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Difference between revisions of "2023 AMC 8 Problems/Problem 23"
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Each square in a 3x3 grid is randomly filled with one of the 4 gray and white tiles shown below on the right. \\
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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.
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<cmath>\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}</cmath>
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==Video Solution Using Cool Probability Technique==
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https://youtu.be/2t_Za0Y2IqY
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~ pi_is_3.14
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==Text Solution==
 
Probability is total favorable outcomes over total outcomes, so we can find these separately to determine the answer.
 
Probability is total favorable outcomes over total outcomes, so we can find these separately to determine the answer.
  

Revision as of 18:24, 24 January 2023

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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