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

(→‎See Also)
(Thoroughly cleaned up the solutions in terms of LaTeX, grammar, formatting, clarity, and readability)
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==Solution 1==
 
==Solution 1==
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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. The probability of both landing in the same bin is <math>\sum_{k=1}^{\infty}2^{-2k}</math>. The sum is equal to <math>\frac{1}{3}</math>. Therefore the other two probabilities have to both be <math>\textbf{(C) } \frac{1}{3}</math>.
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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 <math>\sum_{k=1}^{\infty}{2^{-k} \cdot 2^{-k}} = \sum_{k=1}^{\infty}2^{-2k} = \frac{1}{3}</math> (by the geometric series sum formula). Therefore the other two probabilities have to both be <math>\frac{1-\frac{1}{3}}{2} = \boxed{\textbf{(C) } \frac{1}{3}}</math>.
  
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==Solution 1 Variant==
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==Solution 2==
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We solve for the probability by doing <math>\frac{1-(\text{Probability of Equality})}{2}</math>.
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Suppose the green ball goes in bin <math>i</math>, for some <math>i \ge 1</math>. The probability of this occurring is <math>\frac{1}{2^i}</math>. Given that this occurs, the probability that the red ball goes in a higher-numbered bin is <math>\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}</math> (by the geometric series sum formula). Thus the probability that the green ball goes in bin <math>i</math>, and the red ball goes in a bin greater than <math>i</math>, is <math>\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}</math>. Summing from <math>i=1</math> to infinity, we get
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We see that the probability of equality is the summation of all the probabilities that the balls land in the same container. Thus we have the probability of equality being equal to <math>\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)+\left(\frac{1}{16}\right)\left(\frac{1}{16}\right)...</math>
 
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The summation of this expression is equal to <cmath>\sum_{n=0}^{\infty} (1)\left(\frac{1}{4}\right)^{n}-1</cmath>. Using the geometric sum formula, we obtain the summation of this expression to be <math>\frac{1}{\frac{3}{4}}-1</math> or <math>\frac{1}{3}</math>.
 
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==Solution 2 (variant)==
 
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Suppose the green ball goes in bin <math>i</math>, for some <math>i \ge 1</math>. The probability of this occurring is <math>\frac{1}{2^i}</math>. Given this occurs, the probability that the red ball goes in a higher-numbered bin is <math>\frac{1}{2^{i+1}} + \frac{1}{2^{i+2}} + \ldots = \frac{1}{2^i}</math>. Thus the probability that the green ball goes in bin <math>i</math>, and the red ball goes in a bin greater than <math>i</math>, is <math>\left(\frac{1}{2^i}\right)^2 = \frac{1}{4^i}</math>. Summing from <math>i=1</math> to infinity, we get
 
  
 
<cmath>\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath>
 
<cmath>\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath>
 +
where we again used the geometric series sum formula. (Alternatively, if this sum equals <math>n</math>, then by writing out the terms and multiplying both sides by <math>4</math>, we see <math>4n = n+1</math>, which gives <math>n = \frac{1}{3}</math>.)
  
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(Note: to find this sum, we use the formula <math>\sum_{i=1}^{\infty} r^i = \frac{r}{1-r}</math>. Since in this case <math>r = \frac{1}{4}</math>, the answer is <math>\frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}</math>. If you don't know this formula, you may instead note that if you multiply the sum by <math>4</math>, it is equivalent to adding <math>1</math>. Thus: <math>4n = n+1</math>, which clearly simplifies to <math>n = \frac{1}{3}</math>.
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==Solution 3==
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The probability that the two balls will go into adjacent bins is <math>\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}</math> by the geometric series sum formula. Similarly, the probability that the two balls will go into bins that have a distance of <math>2</math> from each other is <math>\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}</math> (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 <math>\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \cdots</math>, which, by the geometric series sum formula, is <math>\boxed{\textbf{(C) } \frac{1}{3}}</math>.
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==Solution 3 (infinite geometric series)==
 
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The probability that the two balls will go into adjacent bins is <math>\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + ... = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} = \frac{1}{6}</math>. The probability that the two balls will go into bins that have a distance of 2 from each other is <math>\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} = \frac{1}{12}</math>. We can see that each time we add a bin between the two balls, the probability halves. Thus, our answer is <math>\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + ...</math>, which converges into <math>\frac{1}{3}</math>.
 
  
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==Solution 4 (quickest)==
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==Solution 4 (quick, conceptual)==
 
Define a win as a ball appearing in higher numbered box.
 
Define a win as a ball appearing in higher numbered box.
  
 
Start from the first box.  
 
Start from the first box.  
  
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There are 4 possible results in the box: Red, Green, Red and Green, none, with probability of <math>\frac{1}{4}</math> for each. Red win, Green win, Tie all have the same probability of <math>\frac{1}{3}</math>. If none of the balls is in the first box, the game restarts at the second box with the same kind of probability distribution.
+
There are <math>4</math> possible results in the box: Red, Green, Red and Green, or none, with an equal probability of <math>\frac{1}{4}</math> 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 <math>p</math> is the probability that Red wins, we can write <math>p = \frac{1}{4} + \frac{1}{4}p</math>: there is a <math>\frac{1}{4}</math> probability that "Red" wins immediately, a <math>0</math> probability in the cases "Green" or "Red and Green", and in the "None" case (which occurs with <math>\frac{1}{4}</math> probability), we then start again, giving the same probability <math>p</math>. Hence, solving the equation, we get <math>p = \boxed{\textbf{(C) } \frac{1}{3}}</math>.
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So finally, Red win, Green win and Tie all have a probability of <math>\frac{1}{3}</math>
 
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The answer is <math>\boxed{C}</math>
 
  
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==Solution 5(easiest)==
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==Solution 5==
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Write out the infinite geometric series as <math>\frac{1}{2}</math>, <math>\frac{1}{4}</math>, <math>\frac{1}{8}</math>, <math>\frac{1}{16}</math>... To find the probablilty that red scores after, we can simply remove all odd number terms(i.e term 1, term 3...), and then sum the remaining. This works for a similar reason to solution 2 variant(so look at that if you don't understand why I can do this). Writing this out as another infinite geometric sequence,<math>\frac{1}{4}</math>, <math>\frac{1}{16}</math>, <math>\frac{1}{64}</math>..., Summing, we get <cmath>\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath>.
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Write out the infinite geometric series as <math>\frac{1}{2}</math>, <math>\frac{1}{4}, \frac{1}{8}, \frac{1}{16}, \cdots</math>. 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 <math>1</math>, term <math>3</math>, 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 <math>\frac{1}{4}, \frac{1}{16}, \frac{1}{64}, \cdots</math>. Summing, we get <cmath>\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}</cmath>.
  
 
==See Also==
 
==See Also==

Revision as of 23:18, 17 February 2019

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 the other two probabilities have to both 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

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 4 (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 5

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 \[\sum_{i=1}^{\infty} \frac{1}{4^i} = \boxed{\textbf{(C) } \frac{1}{3}}\].

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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