Difference between revisions of "2004 AMC 10A Problems/Problem 5"
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There are <math>\binom{9}{3}</math> ways to choose three points out of the 9 there. There are 8 combinations of dots such that they lie in a straight line: three vertical, three horizontal, and the diagonals. | There are <math>\binom{9}{3}</math> ways to choose three points out of the 9 there. There are 8 combinations of dots such that they lie in a straight line: three vertical, three horizontal, and the diagonals. | ||
| − | <math>\dfrac{8}{\binom{9}{3}}=\dfrac{8}{84}=\dfrac{2}{21} \Rightarrow (C)</math> | + | <math>\dfrac{8}{\binom{9}{3}}=\dfrac{8}{84}=\dfrac{2}{21} \Rightarrow\boxed{\mathrm{(C)}\ \frac{2}{21}}</math> |
== See also == | == See also == | ||
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131315 AoPS topic] | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131315 AoPS topic] | ||
{{AMC10 box|year=2004|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2004|ab=A|num-b=4|num-a=6}} | ||
| + | {{MAA Notice}} | ||
Revision as of 21:16, 9 September 2017
Problem
A set of three points is randomly chosen from the grid shown. Each three point set has the same probability of being chosen. What is the probability that the points lie on the same straight line?
Solution
There are 
ways to choose three points out of the 9 there. There are 8 combinations of dots such that they lie in a straight line: three vertical, three horizontal, and the diagonals.
See also
| 2004 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
