Difference between revisions of "2020 AMC 10B Problems/Problem 11"
m |
m (→Solution) |
||
| Line 9: | Line 9: | ||
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. | ||
| − | The total amount of combinations of books that Betty can select is <math>\binom{ | + | The total amount of combinations of books that Betty can select is <math>\binom{5}{10}=252</math>. |
| − | There are <math>\binom{ | + | 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. |
| − | From the remaining <math>5</math> books that aren't on Harold's list, there are <math>\binom{ | + | 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. |
<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 17:18, 16 February 2020
Contents
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
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
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
