= 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 $P(K=k)=p(1-p)^{k-1}$, $k\ge1$, independently of the exponential interarrival times. For $p>0$, $T=\sum_{j=1}^KX_j$, so conditioning on $K$ gives
$$
M_T(\theta)=\sum_{k\ge1}p(1-p)^{k-1}\left(\frac\lambda{\lambda-\theta}\right)^k=\frac{\lambda p}{\lambda p-\theta}\quad(\theta<\lambda p).
$$
Thus $\boxed{T\sim\operatorname{Exponential}(\lambda p)}$, 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.
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