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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?

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user: Did you write down a solution for (c)? What is the answer you found? – Did Oct 30 '12 at 6:50
I don't have a solution for C yet. although it seems it will be a combination of questions (a) and (b). – user1172558 Oct 31 '12 at 4:05
This is surprising, given that you accepted an answer. – Did Oct 31 '12 at 6:10

a) The probability that the cut is on the shorter end is $P((X \leq Y \cap Y \leq 1/2)\cup (X \geq Y \cap Y \geq 1/2)) = $ by symmetry

$$ 2P((X \leq Y \leq 1/2)) = 2 * 1/8 = 1/4 $$

(draw the unit square here to see this)

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Assume the stick has length 1. Everything below generalizes in the same way for length $L$.


Given the cut occurs at $x$, we have:

$$P(\mbox{in shorter})=\int_0^{1/2}x^2dx+\int_{1/2}^1(1-x)xdx=1/8$$


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:


c) you work this one out

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I'm not convinced for a). If the cut is at 1/4 and the mark is at 3/4, then switching them will give you that the mark is on the longer cut both times – blitzer Oct 30 '12 at 6:01
@blitzer: Whoops. Good point! – Alex R. Oct 30 '12 at 6:02

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