2023 AMC 8 Problem 23

Below is the video solution and professionally curated solution for Problem 23 of the 2023 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 8 solutions, or check the answer key.

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Concepts:basic probabilitymultiplication principle

Difficulty rating: 1840

23.

Each square in a 3×33 \times 3 grid is randomly filled with one of the 44 shaded-and-unshaded tiles shown below on the right.

What is the probability that the tiling will contain a large shaded diamond in one of the smaller 2×22 \times 2 grids? Below is an example of such a tiling.

11024\dfrac{1}{1024}

1256\dfrac{1}{256}

164\dfrac{1}{64}

116\dfrac{1}{16}

14\dfrac{1}{4}

Video solution:
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Written solution:

There are 494^9 possible tilings. There are 44 possible 2×22\times2 grids where a large shaded diamond could appear.

After one of these 2×22\times2 grids is chosen, the four tile orientations inside it are forced, and the other 55 squares can be filled in any way. This gives 454^5 tilings for each chosen 2×22\times2 grid.

Two different 2×22\times2 grids cannot both contain a large shaded diamond, because their overlapping squares would require incompatible tile orientations. Therefore the number of favorable tilings is 445=46.4\cdot4^5=4^6.

The desired probability is then 4649=143=164. \dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \dfrac{1}{64}.

Thus, C is the correct answer.

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