Difference between revisions of "2023 AMC 12A Problems/Problem 7"

(Created page with "Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you ge...")
 
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Hey the solutions will be posted after the contest, most likely around a couple weeks afterwords. We are not going to leak the questions to you, best of luck and I hope you get a good score.
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==Problem==
  
-Jonathan Yu
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Janet rolls a standard <math>6</math>-sided die <math>4</math> times and keeps a running total of the numbers
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she rolls. What is the probability that at some point, her running total will equal <math>3</math>?
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==Solution 1==
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There are <math>4</math> cases where her running total can equal <math>3</math>:
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1. She rolled <math>1</math> for three times consecutively from the beginning. Probability: <math>frac{1}{6^3} = frac{1}{216}
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2.  She rolled a </math>1<math>, then </math>2<math>. Probability: </math>frac{1}{6^2} = frac{1}{36}
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3.  She rolled a <math>2</math>, then <math>1</math>. Probability: <math>frac{1}{6^2} = frac{1}{36}
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4.  She rolled a </math>3<math> at the beginning. Probability: </math>frac{1}{6}
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Add them together to get <math>\boxed{textbf{(B)} frac{49}{216}}</math>

Revision as of 20:56, 9 November 2023

Problem

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 $3$?

Solution 1

There are $4$ cases where her running total can equal $3$: 1. She rolled $1$ for three times consecutively from the beginning. Probability: $frac{1}{6^3} = frac{1}{216} 2.  She rolled a$1$, then$2$. Probability:$frac{1}{6^2} = frac{1}{36} 3. She rolled a $2$, then $1$. Probability: $frac{1}{6^2} = frac{1}{36} 4.  She rolled a$3$at the beginning. Probability:$frac{1}{6}

Add them together to get $\boxed{textbf{(B)} frac{49}{216}}$