A stick has a length of $5$ units. The stick is then broken at two points, chosen at random. What is the probability that all three resulting pieces are shorter than $2$ units?
+600 I will rephrase it:
Given $x+y+z=5, x,y,z\ge 0$, find the probability that $x,y,z\le 2$.
Put $x=2-a, y=2-b, z=2-c$ Then put $a=t_1+\frac13$, etc.
It is now, $t_1+t_2+t_3=0$, $t_1, t_2, t_3\in [-10/3, 5/3]$, find probability that $t_i\in [-1/3, 5/3]$ for all $i$. (this is because $a,b,c\in [-3,2]$ and we want $a,b,c\in [0,2]$ given $a+b+c=1$, it's simply bounding).
Now we do casework:
Case 1: There are 2 negatives and 1 positive. WLOG $t_1\le t_2\le t_3$. Then $t_1+t_2=-t_3$. and $t_3$ is positive. So we need $t_1+t_2\ge -\frac53$ given that they are in $[-10/3, 0]$. We can graph this letting $t_1=x, t_2=y$. Find it here: https://www.desmos.com/calculator/32dlklyibx. We then note that the area of the blue region is $\left(\frac{10}{3}\right)^2$ and the area of the red region in the blue region is $\frac{1}{2}\left(\frac{5}{3}\right)^2$. Using the principle of geometric probability, the probability here is $\frac{1}{8}$.
Case 2: There is 1 positive and 2 negatives. This is the same as the last case (note this by letting $t_i=-\alpha_i$).
All other cases are negligible since they have a probability of $0$ of occurring.
The answer is then $\frac{1}{8}+\frac{1}{8}=\boxed{\frac{1}{4}}$.
Beautiful problem. +1
