**Definition 1**: Given a group $G$, a subset $X \subseteq G$, and a natural number $k$,
we say that $X$ is (left) $k$-generic in $G$ if there are $k$ many left translates of $X$ that cover $G$.
That is, if there exist $a_1, \dots, a_k \in G$ such that $G = \bigcup_i a_i \cdot X$ (where $a_i \cdot X = \{ a_i \cdot x: x \in X \}$).

**Definition 2**: Let $n$ be a natural number and $G$ be a group with $n$ elements.
On the family of subsets of $G$ we put the probability that gives to each subset probability $1/2^n$; equivalently, for every $g \in G$ we toss a coin to decide whether or not $g$ is in a given set.

Let $P_k(G)$ be the probability that a random subset of a group $G$ is $k$-generic in $G$.

**Problem**: I want to show that, for a fixed $k$, $P_k(G) \to 0$ as the cardinality of $G$ goes to infinity, independently of the choice of the group $G$ (actually, I need only the case $G = \mathbb Z / n \mathbb Z$).

I expect that the above is a known fact: I would appreciate a reference for it (my background is in logic).