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

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<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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==Video Solution==
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https://youtu.be/t6yjfKXpwDs
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~IceMatrix
  
 
==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}}

Revision as of 17:33, 7 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 10 Problems and Solutions

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