Difference between revisions of "2004 AMC 10A Problems/Problem 5"
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<math> \mathrm{(A) \ } \frac{1}{21} \qquad \mathrm{(B) \ } \frac{1}{14} \qquad \mathrm{(C) \ } \frac{2}{21} \qquad \mathrm{(D) \ } \frac{1}{7} \qquad \mathrm{(E) \ } \frac{2}{7} </math> | <math> \mathrm{(A) \ } \frac{1}{21} \qquad \mathrm{(B) \ } \frac{1}{14} \qquad \mathrm{(C) \ } \frac{2}{21} \qquad \mathrm{(D) \ } \frac{1}{7} \qquad \mathrm{(E) \ } \frac{2}{7} </math> | ||
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==Solution== | ==Solution== | ||
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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> | ||
| − | == | + | ==Video Solution== |
| − | + | https://youtu.be/jWqX7ruQwr0 | |
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| + | Education, the Study of Everything | ||
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== See also == | == See also == | ||
Latest revision as of 13:14, 21 April 2021
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?
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.
Video Solution
Education, the Study of Everything
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. 
