Attempt
Let $X,Y,Z$ be the points where the three trucks break down on the interval $[0,L]$. We see that all are uniformly distributed over $[0,L]$. There are several cases to consider since if we call $XYZ$ to be the lining of where the trucks break down, then we have two consider the six cases: YXZ, YZX, XYZ, XZY, ZXY, ZYX.
For example, for $YXZ$, we want $|Y-X|<d $ and $|Z-X| < d$. Thus, we want to find
$$ P(|Y-X|<d ) P(|Z-X| < d) = ( \frac{(L-d)^2 }{2L^2} )( \frac{(L-d)^2 }{2L^2} ) = \frac{1}{4} \left( 1 – \frac{L^2}{D^2} \right)^2$$
Now, by symmetry of other cases are the so the required probability is just $6$ times the above.
However, my answer key gives the solution $ \left(1 – \frac{2d}{L} \right)^3 $. What did I do wrong?
Best Answer
Let's start with one case and then you can repeat this on all six variations. First put the middle truck on the road at some particular point $x$. The probability of this is $\frac{1}{L}$. Now, you place the truck in front of it at some distance greater than $d$ from it and the truck behind it at least $d$ distance behind it. For the one in front the distance remaining is $(L-(x+d))$, therefore probability of it being placed in that region is $\frac{(L-(x+d))}{L}$ and similarly for the one behind the length remaining is $(x-d)$ and corresponding probability is $\frac{(x-d)}{L}$. Now multiply all of them to get
$$P(x) = \frac{xL - x^2 -dL +d^2}{L^3}$$
Integrate $P(x)$ with $x$ going from $d$ to $L-d$. The answer will be
$$\frac{1}{6}\left[1 - \frac{2d}{L}\right]^2$$
Multiply with 6 for all the cases (considering the 6 permutations of 3 trucks at specific positions on the road) and you have your answer. Please let me know if you want me to add in the intermediate steps of integration.