Difference between revisions of "2005 AMC 12B Problems/Problem 11"
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==Solution 3== | ==Solution 3== | ||
| − | <math>P(\text{Get at least one twenty}) = 1- \frac{\binom{6}{2}}{\binom{8}{2}} | + | <math>P(\text{Get at least one twenty}) = 1- \frac{\binom{6}{2}}{\binom{8}{2}}=\frac{28-15}{28}=\frac{13}{28}</math> |
== See also == | == See also == | ||
Revision as of 11:58, 22 December 2019
- The following problem is from both the 2005 AMC 12B #11 and 2005 AMC 10B #15, so both problems redirect to this page.
Problem
An envelope contains eight bills: 
ones, fives, tens, and twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $
or more?
Solution 1
The only way to get a total of $ or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of 
ways to choose bills out of 
. There are 
ways to choose a twenty and some other non-twenty bill. There is 
way to choose both twenties, and also way to choose both tens. Adding these up, we find that there are a total of 
ways to attain a sum of or greater, so there is a total probability of 
.
Solution 2
Another way to do this problem is to use complementary counting, i.e. how many ways that the sum is less than 20. Now, you do not have to consider the 2 twenties, so you have 6 bills left. 
ways. However, you counted the case when you have 2 tens, so you need to subtract 1, and you get 14. Finding the ways to get 20 or higher, you subtract 14 from 28 and get 14. So the answer is 
.
Solution 3
See also
| 2005 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
| 2005 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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. 
