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Difference between revisions of "2020 AMC 10B Problems/Problem 11"

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==Problem==
 
==Problem==
  
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Ms. Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?
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Ms. Carr asks her students to read any <math>5</math> of the <math>10</math> books on a reading list. Harold randomly selects <math>5</math> books from this list, and Betty does the same. What is the probability that there are exactly <math>2</math> books that they both select?
  
 
<math>\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}</math>
 
<math>\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}</math>
  
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==Solution==
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==Solution 1==
  
 
We don't care about which books Harold selects. We just care that Betty picks <math>2</math> books from Harold's list and <math>3</math> that aren't on Harold's list.
 
We don't care about which books Harold selects. We just care that Betty picks <math>2</math> books from Harold's list and <math>3</math> that aren't on Harold's list.
  
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The total amount of combinations of books that Betty can select is <math>\binom{5}{10}=252</math>.  
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The total amount of combinations of books that Betty can select is <math>\binom{10}{5}=252</math>.  
  
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There are <math>\binom{2}{5}=10</math> ways for Betty to choose <math>2</math> of the books that are on Harold's list.
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There are <math>\binom{5}{2}=10</math> ways for Betty to choose <math>2</math> of the books that are on Harold's list.
  
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From the remaining <math>5</math> books that aren't on Harold's list, there are <math>\binom{3}{5}=10</math> ways to choose <math>3</math> of them.
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From the remaining <math>5</math> books that aren't on Harold's list, there are <math>\binom{5}{3}=10</math> ways to choose <math>3</math> of them.
  
 
<math>\frac{10\cdot10}{252}=\boxed{\textbf{(D) }\frac{25}{63}}</math> ~quacker88
 
<math>\frac{10\cdot10}{252}=\boxed{\textbf{(D) }\frac{25}{63}}</math> ~quacker88
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 +
==Solution 2==
 +
 +
We can analyze this as two containers with <math>10</math> balls each, with the two people grabbing <math>5</math> balls each. First, we need to find the probability of two of the balls being the same among five: <math>\frac{1}{3} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{5}{7} \cdot \frac{2}{4}</math>. After that we must multiply this probability by <math>{5 \choose 2}</math>, for choosing the 2 balls that are the same chosen among
 +
5 balls. The answer will be <math>\frac{5}{126}*10 = \frac{25}{63}</math>. <math>\boxed{\textbf{(D) }\frac{25}{63}}</math>
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 +
==Solution 3==
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Firstly, we know that the denominator will be <math>\dbinom{10}{5} \cdot \dbinom{10}{5}</math>. To calculate the numerator, or successful events, we first find the number of ways both Betty and Harold can choose the same 2 books. Then we find the number of ways for Betty to choose 3 different other books and the number of ways for Harold to choose 3 different other books. That is <math>\dfrac{\binom{10}{2} \cdot \binom{8}{3} \cdot \binom{5}{3}}{\tbinom{10}{5} \cdot \tbinom{10}{5}} = \dfrac{45 \cdot 56 \cdot 10}{252 \cdot 252}</math>. From here, do not multiply, instead cancel common factors, then simplify. In fact, to make this expression even more manageable, leave the combinations in simplest factored form (for example, <math>\tbinom{10}{2}</math> is <math>5 \cdot 3 \cdot 3</math>). After doing so, we get, <math>\boxed{\textbf{(D) } \dfrac{25}{63}}</math>.
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~BakedPotato66
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==Solution 4==
 +
Say that there is a set <math>S</math> <math>{a,b,c,...,j}</math> which represent the 10 books. We know that intersection of the subsets that Bob and Alice choose has 2 elements. Thus there are <math>\dbinom{10}{2}</math> ways for that to happen. Out of the 8 remaining in the set <math>S</math>, Bob and Alice both need to choose 3 different elements = <math>\dbinom{8}{3} * \dbinom{5}{3}</math>. Thus the number of ways this works is <math>\dbinom{10}{2} * \dbinom{8}{3} * \dbinom{5}{3}</math>. The denominator for the fraction is just <math>\dbinom{10}{5} * \dbinom{10}{5}</math>. Simplifying, you get <math>\boxed{\frac{25}{63}}</math>.
 +
 +
~YBSuburbanTea
 +
 +
==Video Solution (HOW TO CRITICALLY THINK!!!)==
 +
https://youtu.be/wwKJBG5zCAE
 +
 +
~Education, the Study of Everything
 +
 +
 +
 +
 +
 +
== Video Solution ==
 +
https://youtu.be/wopflrvUN2c?t=118
 +
-Sohil Rathi
 +
~Juicer100
  
