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The Poisson thinning theorem, or its multinomial marking version, makes the bin counts at time $n$ independent Poisson variables with mean $n/n=1$. Independence is more than is needed for the following union bound. Fix $ϵ>0$ and put $x_n=(1+ϵ)\log n/\log \log n$. ThenSincethe logarithm of the upper bound is $-ϵ\log n+o(\log n)\to -\infty$. Thus $\boxed{\mathbb{P}(M_n\ge x_n)\to 0}$. The estimate handles the noninteger threshold directly, since the supplied Poisson tail bound holds for real $x$.

In an interval of length $t$, condition on the Poisson process count $X_t=k$. Independently marking its points gives $Y_t\mid X_t=k\sim \mathrm{Bin}(k,p)$. The joint probability generating function of kept and discarded counts isIt factors into the generating functions of independent Poisson variables of means $\lambda pt$ and $\lambda (1-p)t$. The same calculation applies to any interval, and counts and marks in disjoint intervals are independent. Hence $Y$ is a Poisson process of intensity $\boxed{\lambda p}$, and $Y_t$ and $X_t-Y_t$ are independent. This is the Poisson thinning theorem.

For one visitor, letand defineA visitor arriving at time $s\le t$ is in room $j$ at time $t$ exactly when its independent service-time mark satisfies $T_{j-1}\le t-s<T_j$. Thus it receives mark $j$ with probability $p_j(t-s)$; a further mark records that it has already left the gallery.

The Independent marking theorem for Poisson point processes makes the point processes carrying the different room marks independent. Consequently $V_1(t),\dots ,V_{10}(t)$ are independent, and Poisson thinning givesEquivalently,