Difference between revisions of "2021 AMC 12B Problems/Problem 23"
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<math>\textbf{(A) }55 \qquad \textbf{(B) }56 \qquad \textbf{(C) }57\qquad \textbf{(D) }58 \qquad \textbf{(E) }59</math> | <math>\textbf{(A) }55 \qquad \textbf{(B) }56 \qquad \textbf{(C) }57\qquad \textbf{(D) }58 \qquad \textbf{(E) }59</math> | ||
− | ==Solution== | + | ==Solution 1== |
"Evenly spaced" just means the bins form an arithmetic sequence. | "Evenly spaced" just means the bins form an arithmetic sequence. | ||
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<cmath>\begin{align*} | <cmath>\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]\\ | 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[\left(\frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}+\cdots \right) + \left(\frac{1}{8^2} + \frac{1}{8^3} + \frac{1}{8^4} + \cdots \right) + \cdots\right]\\ |
− | &= \frac34\left[\frac17\cdot \left(1 + \frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}\right)\right]\\ | + | &= \frac34\left[\frac17\cdot \left(1 + \frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3} + \cdots \right)\right]\\ |
&= \frac34\cdot \frac{8}{49}\\ | &= \frac34\cdot \frac{8}{49}\\ | ||
&= \frac{6}{49} | &= \frac{6}{49} | ||
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==Solution 2== | ==Solution 2== | ||
− | As in solution 1, note that "evenly spaced" means the bins are in arithmetic sequence. We let the first bin be <math>a</math> and the common difference be <math>d</math>. Further note that each <math>(a, d)</math> pair uniquely determines a set of 3 bins. | + | As in solution 1, note that "evenly spaced" means the bins are in arithmetic sequence. We let the first bin be <math>a</math> and the common difference be <math>d</math>. Further note that each <math>(a, d)</math> pair uniquely determines a set of <math>3</math> bins. |
− | We have <math> | + | We have <math>a\geq1</math> because the leftmost bin in the sequence can be any bin, and <math>d\geq1</math>, because the bins must be distinct. |
This gives us the following sum for the probability: | This gives us the following sum for the probability: | ||
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&= 6 \left( \frac{1}{7} \right) \left( \frac{1}{7} \right) \\ | &= 6 \left( \frac{1}{7} \right) \left( \frac{1}{7} \right) \\ | ||
&= \frac{6}{49} .\end{align*}</cmath> | &= \frac{6}{49} .\end{align*}</cmath> | ||
− | Therefore the answer is <math>6 + 49 = 55</math> | + | Therefore the answer is <math>6 + 49 = \boxed{\textbf{(A) }55}</math>. |
-Darren Yao | -Darren Yao | ||
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&=6\sum_{i=1}^\infty\frac{8}{7}\cdot2^{-3}\cdot2^{-3i}\\ | &=6\sum_{i=1}^\infty\frac{8}{7}\cdot2^{-3}\cdot2^{-3i}\\ | ||
&=\frac{6}{7}\sum_{i=1}^\infty2^{-3i}\\ | &=\frac{6}{7}\sum_{i=1}^\infty2^{-3i}\\ | ||
− | &=\frac{6}{7}\frac{2^{-3}}{1-\tfrac18} = \frac{6}{49}. | + | &=\frac{6}{7}\cdot\frac{2^{-3}}{1-\tfrac18}\\ |
− | \end{align*} | + | &=\frac{6}{49}. |
− | </ | + | \end{align*}</cmath> |
+ | Therefore the answer is <math>6 + 49 = \boxed{\textbf{(A) }55}</math>. | ||
− | ==Solution 4 (Table | + | ==Solution 4 (Table)== |
+ | Based on the value of <math>n,</math> we construct the following table: | ||
<cmath>\begin{array}{c|c|c|c} | <cmath>\begin{array}{c|c|c|c} | ||
+ | & & & \\ [-1.5ex] | ||
\textbf{Exactly }\boldsymbol{n}\textbf{ Spaces Apart} & \textbf{Bin \#s} & \textbf{Expression} & \textbf{Prob. of One Such Perm.} \\ [1ex] | \textbf{Exactly }\boldsymbol{n}\textbf{ Spaces Apart} & \textbf{Bin \#s} & \textbf{Expression} & \textbf{Prob. of One Such Perm.} \\ [1ex] | ||
\hline\hline | \hline\hline | ||
− | & & & \\ [- | + | & & & \\ [-1.5ex] |
n=1 & 1,2,3 & 2^{-1}\cdot2^{-2}\cdot2^{-3} & 2^{-6} \\ [1ex] | n=1 & 1,2,3 & 2^{-1}\cdot2^{-2}\cdot2^{-3} & 2^{-6} \\ [1ex] | ||
& 2,3,4 & 2^{-2}\cdot2^{-3}\cdot2^{-4} & 2^{-9} \\ [1ex] | & 2,3,4 & 2^{-2}\cdot2^{-3}\cdot2^{-4} & 2^{-9} \\ [1ex] | ||
