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| − | ==Problem==
| + | #redirect[[2023 AMC 12A Problems/Problem 5]] |
| − | 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?
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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 ~andyluo
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| − | ==Solution 2 (Slightly different to Solution 1)==
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| − | There are 3 cases where the running total will equal 3.
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| − | Case 1: Rolling a one three times
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| − | Case 2: Rolling a one then a two
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| − | Case 3: Rolling a three immediately
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| − | The probability of Case 1 is <math>1/216</math>, the probability of Case 2 is (<math>1/36 * 2) = 1/18</math>, and the probability of Case 3 is <math>1/6</math>
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| − | Using the rule of sums, adding every case gives the answer <math>\boxed{(D)126}</math>
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| − | ~DRBStudent
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| − | ==See Also==
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| − | {{AMC10 box|year=2023|ab=A|num-b=6|num-a=8}}
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| − | {{MAA Notice}}
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