Difference between revisions of "2021 Fall AMC 10B Problems/Problem 16"
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The "expected value" in the question tips us off to this technique. Consider any ball. The probability it returns to the same position is the probability of being swapped twice plus the probability of never being swapped: <cmath>\frac{2}{5} \cdot \frac{1}{5} + \left(\frac{3}{5}\right)^2 = \frac{11}{5}.</cmath> Multiply by 5 for 5 balls to get <math>\boxed{(\textbf{D}) \: 2.2}.</math> | The "expected value" in the question tips us off to this technique. Consider any ball. The probability it returns to the same position is the probability of being swapped twice plus the probability of never being swapped: <cmath>\frac{2}{5} \cdot \frac{1}{5} + \left(\frac{3}{5}\right)^2 = \frac{11}{5}.</cmath> Multiply by 5 for 5 balls to get <math>\boxed{(\textbf{D}) \: 2.2}.</math> | ||
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| + | ~Dhillonr25 | ||
==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
Revision as of 23:47, 1 November 2022
Contents
Problem
Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris's. What is the expected number of balls that occupy their original positions after these two successive transpositions?
Solution 1
After the first swap, we do casework on the next swap.
Case 1: Silva swaps the two balls that were just swapped
There is only one way for Silva to do this, and it leaves 5 balls occupying their original position.
Case 2: Silva swaps one ball that has just been swapped with one that hasn't swapped
There are two ways for Silva to do this, and it leaves 2 balls occupying their original positions.
Case 3: Silva swaps two balls that have not been swapped
There are two ways for Silva to do this, and it leaves 1 ball occupying their original positions.
Our answer is the average of all 5 possible swaps, so we get ![\[\frac{5+2\cdot2+2\cdot1}{5}=\frac{11}5=\boxed{(\textbf{D}) \: 2.2}.\]](http://latex.artofproblemsolving.com/8/5/a/85a109e07d24364b29df39000cac4b045f0c14d2.png)
~kingofpineapplz
Solution 2 (Linearity of Expectation)
The "expected value" in the question tips us off to this technique. Consider any ball. The probability it returns to the same position is the probability of being swapped twice plus the probability of never being swapped: ![\[\frac{2}{5} \cdot \frac{1}{5} + \left(\frac{3}{5}\right)^2 = \frac{11}{5}.\]](http://latex.artofproblemsolving.com/2/7/7/27708387258bc57b0892376339fbb259aa00515f.png)
Multiply by 5 for 5 balls to get 
~Dhillonr25
Video Solution by Interstigation
https://www.youtube.com/watch?v=0FtXvjn_4y0
~Interstigation
Video Solution
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
