Difference between revisions of "2023 AMC 8 Problems/Problem 23"
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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, SohumUttamchandani, wuwang2002, TaeKim. Cxrupptedpat | ||
Revision as of 18:15, 24 January 2023
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, SohumUttamchandani, wuwang2002, TaeKim. Cxrupptedpat
