Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2020 AMC 10B Problems/Problem 11

Revision as of 22:11, 17 January 2021 by Pi is 3.14 (talk | contribs) (→‎Video Solution)

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: $\frac{1}{3}\frac{4}{9}\frac{3}{8}\frac{5}{7}\frac{2}{4}$. After that we must, multiply this probability by ${5 \choose 2}$, for the 2 balls that are the same are chosen among 5 balls. The answer will be $\frac{5}{126}*10 = \frac{25}{63}$. $\boxed{(D) \frac{25}{63}}$


Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

https://youtu.be/RbKdVmZRxkk

~savannahsolver

Video Solution

https://youtu.be/wopflrvUN2c?t=118

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_11&oldid=142562"