Difference between revisions of "2021 Fall AMC 10B Problems/Problem 16"

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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 balls occupying their original positions.
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There are two ways for Silva to do this, and it leaves 1 ball occupying their original positions.
  
  

Revision as of 21:10, 25 October 2022

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?

$(\textbf{A})\: 1.6\qquad(\textbf{B}) \: 1.8\qquad(\textbf{C}) \: 2.0\qquad(\textbf{D}) \: 2.2\qquad(\textbf{E}) \: 2.4$

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 \[\frac{5+2\cdot2+2\cdot1}{5}=\frac{11}5=\boxed{(\textbf{D}) \: 2.2}.\]


~kingofpineapplz

Video Solution by Interstigation

https://www.youtube.com/watch?v=0FtXvjn_4y0

~Interstigation

Video Solution

https://youtu.be/LLYYvYXl2rw

~Education, the Study of Everything

Video Solution by WhyMath

https://youtu.be/RiD1eoGq36s

~savannahsolver

See Also

2021 Fall AMC 10B (ProblemsAnswer KeyResources)
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

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