Question

掷十二个公平骰子。朝上数字的乘积为质数的概率是多少？

Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?

$\left(\dfrac{1}{12}\right)^{12}$

$\left(\dfrac{1}{6}\right)^{12}$

$2\left(\dfrac{1}{6}\right)^{11}$

$\dfrac52\left(\dfrac{1}{6}\right)^{11}$

$\left(\dfrac{1}{6}\right)^{10}$

答案：E

知识点：骰子（概率）质数乘法原理

难度评级：1480

小提示：

正整数乘积为质数，只可能是一个因数为质数，其余因数都为 $1$ 。

A product of positive integers is prime only when one factor is a prime and the rest are $1$

大提示：

选择哪一个骰子显示质数，再选择是哪一个质数；其余 $11$ 个骰子都必须显示 $1$ 。

Choose which die shows the prime, and which prime; the other $11$ dice must show $1$

解答：

乘积为质数当且仅当一个骰子显示质数（ $2$ 、 $3$ 或 $5$ ），其余十一个骰子都显示 $1$ 。

指定某一个骰子为质数骰时，它显示质数的概率为 $\dfrac36 = \dfrac12$ ，其余每个骰子显示 $1$ 的概率为 $\dfrac16$ 。考虑十二个骰子中哪一个显示质数，所求概率为 $\begin{aligned} 12 \cdot \dfrac12 \cdot \left(\dfrac16\right)^{11} &= 6 \cdot \left(\dfrac16\right)^{11} \\ &= \left(\dfrac16\right)^{10} \end{aligned}\text{。}$

所以正确答案是 E。

The product is prime exactly when one die shows a prime ( $2,$ $3,$ or $5$ ) and the other eleven all show $1.$

The probability that any single die is the prime one is $\dfrac36 = \dfrac12,$ and each of the other eleven shows $1$ with probability $\dfrac16.$ Accounting for which of the twelve dice is prime, the probability is $\begin{aligned} 12 \cdot \dfrac12 \cdot \left(\dfrac16\right)^{11} &= 6 \cdot \left(\dfrac16\right)^{11} \\ &= \left(\dfrac16\right)^{10}. \end{aligned}$

Thus, E is the correct answer.

2005 AMC 10B 第 12 题

答案：E

难度评级：1480

小提示：

大提示：

解答：

其他年份的第 12 题