Difference between revisions of "2023 AMC 8 Problems/Problem 23"

Revision as of 20:54, 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 1 by OmegaLearn (Using Cool Probability Technique)

https://youtu.be/2t_Za0Y2IqY

Solution 1

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

Solution 2 (Linearity of Expectation)

Let $S_1, S_2, S_3$ , and $S_4$ denote the $4$ smaller $2 \times 2$ squares within the $3 \times 3$ square in some order. For each $S_i$ , let $X_i = 1$ if it contains a large gray diamond tiling and $X_i = 0$ otherwise. This means that $\mathbb{E}[X_i]$ is the probability that square $S_i$ has a large gray diamond, so $\mathbb{E}[X_1 + X_2 + X_3 + X_4]$ is our desired probability. However, since there is only one possible way to arrange the squares within every $2 \times 2$ square to form such a tiling, we have $\mathbb{E}[X_i] = (\tfrac{1}{4})^2 = \tfrac{1}{256}$ for all $i$ (as each of the smallest tiles has $4$ possible arrangements), and from the linearity of expectation we get $\mathbb{E}[X_1 + X_2 + X_3 + X_4] = \mathbb{E}[X_1] + \mathbb{E}[X_2] + \mathbb{E}[X_3] + \mathbb{E}[X_4] = \frac{1}{256} + \frac{1}{256} + \frac{1}{256} + \frac{1}{256} = \boxed{\textbf{(C)}\ \frac{1}{64}}$ ~eibc

Remark: This method might be too advanced for the AMC 8, and is probably unnecessary (refer to the other solutions for simpler techniques).

Solution 3

Note that the middle tile can be any of the four tiles. The white part of the middle tile points towards one of the corners, and for the white diamond to appear the three adjacent tiles must all be perfect. Thus, the solution is $\frac14 \cdot \frac14 \cdot \frac14 = \boxed{\text{(C)} \hspace{0.1 in} \frac{1}{64}}$

~aayr

Animated Video Solution

https://youtu.be/f4ffQEG0yUw

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

@@ Line 26: / Line 26: @@
 ==Solution 3==
-Note that the middle tile can be any of the four tiles. The white part of the middle tile points towards one of the corners, and for the white diamond to appear the three adjacent tiles must all be perfect. Thus, the solution is <math>\frac14 \cdot \frac14 \cdot \frac14 = \boxed{\frac{1}{64}}</math>
+Note that the middle tile can be any of the four tiles. The white part of the middle tile points towards one of the corners, and for the white diamond to appear the three adjacent tiles must all be perfect. Thus, the solution is <math>\frac14 \cdot \frac14 \cdot \frac14 = \boxed{\text{(C)} \hspace{0.1 in} \frac{1}{64}}</math>
 ~aayr

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