I've been thinking about this for a while and I couldn't understand how we could apply this to a bigger case. There's a graph with x vertices and y edges. The vertices are randomly put into set A or set B, with a 1/2 chance for getting put in A, and a 1/2 chance for getting put in B. The cut is a set of edges where there is an endpoint in A and one in B. What is the expected number of edges that will be in the cut?
I looked at this starting from a 2 vertices graph. In this case, there could be either 2 vertices in A or 2 vertices in B, or 1 in B and 1 in A (and vice versa). It would be 1/2 (1). But I don't know how this could be applied to a generic x vertices and y edges? Could this involve a binomial distribution?
Label the edges $E_1,\ldots,E_y$. For $i=1,\ldots,y,\;$ define the indicator variable:
$$I_i = \begin{cases} 1, & \text{if $E_i$ is in the cut set} \\ 0, & \text{otherwise} \end{cases}$$
Also let $N$ be the number of edges in the cut set, so that $N = I_1 + \cdots + I_y$. Then, for any $1\leq i\leq y$,
\begin{align} E(I_i) &= P(\text{$E_i$ has one node in $A$ and the other in $B$}) \\ &= \dfrac{1}{2}\cdot \dfrac{1}{2} + \dfrac{1}{2}\cdot \dfrac{1}{2} \\ &= \dfrac{1}{2}. \\ & \\ \therefore\quad E(N) &= E(I_1 + \cdots + I_y) \\ &= E(I_1) + \cdots + E(I_y) \\ &= y/2. \end{align}