TITLE: Probability of obtaining triangle when choosing $3$ points from $3\times3$ array
QUESTION [8 upvotes]: $9$ 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?

REPLY [26 votes]: 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 $\frac{84-8}{84}=\frac{19}{21}$.