Length of broken stick, with a mark I can't seem to solve this problem. Any help is appreciated!
A person randomly put a mark on a specially designed stick of length L. The stick is designed
to be broken into two pieces randomly after 10 years of usage.
(a) What is the chance that the shorter broken piece contains the mark?
(b) What is the expected length of the broken piece that contains the mark?
(c) Given that the shorter piece contains the mark, what is the expected length of that piece?

Assume the stick has length 1. Everything below generalizes in the same way for length $L$.
a) 
Given the cut occurs at $x$, we have:
$$P(\mbox{in shorter})=\int_0^{1/2}x^2dx+\int_{1/2}^1(1-x)xdx=1/8$$
b)
It sounds like we no longer care which piece the mark ends up in. Again it may help to think of first cutting and then placing a mark at random. The chance of cutting at $x$ is $dx$. The chance that the mark ends up in $[0,x]$ is $xdx$, with corresponding length $x$ and similarly ending up in $[x,1]$ with chance $(1-x)dx$ and corresponding length $(1-x)$. Hence, conditioning on the cut location:
$$E[length]=\int_0^1[x^2+(1-x)^2]dx=2/3$$
c) you work this one out
