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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?

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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.

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  • $\begingroup$ I really appreciate your input and I see your point. One thing though, bothers me. The probability would indeed be $0.25$ if the joint distribution is uniform. I would like to be a little more rigorous on that assumption, so would you help me out there? Or in this case, does the problem already imply that the probability density function on that triangle is uniform because the two points are "chosen randomly within the length L"? $\endgroup$ – hyg17 Sep 10 '19 at 0:34
  • $\begingroup$ Yes. The density function is uniform on the triangle. If we remove the constrain that $x_1 \lt x_2$, it is easier to understand now since the density function is uniform in the square now. We could partition the square into to part, one for $x_1 \lt x_2$ and another part for $x_1 \gt x_2$. $\endgroup$ – Zhaohui Du Sep 10 '19 at 2:27
  • $\begingroup$ Thank you! That was very helpful. $\endgroup$ – hyg17 Sep 10 '19 at 6:20

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