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| − | Janet rolls a standard 6-sided die 4 times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal 3?
| + | #redirect[[2023 AMC 12A Problems/Problem 5]] |
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| − | <math>\textbf{(A) }\frac{2}{9}\qquad\textbf{(B) }\frac{49}{216}\qquad\textbf{(C) }\frac{25}{108}\qquad\textbf{(D) }\frac{17}{72}\qquad\textbf{(E) }\frac{13}{54}</math>
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| − | Solution 1
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| − | There are 3 cases where the running total will equal 3; one roll; two rolls; or three rolls:
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| − | Case 1:
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| − | The chance of rolling a running total of <math>3</math> in one roll is <math>1/6</math>.
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| − | Case 2:
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| − | The chance of rolling a running total of <math>3</math> in two rolls is <math>1/6 * 1/6 * 2</math> since the dice rolls are a 2 and a 1 and vice versa.
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| − | Case 3:
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| − | The chance of rolling a running total of 3 in three rolls is <math>1/6 * 1/6 * 1/6</math> since the dice values would have to be three ones.
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| − | Using the rule of sum, <math>1/6 + 1/18 + 1/216 = 49/216</math> <math>\boxed{(B)}</math>.
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| − | ~walmartbrian
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