Difference between revisions of "2020 AMC 10B Problems/Problem 11"
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<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> | ||
| − | ==Solution== | + | ==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. | ||
Revision as of 20:06, 17 January 2021
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?
Solution 1
We don't care about which books Harold selects. We just care that Betty picks 
books from Harold's list and 
that aren't on Harold's list.
The total amount of combinations of books that Betty can select is 
.
There are 
ways for Betty to choose of the books that are on Harold's list.
From the remaining 
books that aren't on Harold's list, there are 
ways to choose of them.
~quacker88
Solution 2
We can analyze this as two containers with 
balls each, with the two people grabbing balls each. First, we need to find the probability of two of the balls being the same among five: 
. After that we must, multiply this probability by 
, for the 2 balls that are the same are chosen among 5 balls. The answer will be 
. 
Video Solution
~IceMatrix
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
