Difference between revisions of "2004 AMC 10A Problems/Problem 5"
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<math>\dfrac{8}{\binom{9}{3}}=\dfrac{8}{84}=\dfrac{2}{21} \Rightarrow\boxed{\mathrm{(C)}\ \frac{2}{21}}</math> | <math>\dfrac{8}{\binom{9}{3}}=\dfrac{8}{84}=\dfrac{2}{21} \Rightarrow\boxed{\mathrm{(C)}\ \frac{2}{21}}</math> | ||
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+ | ==Comment== | ||
+ | This is an exact replica of Mathcounts 1994 State Target #7. | ||
== See also == | == See also == |
Revision as of 10:24, 22 April 2017
Contents
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?
![AMC10 2004A 4.gif](https://wiki-images.artofproblemsolving.com//d/d9/AMC10_2004A_4.gif)
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.
Comment
This is an exact replica of Mathcounts 1994 State Target #7.
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.