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& \cdots & \cdots & \cdots \\ [1ex] | & \cdots & \cdots & \cdots \\ [1ex] | ||
\hline | \hline | ||
− | & & & \\ [- | + | & & & \\ [-1.5ex] |
n=2 & 1,3,5 & 2^{-1}\cdot2^{-3}\cdot2^{-5} & 2^{-9} \\ [1ex] | n=2 & 1,3,5 & 2^{-1}\cdot2^{-3}\cdot2^{-5} & 2^{-9} \\ [1ex] | ||
& 2,4,6 & 2^{-2}\cdot2^{-4}\cdot2^{-6} & 2^{-12} \\ [1ex] | & 2,4,6 & 2^{-2}\cdot2^{-4}\cdot2^{-6} & 2^{-12} \\ [1ex] | ||
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& \cdots & \cdots & \cdots \\ [1ex] | & \cdots & \cdots & \cdots \\ [1ex] | ||
\hline | \hline | ||
− | & & & \\ [- | + | & & & \\ [-1.5ex] |
n=3 & 1,4,7 & 2^{-1}\cdot2^{-4}\cdot2^{-7} & 2^{-12} \\ [1ex] | n=3 & 1,4,7 & 2^{-1}\cdot2^{-4}\cdot2^{-7} & 2^{-12} \\ [1ex] | ||
& 2,5,8 & 2^{-2}\cdot2^{-5}\cdot2^{-8} & 2^{-15} \\ [1ex] | & 2,5,8 & 2^{-2}\cdot2^{-5}\cdot2^{-8} & 2^{-15} \\ [1ex] | ||
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& \cdots & \cdots & \cdots \\ [1ex] | & \cdots & \cdots & \cdots \\ [1ex] | ||
\hline | \hline | ||
− | & & & \\ [- | + | & & & \\ [-1.5ex] |
− | \cdots & \cdots & \cdots & \cdots | + | \cdots & \cdots & \cdots & \cdots \\ [1ex] |
\end{array}</cmath> | \end{array}</cmath> | ||
− | |||
Since three balls have <math>3!=6</math> permutations, the requested probability is | Since three balls have <math>3!=6</math> permutations, the requested probability is | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | 6\left | + | 6\left(\sum_{k=0}^{\infty}2^{-6-3k}+\sum_{k=0}^{\infty}2^{-9-3k}+\sum_{k=0}^{\infty}2^{-12-3k}+\cdots\right)&=6\left(2^{-6}\sum_{k=0}^{\infty}2^{-3k}+2^{-9}\sum_{k=0}^{\infty}2^{-3k}+2^{-12}\sum_{k=0}^{\infty}2^{-3k}+\cdots\right) \\ |
− | &= | + | &=\left(6\sum_{k=0}^{\infty}2^{-3k}\right)\cdot\left(2^{-6}+2^{-9}+2^{-12}+\cdots\right) \\ |
− | &= | + | &=\left(6\sum_{k=0}^{\infty}2^{-3k}\right)\cdot\left(\sum_{k=0}^{\infty}2^{-6-3k}\right) \\ |
− | &=6\left | + | &=\left(6\sum_{k=0}^{\infty}2^{-3k}\right)\cdot\left(2^{-6}\sum_{k=0}^{\infty}2^{-3k}\right) \\ |
+ | &=\frac{6}{1-2^{-3}}\cdot\frac{2^{-6}}{1-2^{-3}} \\ | ||
&=\frac{6}{49} | &=\frac{6}{49} | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | by infinite geometric series, | + | by infinite geometric series, from which the answer is <math>6+49=\boxed{\textbf{(A) }55}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | == Video Solution Using | + | == Video Solution== |
+ | https://youtu.be/x16YUmd0OqY | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/_IvLCWSSDFs | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution Using Infinite Geometric Series == | ||
https://youtu.be/3B-3_nOTIu4 | https://youtu.be/3B-3_nOTIu4 | ||
+ | |||
~hippopotamus1 | ~hippopotamus1 | ||
Latest revision as of 03:25, 28 January 2023
Contents
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 1
"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
The answer is
Solution 2
As in solution 1, note that "evenly spaced" means the bins are in arithmetic sequence. We let the first bin be and the common difference be
. Further note that each
pair uniquely determines a set of
bins.
We have because the leftmost bin in the sequence can be any bin, and
, because the bins must be distinct.
This gives us the following sum for the probability:
Therefore the answer is
.
-Darren Yao
Solution 3
This is a slightly messier variant of solution 2. If the first ball is in bin and the second ball is in bin
, then the third ball is in bin
. Thus the probability is
Therefore the answer is
.
Solution 4 (Table)
Based on the value of we construct the following table:
Since three balls have
permutations, the requested probability is
by infinite geometric series, from which the answer is
~MRENTHUSIASM
Video Solution
~MathProblemSolvingSkills.com
Video Solution by OmegaLearn
~ pi_is_3.14
Video Solution Using Infinite Geometric Series
~hippopotamus1
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.