A point $X$ chosen at random on a line segment $AB$. Show that the
probability that the ratio of the length of the shorter segment to
that of the larger segment is less than $\frac{1}{3}$ is $\frac{1}{2}$
My input
I know the working of this problem and can arrive at exact answer easily with the following steps. But I am not aware of the reasoning behind a few steps. I just know the procedure. So help me out here.
$AB=l$ ; Assuming that $x\sim U(0,l)$
Case $1$.
When the point is chosen lies to the left of the midpoint we have:
$x$ be the length of $AX$ and $l-x$ be the length of $XB$ (shorter).
Case $2$.
When the point is chosen lies to the right of the midpoint we have:
$x$ be the length of $AX$ and $l-x$ be the length of $XB$ (larger).
$P\bigg(\dfrac{x}{l-x} < \dfrac{1}{3} \bigg) \cup P\bigg(\dfrac{l-x}{x} < \dfrac{1}{3} \bigg) $
$P(x<\dfrac{l}{4})+ P(x>\dfrac{3l}{4})$
Well solving this we get $\dfrac{1}{2}$
My doubts are:
In the beginning, I wrote $x\sim U(0,l)$ I am not getting why are we doing this in my head. I have this crammed that all the probability questions related to distances are done with the help of uniform distribution and that's why I just wrote that. I want to know the exact reasoning why we did that. Also in some questions related to stick problem(like breaking stick from middle etc), we assumed that $X \sim U(0,1)$. Why did we do that? Stick length can also be $2$ or $20$ may be. Someone clear this to me, please. Sorry for bad English.
Best Answer
When you pick a point at random, you give every point on the stick the same probability, hence the distribution is uniform.
As you noticed in probability problems with the stick, the actual length of the stick is irrelevant, so you may assume $l=1$ and work with $U(0,1)$ instead of $U(0,l$ or $U(a,b)$. The final result will be the same.