拆下放入袋中，然后一次抽出一个，随机重新贴到两个立方体的面上，每面贴一个数字。随后掷这两个骰子，并把两个顶面上的数字相加。和为 $7$ 的概率是多少？

Two cubical dice each have removable numbers $1$ through $6.$ The twelve numbers on the two dice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is $7?$

$\dfrac{1}{9}$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{2}{11}$

$\dfrac{1}{5}$

答案：D

知识点：骰子（概率）基本概率对称性

难度评级：1820

解答：

随机贴数字再掷骰子，等价于从这十二个数字中随机选两个并相加。

假设第一个顶面显示 $N$ 。若和为 $7$ ，第二个必须是 $7 - N$ ，而在剩余 $11$ 张中恰有 $2$ 张等于 $7 - N$ 。

所以概率为 $\dfrac{2}{11}$ 。

所以正确答案是 D。

Randomly attaching the tiles and then rolling is equivalent to choosing two of the twelve numbers at random and adding them.

Suppose the first top face shows $N.$ For a sum of $7,$ the second must be $7 - N,$ and there are exactly $2$ tiles equal to $7 - N$ among the remaining $11.$

So the probability is $\dfrac{2}{11}.$

Thus, the correct answer is D.

2009 AMC 10A 第 22 题

答案：D

难度评级：1820

解答：

其他年份的第 22 题