Puzzle : Form a triangle

Well,

This is a very easy puzzle for those who know a little bit of maths. This was asked at IIT Kanpur in PG admission test.

Problem : Given a stick of 1 meter, two points are chosen uniformly randomly to cut the stick. What is the probability that 3 pieces so generated would form a triangle?

Solution : The way I solved it is this. Clearly, to form a triangle it must satisfy triangle inequalities, that is if $a, b$ and $c$ are three sides then $a+b > c$, $b+c > a$ and $a+c>b$. Or if we look at it other way, none of the piece should be great or equal to 0.5 meter. So we need to answer, what is the probability that none of the piece measure greater or equal to 0.5?

Let $x < 0.5$(because of symmetry) be the first point from left side where the stick is cut. It is clear that to ensure that none of the piece measures $\geq 0.5$ the second cut-point $y$ must be such that $0.5 > y > 0.5 + x$. So for the second cut point feasible region is of the length $x$.

So,

$2 \int^{0.5} _0 x dx = 2[x^2/2]^{0.5} _0 = 0.5^2 = 1/4$