Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / M / M4 / d / i / Solution

Under the intended independent-marking interpretation, the coach colors are independent also of the arrival process. Let $K$ be the index of the first white coach. Then $P(K=k)=p(1-p{)}^{k-1}$, $k\ge 1$, independently of the exponential interarrival times. For $p>0$, $T={\sum }_{j=1}^K{X}_j$, so conditioning on $K$ givesThus $\boxed{T\sim \mathrm{Exponential}(\lambda p)}$, the waiting time in the thinned Poisson process. If $p=0$, no white coach arrives and $T=+\infty$. Independence of colors from arrival times is needed for this distribution; mutual independence of colors alone would not rule out marks chosen using the interarrival durations.