if $X\sim \operatorname{Bin}(n,p)$ and $Y={X \choose 2 }$ What is the intuition behind $E(Y) = {n \choose 2}p^2$

Question

A possible motivation for what $Y$ represents is: given a graph with $n$ vertices and no edges, each vertex in the graph is painted red with probability $p$. After all vertices are either painted red or not painted, an edge is drawn between every $2$ red vertices. let $Y$ represent the number of edges in the graph.

What i don't understand is: if i were to think, what is the probability that a certain edge appears in the graph, then i would realize that the edge exists iff the two vertices it connects are painted red; and this happens with probability $p^2$. Then if i were to mistakenly consider the drawing of each edge as an independent trial with probability of success equal to $p^2$ (that is the edge being considered is drawn) then i would find that $E(Y)={n \choose 2}p^2$ but clearly the trials in which each edge is either drawn or not drawn are not independent of one another.

Thanks in advance.

Answer 1

Your intuition is correct.

Remember that we have linearity of expectation regardless of independence.

If it helps, you can introduce indicator variable $I_e$ for each edge where it takes value $1$ if there is an edge and $0$ otherwise.

then we have $Y=\sum_e I_e$, hence we have $E[Y]=\sum_e E[I_e]$. Note that there is no interaction terms like $I_{e_1}I_{e_j}$ in our expression.