Difference between revisions of "2005 AMC 12B Problems/Problem 11"
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Case <math>2</math> : <math>P(\text{Get two tens}) = \frac{1}{\binom{8}{2}} = \frac{1}{28}</math> | Case <math>2</math> : <math>P(\text{Get two tens}) = \frac{1}{\binom{8}{2}} = \frac{1}{28}</math> | ||
| − | Summing up our cases, we have <math>\frac{13}{28}+\frac{1}{28}=\frac{14}{28}=\boxed{\ | + | Summing up our cases, we have <math>\frac{13}{28}+\frac{1}{28}=\frac{14}{28}=\boxed{\textbf{(D) } \dfrac{1}{2}}</math> |
==Video Solution by WhyMath== | ==Video Solution by WhyMath== | ||
Revision as of 07:34, 16 December 2021
- 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 . Now, you do not have to consider the twenties, so you have 
bills left. 
ways. However, you counted the case when you have tens, so you need to subtract 1, and you get . Finding the ways to get or higher, you subtract from 
and get . So the answer is 
Solution 3
There are two cases that work, namely getting at least twenty, or getting tens.
Case : 
Case : 
Summing up our cases, we have 
Video Solution by WhyMath
~savannahsolver
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. 
