Difference between revisions of "2005 AMC 12B Problems/Problem 11"
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==Solution 3== | ==Solution 3== | ||
| − | If you think about the problem, it's the same as trying as trying to pick from one dollar bill, one five dollar bill, one ten dollar bill, and one twenty dollar bill and trying to get a single bill that is more than or equal to ten dollars. We simply have the half all the values because the expected value | + | If you think about the problem, it's the same as trying as trying to pick from one dollar bill, one five dollar bill, one ten dollar bill, and one twenty dollar bill and trying to get a single bill that is more than or equal to ten dollars. We simply have the half all the values because the expected value gets halved along with the probability of choosing the bills, hence being the same problem. In doing so, two of the four options listed above are greater than or equal to ten dollars, so the answer is <math>\boxed{\mathrm{(D)}\ \dfrac{1}{2}}</math>. |
== See also == | == See also == | ||
Revision as of 16:55, 16 December 2017
- 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
If you think about the problem, it's the same as trying as trying to pick from one dollar bill, one five dollar bill, one ten dollar bill, and one twenty dollar bill and trying to get a single bill that is more than or equal to ten dollars. We simply have the half all the values because the expected value gets halved along with the probability of choosing the bills, hence being the same problem. In doing so, two of the four options listed above are greater than or equal to ten dollars, so the answer is 
.
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. 
