# Probability of obtaining triangle when choosing $3$ points from $3\times3$ array

$9$ 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? – Andrés E. 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

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 $\frac{84-8}{84}=\frac{19}{21}$.