Question

Kayla 掷四个公平 $6$ 面骰。至少有一个掷出的数大于 $4$ ，且至少有两个掷出的数大于 $2$ 的概率是多少？

Kayla rolls four fair $6$ -sided dice. What is the probability that at least one of the numbers Kayla rolls is greater than $4$ and at least two of the numbers she rolls are greater than $2?$

$\dfrac{2}{3}$

$\dfrac{19}{27}$

$\dfrac{59}{81}$

$\dfrac{61}{81}$

$\dfrac{7}{9}$

答案：D

知识点：骰子（概率）补集计数容斥原理

难度评级：1630

解答：

将每个骰子分类为低档 $\{1,2\}$ 、中档 $\{3,4\}$ 或高档 $\{5,6\}$ ；每类概率都是 $\tfrac13$ ，所以 $3^4 = 81$ 个类别模式等可能。

需要至少一个高骰，即骰面大于 $4$ ，并且至少两个骰面大于 $2$ 也就是中或高。令为至少有一个高骰的事件，为至多一个非低骰的事件。

没有高骰的模式有 $2^4 = 16$ 个，至多一个非低骰的模式有 $9$ 个，两种坏条件同时发生的模式有 $5$ 个。由容斥，可行模式数为 $81 - 16 - 9 + 5 = 61$ 。

概率为 $\dfrac{61}{81}$ 。

所以正确答案是 D。

Sort each die into low $\{1,2\},$ mid $\{3,4\},$ or high $\{5,6\};$ each has probability $\tfrac13,$ so the $3^4 = 81$ category patterns are equally likely.

We need at least one high die (a number greater than $4$ ) and at least two dice that are greater than $2$ (mid or high). The two bad events are having no high die and having at most one non-low die.

There are $2^4 = 16$ patterns of the first kind, $9$ of the second kind, and $5$ in their intersection. By inclusion-exclusion the count of good patterns is $81 - 16 - 9 + 5 = 61.$

The probability is $\dfrac{61}{81}.$

Thus, the correct answer is D.

2022 AMC 12B 第 12 题

答案：D

难度评级：1630

解答：

其他年份的第 12 题