AMC12 2021 B

AMC12 2021 B · Q23

AMC12 2021 B · Q23. It mainly tests Probability (basic), Counting & probability misc.

Three balls are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $i$ is $2^{-i}$ for $i=1,2,3,....$ More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is $\frac pq,$ where $p$ and $q$ are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins $3,17,$ and $10.$) What is $p+q?$

三个球被随机且独立地扔入编号为正整数的箱子中，对于每个球，扔入箱子 $i$ 的概率为 $2^{-i}$，$i=1,2,3,....$ 每个箱子允许多个球。球最终落在不同箱子中且均匀间隔的概率为 $\frac pq$，其中 $p$ 和 $q$ 互质正整数。（例如，如果球扔入箱子 $3,17$ 和 $10$，则均匀间隔。）求 $p+q$？

(A) 55 55

(B) 56 56

(D) 58 58

(E) 59 59

Answer

Correct choice: (A)

正确答案：（A）

Solution

"Evenly spaced" just means the bins form an arithmetic sequence. Suppose the middle bin in the sequence is $x$. There are $x-1$ different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these $3$ bins are chosen is $6\cdot 2^{-3x} = 6\cdot \frac{1}{8^x}$, so the probability $x$ is the middle bin is $6\cdot\frac{x-1}{8^x}$. Then, we want the sum \begin{align*} 6\sum_{x=2}^{\infty}\frac{x-1}{8^x} &= \frac{6}{8}\left[\frac{1}{8} + \frac{2}{8^2} + \frac{3}{8^3}\cdots\right]\\ &= \frac34\left[\left(\frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}+\cdots \right) + \left(\frac{1}{8^2} + \frac{1}{8^3} + \frac{1}{8^4} + \cdots \right) + \cdots\right]\\ &= \frac34\left[\frac17\cdot \left(1 + \frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3} + \cdots \right)\right]\\ &= \frac34\cdot \frac{8}{49}\\ &= \frac{6}{49} \end{align*} The answer is $6+49=\boxed{\textbf{(A) }55}.$

“均匀间隔”仅意味着箱子编号形成等差数列。假设数列中间箱子是 $x$。第一箱有 $x-1$ 种可能，这两个箱子唯一确定最后一个箱子。现在，选择这 $3$ 个箱子的概率是 $6\cdot 2^{-3x} = 6\cdot \frac{1}{8^x}$，因此 $x$ 是中间箱子的概率是 $6\cdot\frac{x-1}{8^x}$。然后，我们要计算和 \begin{align*} 6\sum_{x=2}^{\infty}\frac{x-1}{8^x} &= \frac{6}{8}\left[\frac{1}{8} + \frac{2}{8^2} + \frac{3}{8^3}\cdots\right]\\ &= \frac34\left[\left(\frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}+\cdots \right) + \left(\frac{1}{8^2} + \frac{1}{8^3} + \frac{1}{8^4} + \cdots \right) + \cdots\right]\\ &= \frac34\left[\frac17\cdot \left(1 + \frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3} + \cdots \right)\right]\\ &= \frac34\cdot \frac{8}{49}\\ &= \frac{6}{49} \end{align*} 答案是 $6+49=\boxed{\textbf{(A) }55}$。

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