Difference between revisions of "2005 AMC 10A Problems/Problem 9"
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There are <math>\frac{5!}{2!3!}=10</math> distinct arrangements of three <math>X</math>'s and two <math>O</math>'s. | There are <math>\frac{5!}{2!3!}=10</math> distinct arrangements of three <math>X</math>'s and two <math>O</math>'s. | ||
| − | There is only <math>1</math> distinct arrangement that reads <math>XOXOX</math> | + | There is only <math>1</math> distinct arrangement that reads <math>XOXOX</math>. |
| − | Therefore the desired [[probability]] is <math>\boxed{\frac{1}{10} | + | Therefore the desired [[probability]] is <math>\boxed{\textbf{(B) }\frac{1}{10}}</math> |
==See also== | ==See also== | ||
Revision as of 11:01, 13 December 2021
Problem
Three tiles are marked 
and two other tiles are marked 
. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads 
?
Solution
There are 
distinct arrangements of three 's and two 's.
There is only 
distinct arrangement that reads .
Therefore the desired probability is 
See also
| 2005 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
