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

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==Solution 2==
 
==Solution 2==
  
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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>. Next consider the case where the balls are in bins <math>1,3, and 5</math>. The probability in this case turns out to be <math>\frac{3}{256}</math> by similar logic. We can see every time we increase the distance between 2 consecutive bins by <math>1</math>, the probability reduces by a factor of <math>8</math>, which means the total probability of the given condition for when a ball is in bin <math>1</math> is <math>\frac{3}{32} * (1 + \frac{1}{8} + \frac{1}{8^2} + …) = \frac{3}{28}</math>. Let the balls now be in bins <math>2, 3, and 4</math>. Using the same technique as we previously used, we get the probability for this case is <math>\frac {3}{256}</math>. (Use the probability for each bin, multiply them and the number of ways to orient the balls). We see that this will end up the same way showing that the total probability that the condition holds when one ball is in bin <math>2</math> is <math>\frac{3}{256} * (1 + \frac{1}{8} + \frac{1}{8^2} + …) = \frac{3}{224}</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>. Next consider the case where the balls are in bins <math>1,3,</math> and 5<math>. The probability in this case turns out to be </math>\frac{3}{256}<math> by similar logic. We can see every time we increase the distance between 2 consecutive bins by </math>1<math>, the probability reduces by a factor of </math>8<math>, which means the total probability of the given condition for when a ball is in bin </math>1<math> is </math>\frac{3}{32} * (1 + \frac{1}{8} + \frac{1}{8^2} + …) = \frac{3}{28}<math>. Let the balls now be in bins </math>2, 3, and 4<math>. Using the same technique as we previously used, we get the probability for this case is </math>\frac {3}{256}<math>. We see that this will end up the same way showing that the total probability that the condition holds when one ball is in bin </math>2<math> is </math>\frac{3}{256} * (1 + \frac{1}{8} + \frac{1}{8^2} + …) = \frac{3}{224}$.

Revision as of 16:04, 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 $3! * \frac{1}{2} * \frac{1}{4} * \frac{1}{8} = \frac {3}{32}$. Next consider the case where the balls are in bins $1,3,$ and 5$. The probability in this case turns out to be$\frac{3}{256}$by similar logic. We can see every time we increase the distance between 2 consecutive bins by$1$, the probability reduces by a factor of$8$, which means the total probability of the given condition for when a ball is in bin$1$is$\frac{3}{32} * (1 + \frac{1}{8} + \frac{1}{8^2} + …) = \frac{3}{28}$. Let the balls now be in bins$2, 3, and 4$. Using the same technique as we previously used, we get the probability for this case is$\frac {3}{256}$. We see that this will end up the same way showing that the total probability that the condition holds when one ball is in bin$2$is$\frac{3}{256} * (1 + \frac{1}{8} + \frac{1}{8^2} + …) = \frac{3}{224}$.

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