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Difference between revisions of "2023 AMC 12A Problems/Problem 5"

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If we roll a 1 on the first, the rolls that follow must be 2 or {1,1}, with the following results not mattering. This leaves a probability of <math>\frac{1}{6}\times\frac{1}{6} + \frac{1}{6}\times\frac{1}{6}\times\frac{1}{6} = \frac{7}{216}</math>.
 
If we roll a 1 on the first, the rolls that follow must be 2 or {1,1}, with the following results not mattering. This leaves a probability of <math>\frac{1}{6}\times\frac{1}{6} + \frac{1}{6}\times\frac{1}{6}\times\frac{1}{6} = \frac{7}{216}</math>.
  
−
If we roll a 2 on the first, the roll that follows must be 2, resulting in a probability of <math>\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}</math>.
+
If we roll a 2 on the first, the roll that follows must be 1, resulting in a probability of <math>\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}</math>.
  
−
If we roll a 3 on the first, the rolls that follow do not matter, resulting in a probability of <math>\frac{1}{6}</math>.
+
If we roll a 3 on the first, the following rolls do not matter, resulting in a probability of <math>\frac{1}{6}</math>.
 
Any roll greater than three will result in a running total greater than 3 no matter what, so those cases can be ignored.
 
Any roll greater than three will result in a running total greater than 3 no matter what, so those cases can be ignored.
−
Summing the answers, we have <math>\frac{7}{36} + \frac{1}{216} + \frac{1}{36} = \frac{42+1+6}{216} = \boxed{\textbf{(B) }\frac{49}{216}}</math>.
+
Summing the answers, we have <math>\frac{7}{216} + \frac{1}{36} + \frac{1}{36} = \frac{42+1+6}{216} = \boxed{\textbf{(B) }\frac{49}{216}}</math>.
  
 
~Failure.net
 
~Failure.net

Revision as of 14:35, 10 November 2023

The following problem is from both the 2023 AMC 10A #7 and 2023 AMC 12A #5, so both problems redirect to this page.

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?

$\textbf{(A) }\frac{2}{9}\qquad\textbf{(B) }\frac{49}{216}\qquad\textbf{(C) }\frac{25}{108}\qquad\textbf{(D) }\frac{17}{72}\qquad\textbf{(E) }\frac{13}{54}$


Solution 1 (Casework)

There are $3$ cases where the running total will equal $3$; one roll; two rolls; or three rolls:

Case 1: The chance of rolling a running total of $3$ in one roll is $\frac{1}{6}$.

Case 2: The chance of rolling a running total of $3$ in two rolls is $\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}$ since the dice rolls are a 2 and a 1 and vice versa.

Case 3: The chance of rolling a running total of 3 in three rolls is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}$ since the dice values would have to be three ones.

Using the rule of sum, $\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}$.

~walmartbrian ~andyluo ~DRBStudent

Solution 2 (Brute Force)

Because there is only a maximum of 3 rolls we must count (running total = 3 means there can't be a fourth roll counted), we can simply list out all of the probabilities.

If we roll a 1 on the first, the rolls that follow must be 2 or {1,1}, with the following results not mattering. This leaves a probability of $\frac{1}{6}\times\frac{1}{6} + \frac{1}{6}\times\frac{1}{6}\times\frac{1}{6} = \frac{7}{216}$.

If we roll a 2 on the first, the roll that follows must be 1, resulting in a probability of $\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}$.

If we roll a 3 on the first, the following rolls do not matter, resulting in a probability of $\frac{1}{6}$. Any roll greater than three will result in a running total greater than 3 no matter what, so those cases can be ignored. Summing the answers, we have $\frac{7}{216} + \frac{1}{36} + \frac{1}{36} = \frac{42+1+6}{216} = \boxed{\textbf{(B) }\frac{49}{216}}$.

~Failure.net

Solution 3

Consider sequences of 4 integers with each integer between 1 and 6, the number of permutations of 6 numbers is $6^4=1296$.

The following 4 types of sequences that might generate a running total of the numbers to be equal to 3 (x, y, or z denotes any integer between 1 and 6).

Sequence #1, (1, 1, 1, x): there are $6$ possible sequences.

Sequence #2, (1, 2, x, y): there are $6^2 = 36$ possible sequences.

Sequence #3, (2, 1, x, y): there are $6^2 = 36$ possible sequences.

Sequence #4, (3, x, y, z): there are $6^3 = 216$ possible sequences.

Out of 1296 possible sequences, there are a total of $6 + 36 + 36 + 216 = 294$ sequences that qualify. Hence, the probability is $294 / 1296 = \boxed{\textbf{(B) }\frac{49}{216}}$

~sqroot


Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/cMgngeSmFCY?si=_Q5YRd-S4B1TH79F&t=1426 ~Math-X



See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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