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

(Video Solution)
Line 21: Line 21:
 
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}\frac{4}{9}\frac{3}{8}\frac{5}{7}\frac{2}{4}</math>. After that we must, multiply this probability by <math>{5 \choose 2}</math>, for the 2 balls that are the same are chosen among 5 balls. The answer will be <math>\frac{5}{126}*10 = \frac{25}{63}</math>. <math>\boxed{(D) \frac{25}{63}}</math>
 
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}\frac{4}{9}\frac{3}{8}\frac{5}{7}\frac{2}{4}</math>. After that we must, multiply this probability by <math>{5 \choose 2}</math>, for the 2 balls that are the same are chosen among 5 balls. The answer will be <math>\frac{5}{126}*10 = \frac{25}{63}</math>. <math>\boxed{(D) \frac{25}{63}}</math>
  
 +
==Solution 3==
 +
There are \binom{10}{5} ways for Harold to select 5 books. Of the 5 books Harold selected, Betty wants to select 2 of them. Thus, we have \binom{5}{2} ways for Betty to select 2 books that harold selected. Additionally, we want betty to pick another 3 books out of the remaining 5 books Harold has not selected \binom{5}{3}. The total amount of ways harold and betty can each select 5 books is \binom{10}{5}. Thus, we have \frac{\left (\binom{10}{5}\cdot \binom{5}{2}\cdot \binom{5}{3}  \right )}{\binom{10}{5}\cdot \binom{10}{5}} This is equal to \boxed{\textbf{(D) }\frac{25}{63}}$
 
==Video Solution==
 
==Video Solution==
 
https://youtu.be/t6yjfKXpwDs
 
https://youtu.be/t6yjfKXpwDs

Revision as of 16:58, 22 December 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

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 $\frac{5}{126}*10 = \frac{25}{63}$. $\boxed{(D) \frac{25}{63}}$

Solution 3

There are \binom{10}{5} ways for Harold to select 5 books. Of the 5 books Harold selected, Betty wants to select 2 of them. Thus, we have \binom{5}{2} ways for Betty to select 2 books that harold selected. Additionally, we want betty to pick another 3 books out of the remaining 5 books Harold has not selected \binom{5}{3}. The total amount of ways harold and betty can each select 5 books is \binom{10}{5}. Thus, we have \frac{\left (\binom{10}{5}\cdot \binom{5}{2}\cdot \binom{5}{3} \right )}{\binom{10}{5}\cdot \binom{10}{5}} This is equal to \boxed{\textbf{(D) }\frac{25}{63}}$

Video Solution

https://youtu.be/t6yjfKXpwDs

~IceMatrix

https://youtu.be/RbKdVmZRxkk

~savannahsolver

See Also

2020 AMC 10B (ProblemsAnswer KeyResources)
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