I am working on a probability problem and here is what I have so far.
A stick with fixed length $L$ is broken at two points randomly. Find the probability that the resulting three shorter sticks will form a triangle.
I let $X_1$ and $X_2$ be points on the stick , and without loss of generality I assume $0 < x_1 < x_2 < L$.
Because the points are chosen randomly I let $X_1 \sim U(0,L)$ and $X_2 | X_1 \sim U(x_1,L)$.
Under this assumption I know that
$$f_{X_1}(x_1) = \frac{1}{L}, \quad f_{X_2|X_1}(x_2)=\frac{1}{L-x_1}$$
and
$$f_{X_1,X_2}(x_1,x_2)=\frac{1}{L}\frac{1}{L-x_1}$$
I also want to use the triangle inequality which simplifies to
$$x_2 > \frac{L}{2} > x_2-x_1, \quad \frac{L}{2} > x_1$$
I know how to find these individual probabilities but I do not know how to combine them to get the answer. . .
May I have some assistance?
There's a problem in your model.
By assuming $0\lt x_1 \lt x_2 \lt L$, $X_1$ is not uniformly distributed on (0, L) now. It has a higher probability around 0 and it has a lower probability around L. We could use a 2D Cartesian coordinates. The region for $0\lt x_1 \lt x_2 \lt L$ is a half of an L-by-L square.
The condition to form a triangle is that the length of all three segments is less than $L/2$ so that,
1) $x_2>L/2$
2) $x_2-x_1<L/2$
3) $x_1<L/2$
Draw the picture of those inequality we could find that it forms another triangle whose area is 1/4 of the original one, so that the probability is 1/4.