Question

掷四枚公平的六面骰子。至少三枚骰子显示相同点数的概率是多少？

Four fair six-sided dice are rolled. What is the probability that at least three of the four dice show the same value?

$\dfrac{1}{36}$

$\dfrac{7}{72}$

$\dfrac{1}{9}$

$\dfrac{5}{36}$

$\dfrac{1}{6}$

答案：B

知识点：骰子（概率）分类讨论

难度评级：1420

解答：

共有 $6^4$ 个等可能有序结果。

恰好三枚相同时，重复点数有 $6$ 种，不同点数有 $5$ 种，不同骰子的位置有 $4$ 种，共 $6\cdot5\cdot4=120$ 种。

四枚全相同有 $6$ 种。

所求概率为 $\frac{120+6}{6^4}=\frac{126}{1296}=\frac7{72}$ 。

所以正确答案是 B。

There are $6^4$ equally likely ordered outcomes.

If exactly three dice show the same value, choose the repeated value in $6$ ways, the different value in $5$ ways, and the position of the different die in $4$ ways. This gives $6\cdot5\cdot4=120$ outcomes.

If all four dice match, there are $6$ outcomes.

The probability is $\frac{120+6}{6^4}=\frac{126}{1296}=\frac7{72}$ .

Thus, the correct answer is B .

2014 AMC 10B 第 16 题

答案：B

难度评级：1420

解答：

其他年份的第 16 题