2023 AMC 8 Problems/Problem 23
- 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
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.
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
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 .
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 .
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. .
Video Solution 1 by OmegaLearn (Using Cool Probability Technique)
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by Magic Square
Video Solution by Interstigation
Video Solution by WhyMath
Video Solution by harungurcan
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