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Difference between revisions of "2023 AMC 8 Problems/Problem 23"
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==Solution 4 (Linearity of Expectation)==
 
==Solution 4 (Linearity of Expectation)==
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Let <math>S_1, S_2, S_3</math>, and <math>S_4</math> denote the <math>4</math> smaller <math>2 \times 2</math> squares within the <math>3 \times 3</math> square in some order. For each <math>S_i</math>, let <math>X_i = 1</math> if it contains a large gray diamond tiling and <math>X_i = 0</math> otherwise. This means that <math>\mathbb{E}[X_i]</math> is the probability that square <math>S_i</math> has a large gray diamond, so <math>\mathbb{E}[X_1 + X_2 + X_3 + X_4]</math> is our desired probability. However, since there is only one possible way to arrange the squares within every <math>2 \times 2</math> square to form such a tiling, we have <math>\mathbb{E}[X_i] = (\tfrac{1}{4})^2 = \tfrac{1}{256}</math> for all <math>i</math> (as each of the smallest tiles has <math>4</math> possible arrangements), and from the linearity of expectation we get  
+
Let <math>S_1, S_2, S_3</math>, and <math>S_4</math> denote the <math>4</math> smaller <math>2 \times 2</math> squares within the <math>3 \times 3</math> square in some order. For each <math>S_i</math>, let <math>X_i = 1</math> if it contains a large gray diamond tiling and <math>X_i = 0</math> otherwise. This means that <math>\mathbb{E}[X_i]</math> is the probability that square <math>S_i</math> has a large gray diamond, so <math>\mathbb{E}[X_1 + X_2 + X_3 + X_4]</math> is our desired probability. However, since there is only one possible way to arrange the squares within every <math>2 \times 2</math> square to form such a tiling, we have <math>\mathbb{E}[X_i] = (\tfrac{1}{4})^4 = \tfrac{1}{256}</math> for all <math>i</math> (as each of the smallest tiles has <math>4</math> possible arrangements), and from the linearity of expectation we get  
 
<cmath>\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}}.</cmath>
 
<cmath>\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}}.</cmath>
 
~eibc
 
~eibc

Revision as of 02:35, 24 September 2023

Problem

Each square in a $3 \times 3$ grid is randomly filled with one of the $4$ gray and white tiles shown below on the right. [asy] size(5.663333333cm); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,gray); draw((1,0)--(1,3)--(2,3)--(2,0),gray); draw((0,1)--(3,1)--(3,2)--(0,2),gray); fill((6,.33)--(7,.33)--(7,1.33)--cycle,mediumgray); draw((6,.33)--(7,.33)--(7,1.33)--(6,1.33)--cycle,gray); fill((6,1.67)--(7,2.67)--(6,2.67)--cycle,mediumgray); draw((6,1.67)--(7,1.67)--(7,2.67)--(6,2.67)--cycle,gray); fill((7.33,.33)--(8.33,.33)--(7.33,1.33)--cycle,mediumgray); draw((7.33,.33)--(8.33,.33)--(8.33,1.33)--(7.33,1.33)--cycle,gray); fill((8.33,1.67)--(8.33,2.67)--(7.33,2.67)--cycle,mediumgray); draw((7.33,1.67)--(8.33,1.67)--(8.33,2.67)--(7.33,2.67)--cycle,gray); [/asy] What is the probability that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids? Below is an example of such tiling. [asy] size(2cm); fill((1,0)--(0,1)--(0,2)--(1,1)--cycle,mediumgray); fill((2,0)--(3,1)--(2,2)--(1,1)--cycle,mediumgray); fill((1,2)--(1,3)--(0,3)--cycle,mediumgray); fill((1,2)--(2,2)--(2,3)--cycle,mediumgray); fill((3,2)--(3,3)--(2,3)--cycle,mediumgray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,gray); draw((1,0)--(1,3)--(2,3)--(2,0),gray); draw((0,1)--(3,1)--(3,2)--(0,2),gray); [/asy]

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

Solution 1

There are $4$ cases that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids, as shown below: [asy] size(375); fill((1,1)--(2,2)--(1,3)--(0,2)--cycle,mediumgray); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle,gray); draw((1,0)--(1,3)--(2,3)--(2,0),gray); draw((0,1)--(3,1)--(3,2)--(0,2),gray); fill(shift(7,0)*((1,1)--(2,2)--(1,3)--(0,2)--cycle),mediumgray); draw(shift(6,0)*((0,0)--(3,0)--(3,3)--(0,3)--cycle),gray); draw(shift(6,0)*((1,0)--(1,3)--(2,3)--(2,0)),gray); draw(shift(6,0)*((0,1)--(3,1)--(3,2)--(0,2)),gray); fill(shift(12,-1)*((1,1)--(2,2)--(1,3)--(0,2)--cycle),mediumgray); draw(shift(12,0)*((0,0)--(3,0)--(3,3)--(0,3)--cycle),gray); draw(shift(12,0)*((1,0)--(1,3)--(2,3)--(2,0)),gray); draw(shift(12,0)*((0,1)--(3,1)--(3,2)--(0,2)),gray); fill(shift(19,-1)*((1,1)--(2,2)--(1,3)--(0,2)--cycle),mediumgray); draw(shift(18,0)*((0,0)--(3,0)--(3,3)--(0,3)--cycle),gray); draw(shift(18,0)*((1,0)--(1,3)--(2,3)--(2,0)),gray); draw(shift(18,0)*((0,1)--(3,1)--(3,2)--(0,2)),gray); [/asy] There are $4^5$ ways to decide the $5$ white squares for each case, and the cases do not have any overlap.

So, the requested probability is \[\frac{4\cdot4^5}{4^9} = \frac{4^6}{4^9} = \frac{1}{4^3} = \boxed{\textbf{(C) } \frac{1}{64}}.\] -apex304, TaeKim, MRENTHUSIASM

Solution 2

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 $\frac14 \cdot \frac14 \cdot \frac14 = \boxed{\textbf{(C) } \frac{1}{64}}$.

~aayr

Solution 3

Note that the tiles must be in perfect placement. Because of that, each diamond has a $\left(\frac14\right)^4$ chance of appearing. And since there are 4 placements, our solution $\frac{1}{4^4} \cdot 4 = \boxed{\textbf{(C) } \frac{1}{64}}$.

~Ligonmathkid2

Solution 4 (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 $\mathbb{E}[X_i] = (\tfrac{1}{4})^4 = \tfrac{1}{256}$ 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 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. $X_1 + X_2 + X_3 + X_4 \le 1$.

~numerophile

Video Solution (THINKING CREATIVELY!!!)

https://youtu.be/B5qcEnobmEU

~Education, the Study of Everything


Video Solution 1

https://youtu.be/2t_Za0Y2IqY

Animated Video Solution

https://www.youtube.com/watch?v=wq5Ie6mUxvY&ab_channel=StarLeague

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

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=2405

Video Solution by Interstigation

https://youtu.be/1bA7fD7Lg54?t=2338

Video Solution by WhyMath

https://youtu.be/uJklHght9ds

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1408s

~harungurcan

See Also

2023 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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