Difference between revisions of "2005 AMC 12B Problems/Problem 11"

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<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 ==
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== Solution 1==
 
The only way to get a total of &#36;<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 &#36;<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>.
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== Solution 2==
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TBE
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== 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.

Problem

An envelope contains eight bills: $2$ ones, $2$ fives, $2$ tens, and $2$ twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $$20$ or more?

$\mathrm{(A)}\ {{{\frac{1}{4}}}} \qquad \mathrm{(B)}\ {{{\frac{2}{5}}}} \qquad \mathrm{(C)}\ {{{\frac{3}{7}}}} \qquad \mathrm{(D)}\ {{{\frac{1}{2}}}} \qquad \mathrm{(E)}\ {{{\frac{2}{3}}}}$

Solution 1

The only way to get a total of $$20$ or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of $\dbinom{8}{2}=\dfrac{8\times7}{2\times1}=28$ ways to choose $2$ bills out of $8$. There are $12$ ways to choose a twenty and some other non-twenty bill. There is $1$ way to choose both twenties, and also $1$ way to choose both tens. Adding these up, we find that there are a total of $14$ ways to attain a sum of $20$ or greater, so there is a total probability of $\dfrac{14}{28}=\boxed{\mathrm{(D)}\ \dfrac{1}{2}}$.

Solution 2

TBE

See also

2005 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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