2023 AMC 12A Problems/Problem 5
- The following problem is from both the 2023 AMC 10A #7 and 2023 AMC 12A #5, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Casework)
- 3 Solution 2 (Brute Force)
- 4 Solution 3
- 5 Video Solution (easy to understand) by Power Solve
- 6 Video Solution by Math-X (First understand the problem!!!)
- 7 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 8 Video Solution
- 9 Video Solution (⚡ Just 3 min ⚡)
- 10 See Also
Problem
Janet rolls a standard -sided die times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal ?
Solution 1 (Casework)
There are cases where the running total will equal : one roll; two rolls; or three rolls:
Case 1: The chance of rolling a running total of , namely in exactly one roll is .
Case 2: The chance of rolling a running total of in exactly two rolls, namely and is .
Case 3: The chance of rolling a running total of 3 in exactly three rolls, namely is .
Using the rule of sum we have .
~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 .
If we roll a 2 on the first, the roll that follows must be 1, resulting in a probability of .
If we roll a 3 on the first, the following rolls do not matter, resulting in a probability of . 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 .
~Failure.net
Solution 3
Consider sequences of 4 integers with each integer between 1 and 6, the number of permutations of 6 numbers is .
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 possible sequences.
Sequence #2, (1, 2, x, y): there are possible sequences.
Sequence #3, (2, 1, x, y): there are possible sequences.
Sequence #4, (3, x, y, z): there are possible sequences.
Out of 1296 possible sequences, there are a total of sequences that qualify. Hence, the probability is
~sqroot
Video Solution (easy to understand) by Power Solve
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🔥]
https://www.youtube.com/watch?v=FyVQW-1z60w
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution (⚡ Just 3 min ⚡)
~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.