2021 AMC 12B Problems/Problem 23

Problem 23

Three balls are randomly and independantly 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 $p+q?$

$\textbf{(A) }55 \qquad \textbf{(B) }56 \qquad \textbf{(C) }57\qquad \textbf{(D) }58 \qquad \textbf{(E) }59$

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 $\begin{align*} 6\sum_{x=2}^{\infty}\frac{x-1}{8^x} &= \frac{6}{8}\left[\frac{1}{8} + \frac{2}{8^2} + \frac{3}{8^3}\cdots\right]\\ &= \frac34\left[\left(\frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}\right) + \left(\frac{1}{8^2} + \frac{1}{8^3} + \frac{1}{8^4}\right) + \cdots\right]\\ &= \frac34\left[\frac17\cdot \left(1 + \frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}\right)\right]\\ &= \frac34\cdot \frac{8}{49}\\ &= \frac{6}{49} \end{align*}$ The answer is $6+49=\boxed{\textbf{(A) }55}.$

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2021 AMC 12B Problems/Problem 23

Problem 23

Solution

Solution 2