Difference between revisions of "2023 AMC 8 Problems/Problem 23"
(→Solution 3) |
(→Solution 1) |
||
Line 42: | Line 42: | ||
There are 9 squares and 4 possible designs for each square, giving <math>4^9</math> total outcomes. Thus, our desired probability is <math>\dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \boxed{\text{(C)} \hspace{0.1 in} \dfrac{1}{64}}</math> . | There are 9 squares and 4 possible designs for each square, giving <math>4^9</math> total outcomes. Thus, our desired probability is <math>\dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \boxed{\text{(C)} \hspace{0.1 in} \dfrac{1}{64}}</math> . | ||
− | -apex304 and TaeKim | + | -apex304 and TaeKim |
==Solution 2 (Linearity of Expectation)== | ==Solution 2 (Linearity of Expectation)== |
Revision as of 16:30, 25 January 2023
Contents
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
Probability is total favorable outcomes over total outcomes, so we can find these separately to determine the answer.
There are ways to choose the big diamond location from our
square grid. From our given problem there are
different arrangements of triangles for every square. This implies that from having
diamond we are going to have
distinct patterns outside of the diamond. This gives a total of
favorable cases.
There are 9 squares and 4 possible designs for each square, giving total outcomes. Thus, our desired probability is
.
-apex304 and TaeKim
Solution 2 (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
Solution 3
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
Video Solution 1 by OmegaLearn (Using Cool Probability Technique)
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=2405
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.