Suppose we are talking about graphs with $n$ labeled vertices. Which graphs are more common: connected or non connected?

Connectedness wins, since the complement of any disconnected graph is connected. EDIT: Perhaps you'd like a proof of this. Let G be a disconnected graph, G' its complement. If v and u are in different components of G, then certainly they're connected by an edge in G'. And if they're in the same component of G, then there's some w in another component (since G was disconnected), so vwu is a path in G'. 


For large $n$, not only are the vast majority of graphs on $n$ vertices connected, the vast majority have diameter 2. That is, any two vertices have a neighbor in common. (The standard reference for properties of most graphs on $n$ vertices, for large $n$, is the book "Random Graphs" by Bela Bollobas.) 


Connectedness wins by a knockout: the proportion of disconnected graphs is about $n2^{n+1}$. See Flajolet, Sedgewick "Analytic Combinatorics", p. 138. 


I like Jonah Ostroff's proof, but here is an inductive proof (for the heck of it). Let $c(n)$ and $d(n)$ respectively denote the number of connected and disconnected graph on $n$ vertices. Evidently, $g(n):=c(n)+d(n)$ is the number of graphs on $n$ vertices. As Jonah Ostroff points out $c(4)=38$ and $d(4)=26$. So, inductively assume that $c(n) > d(n)$, let $G$ be a graph with vertex set $[n]$ and consider a new vertex $n+1$. If $G$ is connected, then adding any nonempty subset of edges incident to $n+1$ maintains connectivity. On the other hand, if $G$ is disconnected, then adding all edges incident to $n+1$ results in a connected graph. Therefore, \[ c(n+1) \geq (2^{n}1)c(n)+d(n) = (2^n2)c(n) + g(n). \] By induction, we have $c(n) > g(n)/2$. Substituting yields \[ c(n+1) > 2^{n1} g(n)=g(n+1)/2. \] 


I like Jonah Ostroff short and sweet proof, but the key to it lies in the fact that there is not a bijection between the set $S_1$ of connected graphs and the set $S_2$ of disconnected graphs over $n$ labeled vertices for $n \ge 4$, as follows:
For example, for $n=4$: Take the $12$ possible undrected Hamiltonian paths of length $4$ on a graph over four labeled vertices. The complement of each of these paths is also a hamiltonian path. Since we know that the complement of a disconnected graph is obviously connected for $n>3$, then the number of connected graphs is at least equal to the number of disconnected graphs. Hoewever, since for $n>3$, the complements of at least some of the connected graphs are also connected graphs, that means that there must be more connected graphs than there are unconnected graphs. The $12$ Hamiltonian paths are those connected graphs over $4$ vertices whose complements are also connect: thus the remaining $2^6  12 = 52$ graphs are divided into pairs of complement graphs which are connected and disconnected, yielding a total of $26$ disconnected graphs, and $26+12=38$ connected graphs over the set of $64$ labeled graphs over $4$ labeled vertices. The path graphs of length $n$ on the set of $n$ vertices are the canonical example of connected graphs whose complements are also connected graphs (for $n>3$). 

