$9$ 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?
Probability of obtaining triangle when choosing $3$ points from $3\times3$ array
8
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probability
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0You sure this is the question? – 2011-01-04
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0Yes, I am sure. – 2011-01-04
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0The answer depends on the bivariate distribution of points. Is it uniform? – 2011-01-04
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0Yes, they are uniformly distributed. 3 points each in 3 arrays. – 2011-01-04
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0ok 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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0It is 3 times 3 matrix. – 2011-01-04
1 Answers
26
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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0Does it meant that the answer is 19/27 ? – 2011-01-04
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0How did you get 84? – 2011-01-04