If three trucks break at locals random distributed of a road with lenght $L$, find the probability that $2$ of those trucks are not at a greater distance than $d$, fot $d \leq \frac{L}{2}$
My attempt:
If $X_i$ is the r.v that points the local that the $ith$ truck broke, then $X_i \sim U(0,L)$. Considering that the $3$ trucks break independently, we should calculate the probability that the trucks are not, two by two, at a distance of $d$. That is, $P(|X_1 - X_2| > d, |X_1 - X_3| > d , |X_2 - X_3| > d)$
I'm not sure, but I think we can write it as $(1-P(|X_1 - X_2| \leq d)) (1-P(|X_1 - X_3| \leq d) )(1-P(|X_3 - X_2| \leq d)) $
The answer is $(1-\frac{2d}{L})^3$ and, for example, we have that
$P(|X_1 - X_2| \leq d) = \int_0^L P(X_2 -d \leq X_1 \leq X_2 + d | X_2 = x)f_{X_2}(x) dx = \frac{2d}{L}$, but I think the integration limit must be for $x=d$ to $L-d$ since $0 \leq X_1 \leq L$
Probably my method is wrong.
Thanks in advance!
I think the probability is given by:
\begin{align} \mathbb{P}\left(\big|X_1 - X_2\big| > d \bigcap \big|X_1 - X_3\big| > d \bigcap \big|X_2 - X_3\big| > d\right) &= \frac{\displaystyle 6 \cdot \int_{0}^{L-2\,d}\int_{x+d}^{L-d}\int_{y+d}^{L}dz\, dy\, dx}{\displaystyle \int_0^L \int_0^L \int_0^L dz\, dy\, dx}\\ &=\frac{6\cdot \left({L}^{3}-6\,d\,{L}^{2}+12\,{d}^{2}\,L-8\,{d}^{3}\right)}{L^3} \end{align}
Hence, e.g. when $L=5$ and $d=1$,
$\displaystyle \mathbb{P}=\frac{27}{125}$
when $L=5$ and $d=5/2$,
$\displaystyle \mathbb{P}=0$
when $L=11$ and $d=3$,
$\displaystyle \mathbb{P}=\frac{125}{1331}$
All of which agree with simulations quite well.