AMC10 2023 B

AMC10 2023 B · Q21

AMC10 2023 B · Q21. 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，每个球被随机放入 $3$ 个盒子中。以下哪个是最接近于每个盒子中球的个数均为奇数的概率？

(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 $\boxed{\frac{1}{4}}$.

因为每个盒子中球的个数均为奇数，所以每个盒子至少有一个球。因此，我们可以预先将一个球放入每个盒子。然后，我们可以将成对的 $2$ 个球添加到每个盒子中，这意味着我们现在需要将 $1010$ 对球分配到 $3$ 个盒子中。这将强制每个盒子中的球数为奇数。使用星和杠方法，这等于 $\binom{1012}{2}$。这等于 $\frac{1012\cdot1011}{2}$。总的方式数也使用星和杠方法，即 $\binom{2023+3-1}{3-1} = \binom{2025}{2}$。将两者相除，得到 $\frac{1012 \cdot 1011 \cdot 2}{2 \cdot 2025 \cdot 2024}$。我们可以粗略地消去 $\frac{1012 \cdot 1011}{2025 \cdot 2024}$ 得到 $\frac{1}{4}$。分子分母中的 $2$ 也消去了，所以剩下 $\boxed{\frac{1}{4}}$。

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