2023 AMC 8 Problems/Problem 23
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (Linearity of Expectation)
- 6 Video Solution 1 by OmegaLearn (Using Cool Probability Technique)
- 7 Animated Video Solution
- 8 Video Solution by Magic Square
- 9 Video Solution by Interstigation
- 10 Video Solution by WhyMath
- 11 Video Solution by harungurcan
- 12 See Also
Problem
Each square in a grid is randomly filled with one of the gray and white tiles shown below on the right. What is the probability that the tiling will contain a large gray diamond in one of the smaller grids? Below is an example of such tiling.
Solution 1
There are cases that the tiling will contain a large gray diamond in one of the smaller grids, as shown below: There are ways to decide the white squares for each case, and the cases do not have any overlap.
So, the requested probability is -apex304, TaeKim, MRENTHUSIASM
Solution 2
Note that the middle tile can be any of the four tiles. The gray part of the middle tile points towards one of the corners, and for the gray diamond to appear the three adjacent tiles must all be perfect. Thus, the solution is .
~aayr
Solution 3
Note that the tiles must be in perfect placement. Because of that, each diamond has a chance of appearing. And since there are 4 placements, our solution .
~Ligonmathkid2
Solution 4 (Linearity of Expectation)
Let , and denote the smaller squares within the square in some order. For each , let if it contains a large gray diamond tiling and otherwise. This means that is the probability that square has a large gray diamond, so is our desired probability. However, since there is only one possible way to arrange the squares within every square to form such a tiling, we have for all (as each of the smallest tiles has possible arrangements), and from the linearity of expectation we get ~eibc
Remark 1: This method might be too advanced for the AMC 8, and is probably unnecessary (refer to the other solutions for simpler techniques).
Remark 2: Note that Probability and Expected Value are equivalent in this problem since there will never be two diamonds on one tiling. i.e. .
~numerophile
Video Solution 1 by OmegaLearn (Using Cool Probability Technique)
Animated Video Solution
https://www.youtube.com/watch?v=wq5Ie6mUxvY&ab_channel=StarLeague
~Star League (https://starleague.us)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=2405
Video Solution by Interstigation
https://youtu.be/1bA7fD7Lg54?t=2338
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1408s
~harungurcan
See Also
2023 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.