A stick of length 1 is broken into 3 pieces in the following way:
- We chose random interior point on the stick and break it into two pieces
- After that we choose the longest of the two pieces, choose random point on it and break it again, getting 3 pieces total.
- The task is to find probability that it would be possible to construct a triangle using that pieces.
I've came up with the following idea:
Let $X_1, X_2$ be random variable, uniformly distributed across all points on the stick ($X_1, X_2\in [0,1]$). It is clear, that Probability space may be illustrated as follows (marked with yellow):
And basing on triangle inequality we may highlight areas on the graph, that would suit us in terms of constructing triangle:
So, we get that $|\Omega|=S_{yellow section}=\cfrac{3}{4}$ And probability we are searching for: $Pr(A)=\cfrac{S_{blue section}}{S_{yellow section}}=\cfrac{1}{3}$. So answer here is $\cfrac{1}{3}$, however my mate got very different result, including $log$. Can you help me with this task/find mistake in my solution?
Any response is welcome and would be appreciated a lot.
Your solution does not fulfill the random protocol described in the problem. The second break is not at a random point in $[0,1]$, but at a random point in the longer piece of the first break.
Choose the first breaking point at $x\in\,\bigl]{1\over2},1\bigr]$ and the second at $xy$ with $y\in\,\bigl]{1\over2},1\bigr]$. The variable $(x,y)$ is then uniformly distributed in the square $Q:=\bigl]{1\over2},1\bigr]^2$ of area ${1\over4}$. In this way we obtain three pieces of length $$\ell_1=xy,\quad \ell_2=x-xy=x(1-y),\quad \ell_3=1-x\ .$$ The $\ell_i$ can be the sides of a triangle iff all of them are $<{1\over2}$. For $\ell_2$ and $\ell_3$ this is automatically the case, and we are left with the condition $\ell_1=xy<{1\over2}$. Let $S$ be the corresponding part of $Q$. The required probability then is $$p={{\rm area}(S)\over{\rm area}(Q)}=4\int_{1/2}^1\left({1\over2x}-{1\over2}\right)dx=\ldots=2\log2-1\approx0.3863\ .$$