Difference between revisions of "2005 AMC 12B Problems/Problem 11"
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| − | {{ | + | {{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #11]] and [[2005 AMC 10B Problems|2005 AMC 10B #15]]}} |
== Problem == | == Problem == | ||
| − | {{ | + | An envelope contains eight bills: <math>2</math> ones, <math>2</math> fives, <math>2</math> tens, and <math>2</math> twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $<math>20</math> or more? |
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| + | <math>\mathrm{(A)}\ {{{\frac{1}{4}}}} \qquad \mathrm{(B)}\ {{{\frac{2}{7}}}} \qquad \mathrm{(C)}\ {{{\frac{3}{7}}}} \qquad \mathrm{(D)}\ {{{\frac{1}{2}}}} \qquad \mathrm{(E)}\ {{{\frac{2}{3}}}}</math> | ||
== Solution == | == Solution == | ||
| − | + | The only way to get a total of $<math>20</math> or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of <math>\dbinom{8}{2}=\dfrac{8\times7}{2\times1}=28</math> ways to choose <math>2</math> bills out of <math>8</math>. There are <math>12</math> ways to choose a twenty and some other non-twenty bill. There is <math>1</math> way to choose both twenties, and also <math>1</math> way to choose both tens. Adding these up, we find that there are a total of <math>14</math> ways to attain a sum of <math>20</math> or greater, so there is a total probability of <math>\dfrac{14}{28}=\boxed{\mathrm{(D)}\ \dfrac{1}{2}}</math>. | |
== See also == | == See also == | ||
| − | + | {{AMC10 box|year=2005|ab=B|num-b=14|num-a=16}} | |
| + | {{AMC12 box|year=2005|ab=B|num-b=10|num-a=12}} | ||
Revision as of 17:07, 16 July 2011
- 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
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 
.
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 | |
