Question

在一个标准骰子上，随机去掉一个点，每个点被选中的可能性相同。然后掷这个骰子。朝上一面有奇数个点的概率是多少？

On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

$\dfrac{5}{11}$

$\dfrac{10}{21}$

$\dfrac{1}{2}$

$\dfrac{11}{21}$

$\dfrac{6}{11}$

答案：D

知识点：骰子（概率）条件概率奇偶性

难度评级：1870

解答：

骰子共有 $21$ 个点，所以去掉的点来自 $n$ 点面的概率为 $\dfrac{n}{21}$ 。

若从奇数面去掉一点，朝上一面为奇数点的概率是 $\dfrac{2}{6}=\dfrac{1}{3}$ ；若从偶数面去掉一点，概率是 $\dfrac{4}{6}=\dfrac{2}{3}$ 。去掉点来自奇数面的概率为 $\dfrac{1 + 3 + 5}{21}$ ，来自偶数面的概率为 $\dfrac{2 + 4 + 6}{21}$ 。

因此所求概率为 $\dfrac{1}{3} \cdot \dfrac{9}{21} + \dfrac{2}{3} \cdot \dfrac{12}{21} = \dfrac{33}{63} = \dfrac{11}{21}.$

所以正确答案是 D。

The die has $21$ dots, so a dot is removed from the face with $n$ dots with probability $\dfrac{n}{21}.$

If a dot is removed from an odd face, that face becomes even, leaving two odd faces and hence probability $\dfrac{2}{6}=\dfrac{1}{3}$ of an odd top. If a dot is removed from an even face, that face becomes odd, leaving four odd faces and hence probability $\dfrac{4}{6}=\dfrac{2}{3}.$ The removed dot lies on an odd face with probability $\dfrac{1 + 3 + 5}{21}$ and on an even face with probability $\dfrac{2 + 4 + 6}{21}.$

Hence the answer is $\dfrac{1}{3} \cdot \dfrac{9}{21} + \dfrac{2}{3} \cdot \dfrac{12}{21} = \dfrac{33}{63} = \dfrac{11}{21}.$

Thus, the correct answer is D.

2005 AMC 12A 第 14 题

答案：D

难度评级：1870

解答：

其他年份的第 14 题