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Difference between revisions of "2019 AMC 12B Problems/Problem 13"

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==Solution 7==
 
==Solution 7==
 
This immediately seems like a geometric series problem, so fixing the green ball to fall into bin <math>1</math> gives a probability of <math>\frac{1}{2}(\frac{1}{2^2}+\frac{1}{2^3} +...)</math> for the red ball to fall into a higher bin. Fixing the green ball to fall into bin <math>2</math> gives a probability of <math>\frac{1}{2^2}(\frac{1}{2^3}+\frac{1}{2^4} +...)</math>. Factoring out the denominator of the first fraction in each probability gives <math>\frac{1}{2^3}(1+\frac{1}{2}+\frac{1}{2^2}+..)+\frac{1}{2^5}(1+\frac{1}{2}+\frac{1}{2^2}+...)+...</math> so factoring out <math>(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)</math> results in the probability simplifying to <math>(\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+...)(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)</math> and using the formula <math>\frac{a}{1-r}</math> to find both series, we obtain <math>(\frac{\frac{1}{2^3}}{1-\frac{1}{4}})(\frac{1}{1-\frac{1}{2}})</math> which simplifies to <math>\boxed{\textbf{(C) } \frac{1}{3}}</math> -- OGBooger
 
This immediately seems like a geometric series problem, so fixing the green ball to fall into bin <math>1</math> gives a probability of <math>\frac{1}{2}(\frac{1}{2^2}+\frac{1}{2^3} +...)</math> for the red ball to fall into a higher bin. Fixing the green ball to fall into bin <math>2</math> gives a probability of <math>\frac{1}{2^2}(\frac{1}{2^3}+\frac{1}{2^4} +...)</math>. Factoring out the denominator of the first fraction in each probability gives <math>\frac{1}{2^3}(1+\frac{1}{2}+\frac{1}{2^2}+..)+\frac{1}{2^5}(1+\frac{1}{2}+\frac{1}{2^2}+...)+...</math> so factoring out <math>(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)</math> results in the probability simplifying to <math>(\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+...)(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...)</math> and using the formula <math>\frac{a}{1-r}</math> to find both series, we obtain <math>(\frac{\frac{1}{2^3}}{1-\frac{1}{4}})(\frac{1}{1-\frac{1}{2}})</math> which simplifies to <math>\boxed{\textbf{(C) } \frac{1}{3}}</math> -- OGBooger
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==Solution 8==
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The chance that the green one is below the red one if the red one goes to bin <math>2</math> is <math>\frac{1}{8}</math>. Similarly, if the red goes into bin <math>3</math>, there is a <math>\frac{1}{8} \cdot (\frac{1}{4} + \frac{1}{2})</math>, or <math>\frac{3}{32}</math>, continuing like this, we get this sequence:
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<math>\frac{1}{8}, \frac{3}{32}, \frac{7}{128}, ...</math>
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Let <math>S</math> equal the sum of our sequence:
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<math>S = \frac{1}{8} + \frac{3}{32} + \frac{7}{128} + ...</math>. That means we can write another equation:
 +
<math>\frac{S}{4} = \frac{1}{32} + \frac{3}{128} + ...</math>
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Subtracting <math>\frac{S}{4}</math> from <math>S</math>, yields:
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<math>S - \frac{S}{4} = \frac{1}{8} + \frac{2}{32} + \frac{4}{128} + ...</math>
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We see that the above sequence is a infinite geometric sequence with common ratio <math>\frac{1}{2}</math>. Therefore, the sum of that infinite series is <math>\frac{\frac{1}{8}}{\frac{1}{2}}</math>, which equals <math>\frac{1}{4}</math>. Our equation is now <math>S - \frac{S}{4} = \frac{1}{4}</math>. Solving for <math>S</math> shows that <math>S = \boxed{\frac{1}{3}}</math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 19:41, 22 April 2020

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). \[\Pr(\text{GBR}|\text{RB})=\sum\limits_{k=1}^{\infty}2^{-k}\sum\limits_{i=1}^{k-1}2^{-i}=\sum\limits_{k=1}^{\infty}2^{-k}\cdot\frac{1}{2}(\frac{1-(1/2)^{k-1}}{1-1/2})\] \[\sum\limits_{k=1}^{\infty}\frac{1}{2^{-k}}-2\sum\limits_{k=1}^\infty\frac{1}{(2^2)^{-k}}\implies 1-2/3=\boxed{(\textbf{C}) \frac{1}{3}}\]

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 $\boxed{\textbf{(C) } \frac{1}{3}}$. -fidgetboss_4000

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

Solution 8

The chance that the green one is below the red one if the red one goes to bin $2$ is $\frac{1}{8}$. Similarly, if the red goes into bin $3$, there is a $\frac{1}{8} \cdot (\frac{1}{4} + \frac{1}{2})$, or $\frac{3}{32}$, continuing like this, we get this sequence:

$\frac{1}{8}, \frac{3}{32}, \frac{7}{128}, ...$

Let $S$ equal the sum of our sequence:

$S = \frac{1}{8} + \frac{3}{32} + \frac{7}{128} + ...$. That means we can write another equation: $\frac{S}{4} = \frac{1}{32} + \frac{3}{128} + ...$

Subtracting $\frac{S}{4}$ from $S$, yields:

$S - \frac{S}{4} = \frac{1}{8} + \frac{2}{32} + \frac{4}{128} + ...$

We see that the above sequence is a infinite geometric sequence with common ratio $\frac{1}{2}$. Therefore, the sum of that infinite series is $\frac{\frac{1}{8}}{\frac{1}{2}}$, which equals $\frac{1}{4}$. Our equation is now $S - \frac{S}{4} = \frac{1}{4}$. Solving for $S$ shows that $S = \boxed{\frac{1}{3}}$

Video Solution

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

See Also

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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