Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2 (Linearity of Expectation))
(→‎Solution 2 (Linearity of Expectation))
Line 19: Line 19:
  
 
==Solution 2 (Linearity of Expectation)==
 
==Solution 2 (Linearity of Expectation)==
−
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{3}[X_i] = (\tfrac{1}{4})^2 = \tfrac{1}{256}</math> for all <math>i</math>, 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})^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  
 
<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 19:30, 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

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).

Animated Video Solution

https://youtu.be/f4ffQEG0yUw

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_23&oldid=187494"