Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 12A Problems/Problem 5"
Line 54: Line 54:
  
 
~sqroot
 
~sqroot
 +
 +
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/GP-DYudh5qU?si=xtdXIlJmV_1ivP3T&t=1507
 +
~Math-X
  
 
==Video Solution (easy to understand) by Power Solve==
 
==Video Solution (easy to understand) by Power Solve==
 
https://youtu.be/YXIH3UbLqK8?si=3p9Ap7UQ376Tfi1W&t=312
 
https://youtu.be/YXIH3UbLqK8?si=3p9Ap7UQ376Tfi1W&t=312
−
 
−
==Video Solution by Math-X (First understand the problem!!!)==
 
−
https://youtu.be/cMgngeSmFCY?si=_Q5YRd-S4B1TH79F&t=1426 ~Math-X
 
  
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==

Revision as of 06:46, 6 September 2024

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 $3$, namely $(3)$ in exactly one roll is $\frac{1}{6}$.

Case 2: The chance of rolling a running total of $3$ in exactly two rolls, namely $(1, 2)$ and $(2, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}$.

Case 3: The chance of rolling a running total of 3 in exactly three rolls, namely $(1, 1, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}$.

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

~walmartbrian ~andyluo ~DRBStudent ~MC_ADe

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 $\frac{7}{216} + \frac{1}{36} + \frac{1}{6} = \frac{7+6+36}{216} = \boxed{\textbf{(B) }\frac{49}{216}}$.

~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/GP-DYudh5qU?si=xtdXIlJmV_1ivP3T&t=1507 ~Math-X

Video Solution (easy to understand) by Power Solve

https://youtu.be/YXIH3UbLqK8?si=3p9Ap7UQ376Tfi1W&t=312

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=FyVQW-1z60w

Video Solution

https://youtu.be/bqLA67R_5Bk

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution (⚡ Just 3 min ⚡)

https://youtu.be/8G5FfjUkQkQ

~Education, the Study of Everything

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_5&oldid=226593"