2023 AMC 12B 第 19 题

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}

每个球独立落入箱子；使用奇偶筛，将第

i

个箱子为奇数写作

\tfrac{1-(-1)^{k_i}}{2}

Each ball lands independently; use a parity filter, writing bin

i

odd as

\tfrac{1-(-1)^{k_i}}{2}

对所有分配求和，全奇数的比例为

\dfrac{3^{\,n}-3}{4\cdot 3^{\,n}}

；看它在大

n

时趋近于什么

Summing over all assignments, the all-odd fraction is

\dfrac{3^{\,n}-3}{4\cdot 3^{\,n}};

see what it approaches for large

n

用奇偶筛计数三个箱子全为奇数的分配数，对奇数

n

可得

\scriptsize\frac{1}{8}\sum_{S\subseteq\{1,2,3\}}(-1)^{|S|}(3-2|S|)^n=\frac{3^n-3}{4}

再除以总分配数

3^n

，概率为

\dfrac{3^n-3}{4\cdot 3^n}

，当

n=2023

时极其接近

\tfrac14

。

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.

其他年份的第 19 题