Let $\mathcal G$ be a finite collection of rooted graphs (by a rooted graph we mean a graph which has a distinguished vertex) and $\sigma$ be a probability distribution on $\mathcal G$.
I am struggling to put down the following statement rigorously.
Choose a graph $G$ from $\mathcal G$ randomly according to $\sigma$, and then choose an edge incident with the root of $G$ uniformly at random.
I understand what is meant by choosing a graph from $\mathcal G$ randomly. But I am not able to formalize the choice of an edge incident with a randomly chosen graph.
Let me say something as to what I mean by "formalize".
Suppose we wanted to choose a graph $G$ randomly from $\mathcal G$ according to $\sigma$, and then choose a number from $\{1, 2, 3, \ldots, 10\}$ uniformly at random, independently of the choice of $G$.
We can formalize this by constructing the product probability space $\mathcal G\times \{1, 2, 3, \ldots, 10\}$, where the probability measure is the product of the measure $\sigma$ and the uniform measure.
Another way would be the following: Let $X$ be a random variable valued in $\mathcal G$ whose distribution is same as $\sigma$, and $Y$ be a random variable valued in $\{1, 2, 3, \ldots, 10\}$ distributed according to the uniform distribution, and assume that $X$ and $Y$ are independent.
The shaded is already rigorous though.
Write $\cal{G}$ $=\{G_1,\ldots, G_n\}$ for some $n$ and let $v_j$ be the root of $G_j$. Then let $d_j$ be the degree of $v_j$ in $G_j$. Let $(j,a)$ be the $a$-th edge incident to the root in graph $G_j$; $j \in \{1,\ldots, n\}$ and $a \in \{1,\ldots, d_j\}$. Then the probability of picking the edge $(j,a)$ is precisely $\sigma(j) \times d^{-1}_j$.
This above paragraph is rigorous and follows from the shaded. It is also pedantic and unnecessary; the shaded already said this, with no ambiguity.