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 &#36;<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 &#36;<math>20</math> or more?
  
<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>
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<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 &#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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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{\mathrm{(D)}\ \dfrac{1}{2}}</math>.
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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==
  
If you think about the problem, it's the same as trying as trying to pick from one dollar bill, one five dollar bill, one ten dollar bill, and one twenty dollar bill and trying to get a single bill that is more than or equal to ten dollars. We simply have the half all the values because the expected value remains gets halved along with the probability of choosing the bills, hence being the same problem. In doing so, two of the four options listed above are greater than or equal to ten dollars, so the answer is <math>\boxed{\mathrm{(D)}\ \dfrac{1}{2}}</math>.
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There are two cases that work, namely getting at least <math>1</math> twenty, or getting <math>2</math> tens.
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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>
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Summing up our cases, we have <math>\frac{13}{28}+\frac{1}{28}=\frac{14}{28}=\boxed{\textbf{(D) } \dfrac{1}{2}}</math>
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==Solution 4==
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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).
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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>
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Adding these three probabilities gives us <math>\frac{1}{4}+\frac{3}{14}+\frac{1}{28}=\boxed{\textbf{(D) } \dfrac{1}{2}}</math>
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-Benedict T (countmath1) (edited by AMC_8)
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==Solution 5 (Quick if no time)==
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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>.
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Thus, we can easily see the solution is <math>\boxed{(D) \frac{1}{2}}</math>.
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==Video Solution by WhyMath==
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https://youtu.be/7EOwpzC9C74
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~savannahsolver
  
 
== See also ==
 
== See also ==

Latest revision as of 14:16, 23 June 2024

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?

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

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 $\dfrac{14}{28}=\boxed{\textbf{(D) }\frac{1}{2}}$.

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 $\dfrac{14}{28} = \boxed{\textbf{(D) }\dfrac{1}{2}}$

Solution 3

There are two cases that work, namely getting at least $1$ twenty, or getting $2$ tens.

Case $1$: $P(\text{Get at least one twenty}) = 1-P(\text{Do not get a single twenty})=1- \frac{\binom{6}{2}}{\binom{8}{2}}=\frac{28-15}{28}=\frac{13}{28}$

Case $2$ : $P(\text{Get two tens}) = \frac{1}{\binom{8}{2}} = \frac{1}{28}$

Summing up our cases, we have $\frac{13}{28}+\frac{1}{28}=\frac{14}{28}=\boxed{\textbf{(D) } \dfrac{1}{2}}$

Solution 4

Note that if a twenty is drawn, anything else that is drawn will create a total greater than $20$; The probability of a twenty being drawn first is $\frac{1}{4}.$ 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 $\frac{3}{4}\cdot\frac{2}{7}=\frac{3}{14}.$ Finally, we can draw two tens. The probability of this occuring is $\frac{1}{4}\cdot\frac{1}{7}=\frac{1}{28}.$

Adding these three probabilities gives us $\frac{1}{4}+\frac{3}{14}+\frac{1}{28}=\boxed{\textbf{(D) } \dfrac{1}{2}}$

-Benedict T (countmath1) (edited by AMC_8)

Solution 5 (Quick if no time)

We see that there are $2$ of each bill. We can simplify the question to only drawing $1$ bill out of $4$, and trying to draw $$ 10$. (Originally you need to draw $$ 20$, but the bill is halved, so you need to draw only $$ 10$.) We see that we have $1$ of each bill $$1$, $$5$, $$10$, and $$20$. Thus, we can easily see the solution is $\boxed{(D) \frac{1}{2}}$.

Video Solution by WhyMath

https://youtu.be/7EOwpzC9C74

~savannahsolver

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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