Difference between revisions of "2023 AMC 8 Problems/Problem 23"
(→Solution 3) |
Hsnacademy (talk | contribs) (→Video Solution) |
||
(42 intermediate revisions by 14 users not shown) | |||
Line 33: | Line 33: | ||
<math>\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}</math> | <math>\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}</math> | ||
+ | ==Solution 1== | ||
+ | There are <math>4</math> cases that the tiling will contain a large gray diamond in one of the smaller <math>2 \times 2</math> 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 <math>4^5</math> ways to decide the <math>5</math> white squares for each case, and the cases do not have any overlap. | ||
− | ==Solution | + | So, the requested probability is <cmath>\frac{4\cdot4^5}{4^9} = \frac{4^6}{4^9} = \frac{1}{4^3} = \boxed{\textbf{(C) } \frac{1}{64}}.</cmath> |
− | + | ~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 <math>\frac14 \cdot \frac14 \cdot \frac14 = \boxed{\textbf{(C) } \frac{1}{64}}</math>. | ||
− | + | ~aayr | |
+ | ==Solution 3== | ||
+ | Note that each tile must be in its precise place. Because of that, each diamond has a <math>\left(\frac14\right)^4</math> chance of appearing. And since there are 4 placements, our solution is<math>\frac{1}{4^4} \cdot 4 = \boxed{\textbf{(C) } \frac{1}{64}}</math>. | ||
− | + | ~ligonmathkid2 | |
− | |||
− | ==Solution | + | ==Solution 4 (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{E}[X_i] = (\tfrac{1}{4})^ | + | 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)} | + | <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 | ||
Line 52: | Line 82: | ||
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. <math>X_1 + X_2 + X_3 + X_4 \le 1</math>. | 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. <math>X_1 + X_2 + X_3 + X_4 \le 1</math>. | ||
+ | |||
~numerophile | ~numerophile | ||
− | ==Solution | + | ==Video Solution (Solve under 60 seconds!!!)== |
+ | https://youtu.be/6O5UXi-Jwv4?si=PZJnyfpeDaEY_K2i&t=1070 | ||
+ | |||
+ | ~hsnacademy | ||
+ | |||
+ | ==Video Solution by Math-X (Smart and Simple)== | ||
+ | https://youtu.be/Ku_c1YHnLt0?si=wz7F8E_c7Hxv9GQ5&t=5137 ~Math-X | ||
+ | |||
+ | ==Video Solution (THINKING CREATIVELY!!!)== | ||
− | + | https://youtu.be/B5qcEnobmEU | |
− | ~ | + | ~Education, the Study of Everything |
− | ==Video Solution | + | ==Video Solution== |
https://youtu.be/2t_Za0Y2IqY | https://youtu.be/2t_Za0Y2IqY | ||
==Animated Video Solution== | ==Animated Video Solution== | ||
− | https:// | + | https://www.youtube.com/watch?v=wq5Ie6mUxvY&ab_channel=StarLeague |
~Star League (https://starleague.us) | ~Star League (https://starleague.us) | ||
Line 70: | Line 109: | ||
==Video Solution by Magic Square== | ==Video Solution by Magic Square== | ||
https://youtu.be/-N46BeEKaCQ?t=2405 | https://youtu.be/-N46BeEKaCQ?t=2405 | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/DBqko2xATxs&t=3115 | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/uJklHght9ds | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by harungurcan== | ||
+ | https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1408s | ||
+ | |||
+ | ~harungurcan | ||
+ | |||
+ | == Video Solution== | ||
+ | |||
+ | https://youtu.be/YFLSh4WoZrg | ||
+ | ~please like and subscribe | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=22|num-a=24}} | {{AMC8 box|year=2023|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:52, 21 August 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (Linearity of Expectation)
- 6 Video Solution (Solve under 60 seconds!!!)
- 7 Video Solution by Math-X (Smart and Simple)
- 8 Video Solution (THINKING CREATIVELY!!!)
- 9 Video Solution
- 10 Animated Video Solution
- 11 Video Solution by Magic Square
- 12 Video Solution by Interstigation
- 13 Video Solution by WhyMath
- 14 Video Solution by harungurcan
- 15 Video Solution
- 16 See Also
Problem
Each square in a grid is randomly filled with one of the
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
grids? Below is an example of such tiling.
Solution 1
There are cases that the tiling will contain a large gray diamond in one of the smaller
grids, as shown below:
There are
ways to decide the
white squares for each case, and the cases do not have any overlap.
So, the requested probability is
~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 .
~aayr
Solution 3
Note that each tile must be in its precise place. Because of that, each diamond has a chance of appearing. And since there are 4 placements, our solution is
.
~ligonmathkid2
Solution 4 (Linearity of Expectation)
Let , and
denote the
smaller
squares within the
square in some order. For each
, let
if it contains a large gray diamond tiling and
otherwise. This means that
is the probability that square
has a large gray diamond, so
is our desired probability. However, since there is only one possible way to arrange the squares within every
square to form such a tiling, we have
for all
(as each of the smallest tiles has
possible arrangements), and from the linearity of expectation we get
~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. .
~numerophile
Video Solution (Solve under 60 seconds!!!)
https://youtu.be/6O5UXi-Jwv4?si=PZJnyfpeDaEY_K2i&t=1070
~hsnacademy
Video Solution by Math-X (Smart and Simple)
https://youtu.be/Ku_c1YHnLt0?si=wz7F8E_c7Hxv9GQ5&t=5137 ~Math-X
Video Solution (THINKING CREATIVELY!!!)
~Education, the Study of Everything
Video Solution
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/DBqko2xATxs&t=3115
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1408s
~harungurcan
Video Solution
https://youtu.be/YFLSh4WoZrg ~please like and subscribe
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.