AMC12 2023 B

AMC12 2023 B · Q19

AMC12 2023 B · Q19. It mainly tests Basic counting (rules of product/sum), Probability (basic).

Each of $2023$ balls is randomly 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?

有$2023$个球随机放入$3$个盒子。以下哪个选项最接近每个盒子含有奇数个球的概率？

(A) \frac{2}{3} \frac{2}{3}

(B) \frac{3}{10} \frac{3}{10}

(D) \frac{1}{3} \frac{1}{3}

(E) \frac{1}{4} \frac{1}{4}

Answer

Correct choice: (E)

正确答案：（E）

Solution

Because each bin will have an odd number, they will have at least one ball. So we can put one ball in each bin prematurely. We then can add groups of 2 balls into each bin, meaning we now just have to spread 1010 pairs over 3 bins. This will force every bin to have an odd number of balls. Using stars and bars, we find that this is equal to $\binom{1012}{2}$. This is equal to $\frac{1012\cdot1011}{2}$. The total amount of ways would also be found using stars and bars. That would be $\binom{2023+3-1}{3-1} = \binom{2025}{2}$. Dividing our two quantities, we get $\frac{1012 \cdot 1011 \cdot 2}{2 \cdot 2025 \cdot 2024}$. We can roughly cancel $\frac{1012 \cdot 1011}{2025 \cdot 2024}$ to get $\frac{1}{4}$. The 2 in the numerator and denominator also cancels out, so we're left with $\boxed{\frac{1}{4}}$.

因为每个盒子含有奇数个球，它们至少各有一个球。所以我们可以预先放一个球到每个盒子。然后，我们可以向每个盒子添加成对的2个球，即现在只需将1010对球分配到3个盒子。这将迫使每个盒子含有奇数个球。使用星杠定理，这等于$\binom{2020+3-1}{3-1}=\binom{2022}{2}$。这是等于$\frac{2022\cdot2021}{2}$。总方式数也用星杠定理：$\binom{2023+3-1}{3-1}=\binom{2025}{2}$。两量相除，得$\frac{2022 \cdot 2021}{2025 \cdot 2024}$。粗略取消，得约$\frac{1}{4}$。于是答案为$\boxed{\frac{1}{4}}$。

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