= Solution
For each <hypergraph edge> $W$, let $B_W$ be the event that one of its two colours occupies more than $3r/4$ vertices. By colour symmetry and the <union bound>, the preceding calculation gives
$$
\mathbb P(B_W)<2e^{-r/8}.
$$
Again the dependency <graph> joins intersecting <hypergraph edges> and has $D+1\leq r(\Delta+1)$. The assumed degree bound implies
$$
e\,(2e^{-r/8})(D+1)
\leq2er e^{-r/8}(\Delta+1)<1.
$$
The symmetric <Lovász local lemma> gives positive <probability> that no $B_W$ occurs. Thus \b[there is a two-colouring in which each colour occupies at most $3r/4$ vertices of every <hypergraph edge>]. Equivalently, both colour counts in every <hypergraph edge> lie between $r/4$ and $3r/4$; no rounding difficulty arises because the counts are integers and the bad events use strict inequalities.
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