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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?2011-01-04
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    Yes, I am sure.2011-01-04
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    The answer depends on the bivariate distribution of points. Is it uniform?2011-01-04
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    Yes, they are uniformly distributed. 3 points each in 3 arrays.2011-01-04
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    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?2011-01-04
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    It is 3 times 3 matrix.2011-01-04

1 Answers 1

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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 ?2011-01-04
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    How did you get 84?2011-01-04