Difference between revisions of "2019 AMC 12B Problems/Problem 13"
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==Solution 1== | ==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 <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>. | |
− | ==Solution | + | ==Solution 2== |
− | + | 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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− | 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>.) | ||
− | + | ==Solution 3== | |
− | + | 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>. | |
− | ==Solution 3 | ||
− | 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>. | ||
− | ==Solution 4 ( | + | ==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. | ||
− | There are 4 possible results in the box: Red, Green, Red and Green, none, with probability of <math>\frac{1}{4}</math> for each | + | 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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− | ==Solution 5 | + | ==Solution 5== |
− | Write out the infinite geometric series as <math>\frac{1}{2}</math>, <math>\frac{1}{4} | + | 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 22:18, 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.
Contents
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 is for What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?
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 (by the geometric series sum formula). Therefore the other two probabilities have to both be .
Solution 2
Suppose the green ball goes in bin , for some . The probability of this occurring is . Given that this occurs, the probability that the red ball goes in a higher-numbered bin is (by the geometric series sum formula). Thus the probability that the green ball goes in bin , and the red ball goes in a bin greater than , is . Summing from to infinity, we get
where we again used the geometric series sum formula. (Alternatively, if this sum equals , then by writing out the terms and multiplying both sides by , we see , which gives .)
Solution 3
The probability that the two balls will go into adjacent bins is by the geometric series sum formula. Similarly, the probability that the two balls will go into bins that have a distance of from each other is (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 , which, by the geometric series sum formula, is .
Solution 4 (quick, conceptual)
Define a win as a ball appearing in higher numbered box.
Start from the first box.
There are possible results in the box: Red, Green, Red and Green, or none, with an equal probability of 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 is the probability that Red wins, we can write : there is a probability that "Red" wins immediately, a probability in the cases "Green" or "Red and Green", and in the "None" case (which occurs with probability), we then start again, giving the same probability . Hence, solving the equation, we get .
Solution 5
Write out the infinite geometric series as , . 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 , term , 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 . Summing, we get .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.