2021 AMC 12B Problems/Problem 23

Revision as of 21:25, 22 August 2022 by MRENTHUSIASM (talk | contribs) (Video Solution Using Infinite Geometric Series)

Problem

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 1

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

Solution 2

As in solution 1, note that "evenly spaced" means the bins are in arithmetic sequence. We let the first bin be $a$ and the common difference be $d$. Further note that each $(a, d)$ pair uniquely determines a set of $3$ bins.

We have $a\geq1$ because the leftmost bin in the sequence can be any bin, and $d\geq1$, because the bins must be distinct.

This gives us the following sum for the probability: \begin{align*} 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a-3d} &= 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a} \cdot 2^{-3d} \\ &= 6 \left( \sum_{a=1}^{\infty} 2^{-3a} \right) \left( \sum_{d=1}^{\infty} 2^{-3d} \right) \\ &= 6 \left( \sum_{a=1}^{\infty} 8^{-a} \right) \left( \sum_{d=1}^{\infty} 8^{-d} \right) \\ &= 6 \left( \frac{1}{7} \right) \left( \frac{1}{7} \right) \\ &= \frac{6}{49} .\end{align*} Therefore the answer is $6 + 49 = \boxed{\textbf{(A) }55}$.

-Darren Yao

Solution 3

This is a slightly messier variant of solution 2. If the first ball is in bin $i$ and the second ball is in bin $j>i$, then the third ball is in bin $2j-i$. Thus the probability is \begin{align*} 6\sum_{i=1}^{\infty}\sum_{j=i+1}^\infty2^{-i}2^{-j}2^{-2j+i}&=6\sum_{i=1}^{\infty}\sum_{j=i+1}^\infty2^{-3j}\\ &=6\sum_{i=1}^{\infty}\left(\frac{2^{-3(i+1)}}{1-\tfrac{1}{8}}\right)\\ &=6\sum_{i=1}^\infty\frac{8}{7}\cdot2^{-3}\cdot2^{-3i}\\ &=\frac{6}{7}\sum_{i=1}^\infty2^{-3i}\\ &=\frac{6}{7}\cdot\frac{2^{-3}}{1-\tfrac18}\\ &=\frac{6}{49}. \end{align*} Therefore the answer is $6 + 49 = \boxed{\textbf{(A) }55}$.

Solution 4 (Table)

Based on the value of $n,$ we construct the following table: \[\begin{array}{c|c|c|c}  & & & \\ [-1.5ex] \textbf{Exactly }\boldsymbol{n}\textbf{ Spaces Apart} & \textbf{Bin \#s} & \textbf{Expression} & \textbf{Prob. of One Such Perm.} \\ [1ex]  \hline\hline   & & & \\ [-1.5ex]  n=1 & 1,2,3 & 2^{-1}\cdot2^{-2}\cdot2^{-3} & 2^{-6} \\ [1ex]   & 2,3,4 & 2^{-2}\cdot2^{-3}\cdot2^{-4} & 2^{-9} \\ [1ex]   & 3,4,5 & 2^{-3}\cdot2^{-4}\cdot2^{-5} & 2^{-12} \\ [1ex]   & 4,5,6 & 2^{-4}\cdot2^{-5}\cdot2^{-6} & 2^{-15} \\ [1ex]   & \cdots & \cdots & \cdots \\ [1ex]  \hline  & & & \\ [-1.5ex]  n=2 & 1,3,5 & 2^{-1}\cdot2^{-3}\cdot2^{-5} & 2^{-9} \\ [1ex]   & 2,4,6 & 2^{-2}\cdot2^{-4}\cdot2^{-6} & 2^{-12} \\ [1ex]   & 3,5,7 & 2^{-3}\cdot2^{-5}\cdot2^{-7} & 2^{-15} \\ [1ex]   & 4,6,8 & 2^{-4}\cdot2^{-6}\cdot2^{-8} & 2^{-18} \\ [1ex]   & \cdots & \cdots & \cdots \\ [1ex]  \hline  & & & \\ [-1.5ex]  n=3 & 1,4,7 & 2^{-1}\cdot2^{-4}\cdot2^{-7} & 2^{-12} \\ [1ex]   & 2,5,8 & 2^{-2}\cdot2^{-5}\cdot2^{-8} & 2^{-15} \\ [1ex]   & 3,6,9 & 2^{-3}\cdot2^{-6}\cdot2^{-9} & 2^{-18} \\ [1ex]  & 4,7,10 & 2^{-4}\cdot2^{-7}\cdot2^{-10} & 2^{-21} \\ [1ex]  & \cdots & \cdots & \cdots \\ [1ex]  \hline   & & & \\ [-1.5ex]    \cdots & \cdots & \cdots & \cdots \\ [1ex] \end{array}\] Since three balls have $3!=6$ permutations, the requested probability is \begin{align*} 6\left(\sum_{k=0}^{\infty}2^{-6-3k}+\sum_{k=0}^{\infty}2^{-9-3k}+\sum_{k=0}^{\infty}2^{-12-3k}+\cdots\right)&=6\left(2^{-6}\sum_{k=0}^{\infty}2^{-3k}+2^{-9}\sum_{k=0}^{\infty}2^{-3k}+2^{-12}\sum_{k=0}^{\infty}2^{-3k}+\cdots\right) \\ &=\left(6\sum_{k=0}^{\infty}2^{-3k}\right)\cdot\left(2^{-6}+2^{-9}+2^{-12}+\cdots\right) \\ &=\left(6\sum_{k=0}^{\infty}2^{-3k}\right)\cdot\left(\sum_{k=0}^{\infty}2^{-6-3k}\right) \\ &=\left(6\sum_{k=0}^{\infty}2^{-3k}\right)\cdot\left(2^{-6}\sum_{k=0}^{\infty}2^{-3k}\right) \\ &=\frac{6}{1-2^{-3}}\cdot\frac{2^{-6}}{1-2^{-3}} \\ &=\frac{6}{49} \end{align*} by infinite geometric series, from which the answer is $6+49=\boxed{\textbf{(A) }55}.$

~MRENTHUSIASM

Video Solution Using Infinite Geometric Series

https://youtu.be/3B-3_nOTIu4

~hippopotamus1

Video Solution

https://youtu.be/x16YUmd0OqY

~MathProblemSolvingSkills.com

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 12 Problems and Solutions

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