Difference between revisions of "2021 Fall AMC 10B Problems/Problem 16"
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In probability and statistics, you learn that the mean and expected value are fundamentally equivalent values. | In probability and statistics, you learn that the mean and expected value are fundamentally equivalent values. | ||
| − | The total number of outcomes of ball picking is <math> | + | The total number of outcomes of ball picking is <math>{5 \choose 2}^2 = 100</math>. |
Now there are three cases either Chris and Silva choose the same two balls (C1), they choose two different balls (C2), or they choose one overlapping ball and three nonoverlapping (C3). | Now there are three cases either Chris and Silva choose the same two balls (C1), they choose two different balls (C2), or they choose one overlapping ball and three nonoverlapping (C3). | ||
C1: There are <math>{5 \choose 2} = 10</math> possible placements of choosing <math>2</math> balls. This gives a remaining <math>5</math> unarranged balls. This gives a weighted mean pair of <math>10(5) = 50</math>. | C1: There are <math>{5 \choose 2} = 10</math> possible placements of choosing <math>2</math> balls. This gives a remaining <math>5</math> unarranged balls. This gives a weighted mean pair of <math>10(5) = 50</math>. | ||
| + | |||
C2: There are <math>5</math> ways of choosing <math>4</math> balls adjacently, and due to Chris and Silva each choosing <math>2</math> gives <math>10</math> ways due to symmetry. This gives a weighted mean pair of <math>10(1) = 10</math>. | C2: There are <math>5</math> ways of choosing <math>4</math> balls adjacently, and due to Chris and Silva each choosing <math>2</math> gives <math>10</math> ways due to symmetry. This gives a weighted mean pair of <math>10(1) = 10</math>. | ||
| + | |||
C3: There are now <math>100 - 10 - 10 = 80</math> ways for C3 to occur. This gives <math>2</math> balls that remain in their original positions for a weighted mean pair of <math>80(2) = 160</math>. | C3: There are now <math>100 - 10 - 10 = 80</math> ways for C3 to occur. This gives <math>2</math> balls that remain in their original positions for a weighted mean pair of <math>80(2) = 160</math>. | ||
Latest revision as of 03:56, 25 August 2024
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 been 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 position.
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: ![\[\frac{2}{5} \cdot \frac{1}{5} + \left(\frac{3}{5}\right)^2 = \frac{11}{25}.\]](http://latex.artofproblemsolving.com/b/8/8/b889a9a933975e43ae0744eeccc20abbdd5dfd6f.png)
Multiply by 5 for 5 balls to get 
~Dhillonr25
Solution 3 (Calculating-Mean)
In probability and statistics, you learn that the mean and expected value are fundamentally equivalent values.
The total number of outcomes of ball picking is 
.
Now there are three cases either Chris and Silva choose the same two balls (C1), they choose two different balls (C2), or they choose one overlapping ball and three nonoverlapping (C3).
C1: There are 
possible placements of choosing 
balls. This gives a remaining 
unarranged balls. This gives a weighted mean pair of 
.
C2: There are ways of choosing 
balls adjacently, and due to Chris and Silva each choosing gives 
ways due to symmetry. This gives a weighted mean pair of 
.
C3: There are now 
ways for C3 to occur. This gives balls that remain in their original positions for a weighted mean pair of 
.
Calculating the mean gives 
.
~PeterDoesPhysics
Video Solution by OmegaLearn
https://youtu.be/EE-TtptBHeI?t=174
~ pi_is_3.14
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
Video Solution by TheBeautyofMath
~IceMatrix
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. 
