Question

A stick is broken at a point, chosen at random, along its length. Find the probability that the ratio, R, of the length of the shorter piece to the length of the longer piece is less than r, where r is a given positive number.

Answer 1

Solution :

Let the distance of the stick from one break be X

And let us assume that $X \leq l/2$.

Here, l = length of stick

Therefore, $P(X<x) =\frac{x}{l/2}$

We know that, $R=\frac{X}{l-X}$  ,  so by definition we get

$X =\frac{lR}{1+R}$

The cumulative distribution function for R is

$P(R<r) = P (\frac{X}{l-X}<r) = P(X<\frac{lr}{1+r})=\frac{lr}{(1+r)(l/2)}=\frac{2r}{1+r}$

When it starts at zero, then r =0. It ends at one when the r has a maximum

value of one.

The probability density function is given by

$\frac{d}{dr}(\frac{2r}{1+r})= \frac{2}{(r+1)^2}$

Now integrating, we find E(R) and $E(R)^2$ gives :

$E(R) =\int\limits^1_0 \frac{2r}{(1+r)^2} \, dr = 2 \ln 2-1$

$E(R)^2= 3 - 4\ln 2$

Therefore, Var(R)= $2-4(\ln \ 2)^2$