Difference between revisions of "2023 AMC 8 Problems/Problem 23"
(Ayo, get back to grinding stop trying to cheat. Become one with the Sigma :).) |
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| − | + | Probability is total favorable outcomes over total outcomes, so we can find these separately to determine the answer. | |
| − | < | + | There are <math>4</math> ways to choose the big diamond location from our <math>9</math> square grid. From our given problem there are <math>4</math> different arrangements of triangles for every square. This implies that from having <math>1</math> diamond we are going to have <math>4^5</math> distinct patterns outside of the diamond. This gives a total of <math>4\cdot 4^5 = 4^6</math> favorable cases. |
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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> . | |
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Revision as of 19:15, 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 
.
