Question

将 $2023$ 个球分别放入 $3$ 个盒子中的一个。下列哪一项最接近每个盒子中球数都是奇数的概率？

Each of $2023$ balls is placed into 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：2120

解答：

所有 $3^{2023}$ 种放法等可能。用符号筛计数每个盒子都是奇数的放法： $\frac{1}{8}\sum_{s \in \{\pm 1\}^3}(s_1 s_2 s_3)$ $\cdot (s_1 + s_2 + s_3)^{2023}$ 。当 $s = (1,1,1)$ 和 $s = (-1,-1,-1)$ 时，两项都等于 $3^{2023}$ 。对于其余六种符号选择，括号内的和为 $1$ 或 $-1$ ，而整项都等于 $-1$ ，所以这六项合计为 $-6$ 。因此计数为 $\frac{2 \cdot 3^{2023} - 6}{8} = \frac{3^{2023} - 3}{4}$ 。除以总数，概率为 $\frac{3^{2023} - 3}{4 \cdot 3^{2023}} = \frac{1}{4} - \frac{1}{4 \cdot 3^{2022}}$ ，略小于 $\frac{1}{4}$ 。所以正确答案是 E。

All $3^{2023}$ assignments are equally likely. A sign filter counts the ones with every bin odd: $\frac{1}{8}\sum_{s \in \{\pm 1\}^3}(s_1 s_2 s_3)$ $\cdot (s_1 + s_2 + s_3)^{2023}.$ For $s = (1,1,1)$ and $s = (-1,-1,-1),$ both terms equal $3^{2023}.$ For each other sign choice, the sum in parentheses is $1$ or $-1,$ and the full term equals $-1,$ so these six terms total $-6.$ Thus the count is $\frac{2 \cdot 3^{2023} - 6}{8} = \frac{3^{2023} - 3}{4}.$ Dividing, the probability is $\frac{3^{2023} - 3}{4 \cdot 3^{2023}} = \frac{1}{4} - \frac{1}{4 \cdot 3^{2022}},$ a hair under $\frac{1}{4}.$ Thus, E is the correct answer.

2023 AMC 10B 第 21 题

答案：E

难度评级：2120

解答：

其他年份的第 21 题