Question

将 $2023$ 个球中的每一个放入 $3$ 个箱子之一。每个箱子中球数都是奇数的概率最接近以下哪一个？

Each of $2023$ balls is placed in one of $3$ bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?

$\dfrac{2}{3}$

$\dfrac{3}{10}$

$\dfrac{1}{2}$

$\dfrac{1}{3}$

$\dfrac{1}{4}$

答案：E

知识点：单位根奇偶性基本概率

难度评级：1990

解答：

用奇偶筛计数三个箱子全为奇数的分配数，对奇数 $n$ 可得除以总分配数 $3^n$ ，概率为 $\dfrac{3^n-3}{4\cdot 3^n}$ ，当 $n=2023$ 时极其接近 $\tfrac14$ 。 $\scriptsize\frac{1}{8}\sum_{S\subseteq\{1,2,3\}}(-1)^{|S|}(3-2|S|)^n=\frac{3^n-3}{4}$

因此，正确答案是 E。

Counting assignments where all three bins are odd with the parity filter gives $\scriptsize\frac{1}{8}\sum_{S\subseteq\{1,2,3\}}(-1)^{|S|}(3-2|S|)^n=\frac{3^n-3}{4}$ for odd $n.$ Dividing by the $3^n$ total assignments, the probability is $\dfrac{3^n-3}{4\cdot 3^n},$ which for $n=2023$ is extremely close to $\tfrac14.$

Thus, the correct answer is E.

2023 AMC 12B 第 19 题

答案：E

难度评级：1990

解答：

其他年份的第 19 题