Difference between revisions of "2004 AMC 10A Problems/Problem 5"

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==Video Solution==
 
https://youtu.be/jWqX7ruQwr0
 
 
Education, the Study of Everything
 
 
 
 
 
==Problem==
 
==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?
 
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?
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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
 +
 +
Education, the Study of Everything
 +
  
 
== See also ==
 
== See also ==

Revision as of 14:14, 21 April 2021

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

$\mathrm{(A) \ } \frac{1}{21} \qquad \mathrm{(B) \ } \frac{1}{14} \qquad \mathrm{(C) \ } \frac{2}{21} \qquad \mathrm{(D) \ } \frac{1}{7} \qquad \mathrm{(E) \ } \frac{2}{7}$

Solution

There are $\binom{9}{3}$ 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.

$\dfrac{8}{\binom{9}{3}}=\dfrac{8}{84}=\dfrac{2}{21} \Rightarrow\boxed{\mathrm{(C)}\ \frac{2}{21}}$

Video Solution

https://youtu.be/jWqX7ruQwr0

Education, the Study of Everything


See also

2004 AMC 10A (ProblemsAnswer KeyResources)
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

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