probability of an edge capacity if we have information about the vertices..? Assume you have a graph with $n$ nodes/vertices and we can assign to each node a "type" : type $0$ or type 1. The types are independent. The probability of type $0$ is $1 - \lambda \in (0,1)$ and the probability of type $1$ is $\lambda \in (0,1)$. We are interested in knowing the type because it gives the capacity of the edge. If an edge connecting node $i$ to node $j$ both of type $0$ then we denote the capacity by $\alpha_{00} \in \mathbb{R}_+ ^*$. If an edge connects node $i$ to $j$ where $i$ is of type of $0$ and $j$ of type $1$ then we denote the capacity by $\alpha_{01}$ and so on.. 
Now my question is the following: is the probability of having an edge of capacity $\alpha_{11}$ simply writes $\lambda^2$ ? 
Thank you

The probability of having an edge capacity $\alpha_{11} = \frac{1}{4}$ Let the vertices be v1,v2,v3 and that each vertex could be of type 0 or 1 with equal probability.  Then ther are 2^3 different ways that these vertices could make out  to be. An edge is defined as the link between any two vertices.  For n =3, there are $\boxed{6}$ such edges which are connected by vertices of type 1 and 1.  In each configuration there are ${3\choose2}$ edges.  Thus the total number of possible edges with types (0,0),(0,1),(1,1) = $\boxed{{3\choose2}2^3} = \boxed{24}$.  Thus the prbability that an edge will be connected by vertices of type 1 and 1 = $\frac{6}{24} = \boxed{\frac{1}{4}}$.  This is the same for n= 4 and n=5