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/t6yjfKXpwDs
 
https://youtu.be/t6yjfKXpwDs
  
−
~IceMatrix
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 +
 
 +
https://youtu.be/RbKdVmZRxkk
 +
 
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2020|ab=B|num-b=10|num-a=12}}
 
{{AMC10 box|year=2020|ab=B|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:32, 31 August 2024

Problem

Ms. Carr asks her students to read any $5$ of the $10$ books on a reading list. Harold randomly selects $5$ books from this list, and Betty does the same. What is the probability that there are exactly $2$ books that they both select?

$\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}$

Solution 1

We don't care about which books Harold selects. We just care that Betty picks $2$ books from Harold's list and $3$ that aren't on Harold's list.

The total amount of combinations of books that Betty can select is $\binom{10}{5}=252$.

There are $\binom{5}{2}=10$ ways for Betty to choose $2$ of the books that are on Harold's list.

From the remaining $5$ books that aren't on Harold's list, there are $\binom{5}{3}=10$ ways to choose $3$ of them.

$\frac{10\cdot10}{252}=\boxed{\textbf{(D) }\frac{25}{63}}$ ~quacker88

Solution 2

We can analyze this as two containers with $10$ balls each, with the two people grabbing $5$ balls each. First, we need to find the probability of two of the balls being the same among five: $\frac{1}{3} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{5}{7} \cdot \frac{2}{4}$. After that we must multiply this probability by ${5 \choose 2}$, for choosing the 2 balls that are the same chosen among 5 balls. The answer will be $\frac{5}{126}*10 = \frac{25}{63}$. $\boxed{\textbf{(D) }\frac{25}{63}}$

Solution 3

Firstly, we know that the denominator will be $\dbinom{10}{5} \cdot \dbinom{10}{5}$. To calculate the numerator, or successful events, we first find the number of ways both Betty and Harold can choose the same 2 books. Then we find the number of ways for Betty to choose 3 different other books and the number of ways for Harold to choose 3 different other books. That is $\dfrac{\binom{10}{2} \cdot \binom{8}{3} \cdot \binom{5}{3}}{\tbinom{10}{5} \cdot \tbinom{10}{5}} = \dfrac{45 \cdot 56 \cdot 10}{252 \cdot 252}$. From here, do not multiply, instead cancel common factors, then simplify. In fact, to make this expression even more manageable, leave the combinations in simplest factored form (for example, $\tbinom{10}{2}$ is $5 \cdot 3 \cdot 3$). After doing so, we get, $\boxed{\textbf{(D) } \dfrac{25}{63}}$.

~BakedPotato66

Solution 4

Say that there is a set $S$ ${a,b,c,...,j}$ which represent the 10 books. We know that intersection of the subsets that Bob and Alice choose has 2 elements. Thus there are $\dbinom{10}{2}$ ways for that to happen. Out of the 8 remaining in the set $S$, Bob and Alice both need to choose 3 different elements = $\dbinom{8}{3} * \dbinom{5}{3}$. Thus the number of ways this works is $\dbinom{10}{2} * \dbinom{8}{3} * \dbinom{5}{3}$. The denominator for the fraction is just $\dbinom{10}{5} * \dbinom{10}{5}$. Simplifying, you get $\boxed{\frac{25}{63}}$.

~YBSuburbanTea

Video Solution (HOW TO CRITICALLY THINK!!!)

https://youtu.be/wwKJBG5zCAE

~Education, the Study of Everything



Video Solution

https://youtu.be/wopflrvUN2c?t=118 -Sohil Rathi ~Juicer100

Video Solution

https://youtu.be/t6yjfKXpwDs


https://youtu.be/RbKdVmZRxkk

~savannahsolver

See Also

2020 AMC 10B (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 10 Problems and Solutions

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