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

(Solution 3 is incorrect. Cannot simply apply expected value arguments like that.)
(Solution 2)
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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>.
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==Solution 3==
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<math>P(\text{Get at least </math>1<math> twenty})</math> = <math>1-\frac{\binom{6}{2}}{\binom{8}{2}</math>=\frac{28-15}{28}=\frac{13}{28}$
  
 
== See also ==
 
== See also ==

Revision as of 12:56, 22 December 2019

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

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. $\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15$ 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 $\dfrac{14}{28} = \boxed{\mathrm{(D)}\ \dfrac{1}{2}}$.

Solution 3

$P(\text{Get at least$ (Error compiling LaTeX. ! File ended while scanning use of \text@.)1$twenty})$ (Error compiling LaTeX. ! Extra }, or forgotten $.) = $1-\frac{\binom{6}{2}}{\binom{8}{2}$ (Error compiling LaTeX. ! File ended while scanning use of \frac .)=\frac{28-15}{28}=\frac{13}{28}$

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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