Probability and Straight line Two points are taken at random on the given straight line segment $AB$ of length $L$. Determine the probability that the length of none of the $3$ parts exceeds the given value $c$, where $L/3 \le c < L$.
I have tried dividing the line into segments for example, $AX = x$, the distance between $A$ and $X$.
and $XY = y$, the distance between the first and second arbitrary point and $YB = z$ the distance between $Y$ and the tip of the straight line.
Intuitively, I think the distance should be $1/3$ between them. But I don't know how to solve my problem.
 A: let $L=1$
Let $x$ and $y$ be the locations of the 2 cuts
Let $S$ be the region defined by $0<y<x<1$
The three pieces will have lengths of $y, x-y$, and $1-x$
the condition that all three pieces have length less than $c$ is that the point $(x,y)$ is in the region $T$ of points satisfying$$(y<c) \cap(y>x-c)\cap ( x>1-c)$$
Then the probability that all three pieces have length less than $c$ is given by the ratio of the areas...
$$ P = \frac {A_{S\cap T}}{A_S}$$
When $\frac 13 < c < \frac 12$ a plot of the regions looks as in the image below where the region $T$ is shown in red...

$T$ is a single isosceles right triangle with two sides of length $3c-1$
so $$P=(3c-1)^2$$
When When $\frac 12 < c < 1$ a plot of the regions looks as in the image below ...

The 3 blue isosceles right triangles each have two sides of length $1-c$, so the probability in this regime is $$P = 1 - 3(1-c)^2 $$
Putting it all together ...
$$ P(c) = \begin{cases}
(3c-1)^2  & \frac 13 \le c < \frac 12
\\ 1-3(1-c)^2  &  \frac 12 \le c \le 1 \end{cases}  $$
