= Solution
At level $n$, the $N_n$ dyadic cells fully contained in $J$ are disjoint. Their occupied indicators are independent by the preceding part, and every cell has length $2^{-n}$, so each has success <probability>
$$
p_n=\mathbb P(M(D_k^n)>0)=1-e^{-2^{-n}}.
$$
Their sum therefore satisfies
$$
\boxed{M_n(J)\sim\operatorname{Binomial}(N_n,1-e^{-2^{-n}}).}
$$
If there are no fully contained cells, $N_n=0$ and this means the deterministic zero count. The binomial law follows directly from <independence> of occupation indicators, without assuming that the full counts $M(D_k^n)$ are already Poisson.
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