Difference between revisions of "2005 AMC 12B Problems/Problem 11"
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== Solution 2== | == 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. <math>\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15</math> 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 <math>\dfrac{14}{28} = \boxed{\mathrm{(D)}\ \dfrac{1}{2}}</math>. | 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. <math>\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15</math> 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 <math>\dfrac{14}{28} = \boxed{\mathrm{(D)}\ \dfrac{1}{2}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | There are two cases that work, namely getting at least <math>1</math> twenty, or getting <math>2</math> tens. | ||
+ | |||
+ | Case <math>1</math>: <math>P(\text{Get at least one twenty}) = 1-P(\text{Do not get a single twenty})=1- \frac{\binom{6}{2}}{\binom{8}{2}}=\frac{28-15}{28}=\frac{13}{28}</math> | ||
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+ | 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{\text{(D)} \dfrac{1}{2}}</math> | ||
+ | |||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/7EOwpzC9C74 | ||
+ | |||
+ | ~savannahsolver | ||
== See also == | == See also == |
Latest revision as of 22:01, 7 May 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 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
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.