= Solution
Identify a configuration with its set $S$ of open <edges>. Let $k(S)$ count the <connected components of a graph> on the full <graph vertex> set, including isolated <graph vertices>. The <random-cluster measure> is
$$
\mu_{p,q}(S)=\frac1{Z_{p,q}}p^{|S|}(1-p)^{|E|-|S|}q^{k(S)}.
$$
For $p\in(0,1)$ and finite $q\ge1$ all its weights are positive. We compare $1\le q_1\le q_2$, taking $\mu_1=\mu_{p,q_1}$ and $\mu_2=\mu_{p,q_2}$.
The key fact is <supermodularity of graph component count>:
$$
k(S\cap T)+k(S\cup T)\ge k(S)+k(T).
$$
To prove it, orient the graph <edges> arbitrarily and let $U_S$ be the real span of their incidence vectors for <edges> in $S$. Its <dimension of a vector space> is $r(S)=|V|-k(S)$: on each <connected component of a graph> these vectors span the vectors whose coordinates sum to zero. Since $U_{S\cup T}=U_S+U_T$ and $U_{S\cap T}\subset U_S\cap U_T$, the <dimension formula for a sum of subspaces> gives
$$
r(S\cup T)+r(S\cap T)\le r(S)+r(T),
$$
which is exactly the component-count inequality.
Put $a=k(S)-k(S\cup T)\ge0$ and $b=k(S\cap T)-k(T)\ge0$. The preceding inequality gives $b\ge a$. In the ratio of the two sides of <Holley's condition>, the normalization constants and Bernoulli <edge> factors cancel, because $|S\cup T|+|S\cap T|=|S|+|T|$. The ratio is therefore
$$
\frac{\mu_1(S\cup T)\mu_2(S\cap T)}{\mu_1(S)\mu_2(T)}
=q_1^{-a}q_2^b
=\left(\frac{q_2}{q_1}\right)^a q_2^{b-a}\ge1.
$$
By the <Holley inequality>, $\mu_{p,q_1}$ <stochastically dominates> $\mu_{p,q_2}$. Thus \b[increasing <cluster weight> suppresses every <increasing event>]:
$$
\boxed{\mu_{p,q_1}(A)\ge\mu_{p,q_2}(A)\quad(q_1\le q_2).}
$$
The <random-cluster single-edge conditional probability> confirms the mechanism: it is $p$ when the endpoints are already connected, and $p/(p+q(1-p))$ otherwise. Larger $q$ favours configurations with more separate components. The condition $q\ge1$ enters explicitly in the last factor $q_2^{b-a}$.
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