Difference between revisions of "2021 Fall AMC 10B Problems/Problem 16"
Sillycow21 (talk | contribs) (→Solution) |
|||
| Line 21: | Line 21: | ||
Case 3: Silva swaps two balls that have not been swapped | Case 3: Silva swaps two balls that have not been swapped | ||
| − | There are two ways for Silva to do this, and it leaves 1 | + | There are two ways for Silva to do this, and it leaves 1 ball occupying their original positions. |
Revision as of 22:10, 25 October 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
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
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. 
