= Solution
An arrival at time $u\in[0,t]$ is still present at $t$ exactly when its <independent> service time exceeds $t-u$. Thinning the marked <Poisson process> with this retention probability gives a Poisson count with parameter
$$
m_t=\lambda\int_0^t\mathbb P(S>t-u)\,du
=\lambda\int_0^t\mathbb P(S>v)\,dv.
$$
Thus
$$
\boxed{Q(t)\sim\operatorname{Poisson}(m_t).}
$$
This reasoning allows atoms in the service-time distribution; the strict survival inequality correctly treats a customer whose service has ended.
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