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Difference between revisions of "2005 AMC 12B Problems/Problem 11"
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-Benedict T (countmath1) (edited by AMC_8)
 
-Benedict T (countmath1) (edited by AMC_8)
  
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==Solution 5==
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==Solution 5 (Quick)==
  
 
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>.
 
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{(C) \frac{1}{2}}</math>.
 
Thus, we can easily see the solution is <math>\boxed{(C) \frac{1}{2}}</math>.
 +
 
==Video Solution by WhyMath==
 
==Video Solution by WhyMath==
 
https://youtu.be/7EOwpzC9C74
 
https://youtu.be/7EOwpzC9C74

Revision as of 19:52, 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.

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)

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 $\boxed{(C) \frac{1}{2}}$.

Video Solution by WhyMath

https://youtu.be/7EOwpzC9C74

~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

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