2019 AMC 12B Problems/Problem 13

The following problem is from both the 2019 AMC 10B #17 and 2019 AMC 12B #13, so both problems redirect to this page.

Problem

A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k = 1,2,3....$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?
$\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}$

Solution 1

By symmetry, the probability of the red ball landing in a higher-numbered bin is the same as the probability of the green ball landing in a higher-numbered bin. Clearly, the probability of both landing in the same bin is $\sum_{k=1}^{\infty}{2^{-k} \cdot 2^{-k}} = \sum_{k=1}^{\infty}2^{-2k} = \frac{1}{3}$ (by the geometric series sum formula). Therefore, since the other two probabilities have to both the same, they have to be $\frac{1-\frac{1}{3}}{2} = \boxed{\textbf{(C) } \frac{1}{3}}$ .

Solution 2

Suppose the green ball goes in bin $i$ , for some $i \ge 1$ . The probability of this occurring is $\frac{1}{2^i}$ . Given that this occurs, the probability that the red ball goes in a higher-numbered bin is $\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}$ (by the geometric series sum formula). Thus the probability that the green ball goes in bin $i$ , and the red ball goes in a bin greater than $i$ , is $\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}$ . Summing from $i=1$ to infinity, we get

$\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}$ where we again used the geometric series sum formula. (Alternatively, if this sum equals $n$ , then by writing out the terms and multiplying both sides by $4$ , we see $4n = n+1$ , which gives $n = \frac{1}{3}$ .)

Solution 3

For red ball in bin $k$ , $\Pr(\text{Green Below Red})=\sum\limits_{i=1}^{k-1}2^{-i}$ (GBR) and $\Pr(\text{Red in Bin k}=2^{-k}$ (RB). $\Pr(\text{GBR}|\text{RB})=\sum\limits_{k=1}^{\infty}2^{-k}\sum\limits_{i=1}^{k-1}2^{-i}=\sum\limits_{k=1}^{\infty}2^{-k}\cdot\frac{1}{2}(\frac{1-(1/2)^{k-1}}{1-1/2})$ $\sum\limits_{k=1}^{\infty}\frac{1}{2^{-k}}-2\sum\limits_{k=1}^\infty\frac{1}{(2^2)^{-k}}\implies 1-2/3=\boxed{\textbf{D}) \frac{1}{3}}$

Solution 4

The probability that the two balls will go into adjacent bins is $\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + ... = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \cdots = \frac{1}{6}$ by the geometric series sum formula. Similarly, the probability that the two balls will go into bins that have a distance of $2$ from each other is $\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} + \cdots = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots = \frac{1}{12}$ (again recognizing a geometric series). We can see that each time we add a bin between the two balls, the probability halves. Thus, our answer is $\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \cdots$ , which, by the geometric series sum formula, is $\boxed{\textbf{(C) } \frac{1}{3}}$ .

Solution 5 (quick, conceptual)

Define a win as a ball appearing in higher numbered box.

Start from the first box.

There are $4$ possible results in the box: Red, Green, Red and Green, or none, with an equal probability of $\frac{1}{4}$ for each. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution, so if $p$ is the probability that Red wins, we can write $p = \frac{1}{4} + \frac{1}{4}p$ : there is a $\frac{1}{4}$ probability that "Red" wins immediately, a $0$ probability in the cases "Green" or "Red and Green", and in the "None" case (which occurs with $\frac{1}{4}$ probability), we then start again, giving the same probability $p$ . Hence, solving the equation, we get $p = \boxed{\textbf{(C) } \frac{1}{3}}$ .

Solution 6

Write out the infinite geometric series as $\frac{1}{2}$ , $\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \cdots$ . To find the probablilty that red goes in a higher-numbered bin than green, we can simply remove all odd-index terms (i.e term $1$ , term $3$ , etc.), and then sum the remaining terms - this is in fact precisely equivalent to the method of Solution 2. Writing this out as another infinite geometric sequence, we are left with $\frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \cdots$ . Summing, we get $\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}$

Solution 7

This immediately seems like a geometric series problem, so fixing the green ball to fall into bin $1$ gives a probability of $\frac{1}{2}(\frac{1}{2^2}+\frac{1}{2^3} +...)$ for the red ball to fall into a higher bin. Fixing the green ball to fall into bin $2$ gives a probability of $\frac{1}{2^2}(\frac{1}{2^3}+\frac{1}{2^4} +...)$ . Factoring out the denominator of the first fraction in each probability gives $\frac{1}{2^3}(1+\frac{1}{2}+\frac{1}{2^2}+..)+\frac{1}{2^5}(1+\frac{1}{2}+\frac{1}{2^2}+...)+...$ so factoring out $(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)$ results in the probability simplifying to $(\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+...)(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)$ and using the formula $\frac{a}{1-r}$ to find both series, we obtain $(\frac{\frac{1}{2^3}}{1-\frac{1}{4}})(\frac{1}{1-\frac{1}{2}})$ which simplifies to $\boxed{\textbf{(C) } \frac{1}{3}}$ -- OGBooger

Video Solution

For those who want a video solution: https://youtu.be/VP7ltu-XEq8

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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