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

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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

Revision as of 11:05, 17 February 2020

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

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

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

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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