A 10-cm stick has a mark at each centimeter. By breaking the stick at two of these nine marks at random, the stick is split into three pieces, each of integer length. What is the probability that the three lengths could be the three side lengths of a triangles?

Total count is 9C2=36; but I don't know how to go with using the triangle inequality to count the valid ones for a triangle. Would appreciate any help.

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    $\begingroup$ $36$ cases can be done by hand $\endgroup$ – Vasya Mar 3 at 0:53

If one piece has length $5$ or greater, you can’t form a triangle. On the other hand, at least one piece has length $4$ or greater. Thus the admissible combinations are exactly the ones where the maximal length is $4$. But then the lengths must be $3,3,4$ or $2,4,4$, each in one of $3$ orders. Thus the probability is $\frac6{\binom92}=\frac16$.

Note what a difference the discreteness makes – for the same problem with a continuous uniform distribution of the marks, the probability is $\frac14$ (see this question on MO).

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