= Solution
Let $e=\{a,b\}$ be an <edge> of $G$. Apply the preceding result with source $a$ and sink $b$. A <tree> contains $e$ exactly when its unique path from $a$ to $b$ is the single <edge> $e$, in the positive direction. Consequently the unit electrical current on $e$ is $\mathbb P(e\in T)$. By <Ohm's law> and the definition of <effective resistance>,
$$
\boxed{\mathbb P(e\in T)=R_{\mathrm{eff}}^G(a,b),}
$$
since this <edge> has unit resistance. This is the unit-conductance case of the <edge-inclusion formula for a uniform spanning tree>.
The general principle needed is <Rayleigh monotonicity principle>: adding <edges>, or decreasing their resistances, cannot increase the <effective resistance> between fixed terminals. One precise variational formulation is the <Thomson principle>:
$$
R_{\mathrm{eff}}(a,b)=\min\left\{\sum_e r_ej_e^2:j\text{ is a unit flow from }a\text{ to }b\right\},
$$
with one orientation chosen for each unoriented <edge>. To see this, write any <unit flow> as $i+k$, where $i$ is the electrical unit current and $k$ has zero divergence. Since $r_ei_e$ is a potential difference, summation by parts gives $\sum_er_ei_ek_e=0$. Thus its energy is $\sum_er_ei_e^2+\sum_er_ek_e^2$, minimized by $i$. The energy of $i$ equals its terminal voltage drop, hence the <effective resistance>.
Extend any <unit flow> on $G$ to $G'$ by zero on all new <edges>. It is still a <unit flow>, including at new <graph vertices>, and its energy is unchanged. The variational minimum on $G'$ is therefore no larger. Using the edge-inclusion formula in both <graphs> gives
$$
\boxed{\mathbb P(e\in T)\ge\mathbb P(e\in T').}
$$
The <graphs> are understood to be finite and connected, as required for their uniform spanning-tree laws. The comparison allows added <graph vertices> as well as added <edges>.
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