Difference between revisions of "2023 AMC 12A Problems/Problem 5"
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~Failure.net | ~Failure.net | ||
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| + | ==Solution 3 (same as ~walmartbrian ~andyluo ~DRBStudent's first solution with slightly different explanation) == | ||
| + | * Can we get a 3 in 1 roll? Sure but only if we get a 3, which has probability 1/6 | ||
| + | * Can we get a 3 in 2 rolls? There are 36 (6x6) possible rolls, 2 of which (1 and 2, 2 and 1) make 3, so probability is 2/36 = 1/18 | ||
| + | * Can we get a 3 in 3 rolls? There's only 1 way - to get 3 1s, so probability is (1/6)**3 | ||
| + | * Can we get a 3 in 4 rolls? No - the minimum roll would be 1+1+1+1 = 4 | ||
| + | |||
| + | So our answer is 1/6 (1 + 1/3 + 1/36) = 1/216 (36+12+1) = <math>\boxed{\textbf{(B) 49/216}}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 02:01, 10 November 2023
Contents
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
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 3 (same as ~walmartbrian ~andyluo ~DRBStudent's first solution with slightly different explanation)
- Can we get a 3 in 1 roll? Sure but only if we get a 3, which has probability 1/6
- Can we get a 3 in 2 rolls? There are 36 (6x6) possible rolls, 2 of which (1 and 2, 2 and 1) make 3, so probability is 2/36 = 1/18
- Can we get a 3 in 3 rolls? There's only 1 way - to get 3 1s, so probability is (1/6)**3
- Can we get a 3 in 4 rolls? No - the minimum roll would be 1+1+1+1 = 4
So our answer is 1/6 (1 + 1/3 + 1/36) = 1/216 (36+12+1) = 
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. 
