Difference between revisions of "2005 AMC 12B Problems/Problem 11"
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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? | 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? | ||
| − | <math>\ | + | <math>\textbf{(A) }\ {{{\frac{1}{4}}}} \qquad \textbf{(B) }\ {{{\frac{2}{5}}}} \qquad \textbf{(C) }\ {{{\frac{3}{7}}}} \qquad \textbf{(D) }\ {{{\frac{1}{2}}}} \qquad \textbf{(E) }\ {{{\frac{2}{3}}}}</math> |
== Solution 1== | == 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{\ | + | 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{\textbf{(D) }\frac{1}{2}}</math>. |
== 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{\ | + | Another way to do this problem is to use [[complementary counting]], i.e. how many ways that the sum is less than <math>20</math>. Now, you do not have to consider the <math>2</math> twenties, so you have <math>6</math> bills left. <math>\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15</math> ways. However, you counted the case when you have <math>2</math> tens, so you need to subtract 1, and you get <math>14</math>. Finding the ways to get <math>20</math> or higher, you subtract <math>14</math> from <math>28</math> and get <math>14</math>. So the answer is <math>\dfrac{14}{28} = \boxed{\textbf{(D) }\dfrac{1}{2}}</math> |
==Solution 3== | ==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> | ||
| + | |||
| + | 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{\textbf{(D) } \dfrac{1}{2}}</math> | ||
| + | |||
| + | ==Solution 4== | ||
| + | |||
| + | Note that if a twenty is drawn, anything else that is drawn will create a total greater than <math>20</math>; The probability of a twenty being drawn first is <math>\frac{1}{4}.</math> The same could be said for drawing anything, and then drawing a twenty. However, we can only draw something that isn't a twenty first (since we've already accounted for the probability of drawing two twenties). | ||
| + | |||
| + | The probability of drawing a non-twenty first, then a twenty second is <math>\frac{3}{4}\cdot\frac{2}{7}=\frac{3}{14}.</math> Finally, we can draw two tens. The probability of this occuring is <math>\frac{1}{4}\cdot\frac{1}{7}=\frac{1}{28}.</math> | ||
| + | |||
| + | Adding these three probabilities gives us <math>\frac{1}{4}+\frac{3}{14}+\frac{1}{28}=\boxed{\textbf{(D) } \dfrac{1}{2}}</math> | ||
| + | |||
| + | -Benedict T (countmath1) (edited by AMC_8) | ||
| + | |||
| + | ==Solution 5 (Quick if no time)== | ||
| + | |||
| + | We see that there are <math>2</math> of each bill. We can simplify the question to only drawing <math>1</math> bill out of <math>4</math>, and trying to draw <math>\$ 10</math>. (Originally you need to draw <math>\$ 20</math>, but the bill is halved, so you need to draw only <math> \$ 10</math>.) We see that we have <math>1</math> of each bill <math>\$1</math>, <math>\$5</math>, <math>\$10</math>, and <math>\$20</math>. | ||
| + | Thus, we can easily see the solution is <math>\boxed{(D) \frac{1}{2}}</math>. | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/7EOwpzC9C74 | ||
| + | |||
| + | ~savannahsolver | ||
== See also == | == See also == | ||
Latest revision as of 21:01, 20 May 2023
- 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
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 
Solution 4
Note that if a twenty is drawn, anything else that is drawn will create a total greater than ; The probability of a twenty being drawn first is 
The same could be said for drawing anything, and then drawing a twenty. However, we can only draw something that isn't a twenty first (since we've already accounted for the probability of drawing two twenties).
The probability of drawing a non-twenty first, then a twenty second is 
Finally, we can draw two tens. The probability of this occuring is 
Adding these three probabilities gives us 
-Benedict T (countmath1) (edited by AMC_8)
Solution 5 (Quick if no time)
We see that there are of each bill. We can simplify the question to only drawing bill out of 
, and trying to draw 
. (Originally you need to draw 
, but the bill is halved, so you need to draw only .) We see that we have of each bill 
, 
, 
, and 
.
Thus, we can easily see the solution is 
.
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. 
