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2023 AMC 8 Problems/Problem 23

Revision as of 18:24, 24 January 2023 by Pi is 3.14 (talk | contribs)

Each square in a 3x3 grid is randomly filled with one of the 4 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 2x2 grids? Below is an example of such a tiling.

\[\textbf{(A) } \frac{1}{1024} \qquad \textbf{(B) } \frac{1}{256} \qquad \textbf{(C) } \frac{1}{64} \qquad \textbf{(D) } \frac{1}{16} \qquad \textbf{(E) } ~\frac{1}{4}\]

Video Solution Using Cool Probability Technique

https://youtu.be/2t_Za0Y2IqY ~ pi_is_3.14

Text Solution

Probability is total favorable outcomes over total outcomes, so we can find these separately to determine the answer.

There are $4$ ways to choose the big diamond location from our $9$ square grid. From our given problem there are $4$ different arrangements of triangles for every square. This implies that from having $1$ diamond we are going to have $4^5$ distinct patterns outside of the diamond. This gives a total of $4\cdot 4^5 = 4^6$ favorable cases.


There are 9 squares and 4 possible designs for each square, giving $4^9$ total outcomes. Thus, our desired probability is $\dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \boxed{\text{(C)} \hspace{0.1 in} \dfrac{1}{64}}$ . -apex304, SohumUttamchandani, wuwang2002, TaeKim. Cxrupptedpat

Animated Video Solution

https://youtu.be/f4ffQEG0yUw

~Star League (https://starleague.us)

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