Difference between revisions of "2021 AMC 12B Problems/Problem 23"
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| − | ==Problem | + | ==Problem== |
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 <math>i</math> is <math>2^{-i}</math> for <math>i=1,2,3,....</math> More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is <math>\frac pq,</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins <math>3,17,</math> and <math>10.</math>) What is <math>p+q?</math> | 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 <math>i</math> is <math>2^{-i}</math> for <math>i=1,2,3,....</math> More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is <math>\frac pq,</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins <math>3,17,</math> and <math>10.</math>) What is <math>p+q?</math> | ||
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\end{align*}</cmath> | \end{align*}</cmath> | ||
The answer is <math>6+49=\boxed{\textbf{(A) }55}.</math> | The answer is <math>6+49=\boxed{\textbf{(A) }55}.</math> | ||
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| + | ==See Also== | ||
| + | {{AMC12 box|year=2021|ab=B|num-b=22|num-a=24}} | ||
| + | {{MAA Notice}} | ||
Revision as of 20:01, 11 February 2021
Problem
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 
is 
for 
More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is 
where 
and 
are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins 
and 
) What is 
Solution
"Evenly spaced" just means the bins form an arithmetic sequence.
Suppose the middle bin in the sequence is 
. There are 
different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these 
bins are chosen is 
, so the probability is the middle bin is 
. 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*}](http://latex.artofproblemsolving.com/b/5/3/b53c2ce3c14c76f8cd9336f940fddc695f69f106.png)
The answer is 
See Also
| 2021 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 22 |
Followed by Problem 24 |
| 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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