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Difference between revisions of "2021 AMC 12B Problems/Problem 23"

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Let us first consider the cases where the balls are in bins <math>1, 2,</math> and <math>3</math>. The probability for the first ball is <math>\frac{1}{2}</math>, the probability for the second ball is <math>\frac{1}{4}</math> and the probability for the third ball is <math>\frac{1}{8}</math>.
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Let us first consider the cases where the balls are in bins <math>1, 2,</math> and <math>3</math>. The probability for the first ball is <math>\frac{1}{2}</math>, the probability for the second ball is <math>\frac{1}{4}</math> and the probability for the third ball is <math>\frac{1}{8}</math>. There are <math>3!</math> ways to orient which ball is in which bin since there are 3 colors. Multiply all of them to get <math>3! * \frac{1}{2} * \frac{1}{4} * \frac{1}{8} = \frac {3}{32}</math>.

Revision as of 15:53, 11 February 2021

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 $6+49=\boxed{\textbf{(A) }55}.$


Solution 2

Let us first consider the cases where the balls are in bins $1, 2,$ and $3$. The probability for the first ball is $\frac{1}{2}$, the probability for the second ball is $\frac{1}{4}$ and the probability for the third ball is $\frac{1}{8}$. There are $3!$ ways to orient which ball is in which bin since there are 3 colors. Multiply all of them to get $3! * \frac{1}{2} * \frac{1}{4} * \frac{1}{8} = \frac {3}{32}$.

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