Question

An ambulance travels back and forth at a constant speed along a road of length L. At a certain moment of time, an accident occurs at a point uniformly distributed on the road. [That is, the distance of the point from one of the fixed ends of the road is uniformly distributed over (0, L).] Assuming that the ambulance’s location at the moment of the accident is also uniformly distributed, and assuming independence of the variables, compute the distribution of the distance of the ambulance from the accident.

Short answer

Probability is$=\frac{a}{L}\left(2-\frac{a}{L}\right),0<a<L$

Step-by-step solution

Assumption

Let X and Ydenoted respectively the locations of the ambulance and the accident of the moment the accident occurs.

Calculation

$$\begin{gathered} P\left(\right|X-Y|<a)=P(Y<X<Y+a)+P(X<Y<X+a) \\ =\frac{2}{L^2}{\int }_0^L{\int }_y^{min(y+a,L)}dxdy \\ =\frac{2}{L^2}\left[{\int }_0^{L-a}{\int }_y^{y+a}dxdy+{\int }_{L-a}^L{\int }_y^{L}dxdy\right] \\ =1-\frac{L-a}{L}+\frac{a}{L^2}(L-a) \\ =\frac{a}{L}\left(2-\frac{a}{L}\right),0<a<L \end{gathered}$$

which is the required probability.