Difference between revisions of "2005 AMC 12B Problems/Problem 11"
m (→Problem) |
(→Solution) |
||
Line 5: | Line 5: | ||
<math>\mathrm{(A)}\ {{{\frac{1}{4}}}} \qquad \mathrm{(B)}\ {{{\frac{2}{5}}}} \qquad \mathrm{(C)}\ {{{\frac{3}{7}}}} \qquad \mathrm{(D)}\ {{{\frac{1}{2}}}} \qquad \mathrm{(E)}\ {{{\frac{2}{3}}}}</math> | <math>\mathrm{(A)}\ {{{\frac{1}{4}}}} \qquad \mathrm{(B)}\ {{{\frac{2}{5}}}} \qquad \mathrm{(C)}\ {{{\frac{3}{7}}}} \qquad \mathrm{(D)}\ {{{\frac{1}{2}}}} \qquad \mathrm{(E)}\ {{{\frac{2}{3}}}}</math> | ||
− | == Solution == | + | == Solution 1== |
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>. | 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>. | ||
+ | |||
+ | == Solution 2== | ||
+ | TBE | ||
+ | |||
== See also == | == See also == | ||
{{AMC10 box|year=2005|ab=B|num-b=14|num-a=16}} | {{AMC10 box|year=2005|ab=B|num-b=14|num-a=16}} | ||
{{AMC12 box|year=2005|ab=B|num-b=10|num-a=12}} | {{AMC12 box|year=2005|ab=B|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:49, 6 January 2016
- The following problem is from both the 2005 AMC 12B #11 and 2005 AMC 10B #15, so both problems redirect to this page.
Contents
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
TBE
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.