Difference between revisions of "2023 AMC 12A Problems/Problem 5"
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There are <math>3</math> cases where the running total will equal <math>3</math>; one roll; two rolls; or three rolls: | There are <math>3</math> cases where the running total will equal <math>3</math>; one roll; two rolls; or three rolls: |
Revision as of 09:02, 10 November 2023
Contents
[hide]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 (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 in one roll is .
Case 2: The chance of rolling a running total of in two rolls is 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 since the dice values would have to be three ones.
Using the rule of sum, .
~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 .
If we roll a 2 on the first, the roll that follows must be 2, resulting in a probability of .
If we roll a 3 on the first, the rolls that follow 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 4
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
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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