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$9$ points are placed in a $3\times3$ array. If $3$ points are randomly selected, what is the probability that they are the vertices of a triangle?

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You sure this is the question? –  Andres Caicedo Jan 4 '11 at 2:53
    
Yes, I am sure. –  Ranga Jan 4 '11 at 2:54
    
The answer depends on the bivariate distribution of points. Is it uniform? –  mpiktas Jan 4 '11 at 2:59
    
Yes, they are uniformly distributed. 3 points each in 3 arrays. –  Ranga Jan 4 '11 at 3:06
    
ok I misunderstood the question. I thought the points are placed in a square $[0,1]^2$. Just to make sure by square array you mean square $n\times n$ matrix? What is $n$ then? –  mpiktas Jan 4 '11 at 3:15
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1 Answer

up vote 26 down vote accepted

Any $3$ points would form a triangle unless they are collinear. By considering horizontal, vertical and diagonal lines, we see that there are exactly $8$ cases of collinearity. Now there are $\binom{9}{3}=84$ ways to choose $3$ points out of $9$. Hence the probability is $\frac{84-8}{84}=\frac{19}{21}$.

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Does it meant that the answer is 19/27 ? –  Ranga Jan 4 '11 at 2:59
    
How did you get 84? –  Ranga Jan 4 '11 at 3:32
    
Thank you very much! –  Ranga Jan 4 '11 at 3:51
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