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

Revision as of 02:38, 14 February 2021

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 $p+q?$

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

Video Solution Using infinite Geometric Series

https://youtu.be/3B-3_nOTIu4 ~hippopotamus1

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

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 $1 \leq a \leq \infty$ because the leftmost bin in the sequence can be any bin, and $1 \leq d \leq \infty$ , 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 $6 + 49 = 55$ , which is choice (A).

-Darren Yao

@@ Line 28: / Line 28: @@
 This gives us the following sum for the probability:
+<cmath>\begin{align*} 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a-3d}
-<cmath> 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a-3d} </cmath>
+&= 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) \\
-<cmath> = 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a} \cdot 2^{-3d} </cmath>
+&= 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) \\
-<cmath> = 6 \left( \sum_{a=1}^{\infty} 2^{-3a} \right) \left( \sum_{d=1}^{\infty} 2^{-3d} \right) </cmath>
+&= \frac{6}{49} .\end{align*}</cmath>
+Therefore the answer is <math>6 + 49 = 55</math>, which is choice (A).
-<cmath> = 6 \left( \sum_{a=1}^{\infty} 8^{-a} \right) \left( \sum_{d=1}^{\infty} 8^{-d} \right) </cmath>
-<cmath> = 6 \left( \frac{1}{7} \right) \left( \frac{1}{7} \right) </cmath>
-<cmath> = \frac{6}{49} </cmath>
-<cmath> \to \boxed{ (A) 55}. </cmath>
 -Darren Yao

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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