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

(→‎Solution 1)
(→‎Solution 1)
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There are <math>4</math> cases where her running total can equal <math>3</math>:
 
There are <math>4</math> cases where her running total can equal <math>3</math>:
−
1. She rolled <math>1</math> for three times consecutively from the beginning. Probability: <math>frac{1}{6^3} = frac{1}{216}</math>
+
1. She rolled <math>1</math> for three times consecutively from the beginning. Probability: <math>\frac{1}{6^3} = \frac{1}{216}</math>
−
2.  She rolled a <math>1</math>, then <math>2</math>. Probability: <math>frac{1}{6^2} = frac{1}{36}</math>
+
2.  She rolled a <math>1</math>, then <math>2</math>. Probability: <math>\frac{1}{6^2} = \frac{1}{36}</math>
−
3.  She rolled a <math>2</math>, then <math>1</math>. Probability: <math>frac{1}{6^2} = frac{1}{36}</math>
+
3.  She rolled a <math>2</math>, then <math>1</math>. Probability: <math>\frac{1}{6^2} = \frac{1}{36}</math>
−
4.  She rolled a <math>3</math> at the beginning. Probability: <math>frac{1}{6}</math>
+
4.  She rolled a <math>3</math> at the beginning. Probability: <math>\frac{1}{6}</math>
  
 
Add them together to get <math>\boxed{\textbf{(B)} \frac{49}{216}}.</math>
 
Add them together to get <math>\boxed{\textbf{(B)} \frac{49}{216}}.</math>
  
 
~d_code
 
~d_code

Revision as of 20:58, 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 $\boxed{\textbf{(B)} \frac{49}{216}}.$

~d_code

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