Number of connections for a graph Suppose we have a graph $G(12,0.7)$ where 12 is the number of nodes and 0.7 is the probability of an edge being present. So total number of edges = $\binom{12}{2} = 66$
Q1 (SOLVED): What is the expected number of friends for Node A (let's say they are labeled A through L)?
My approach is no friends + 1 friend + 2 friends + ... + 12 friends
= $0.3*66 + 0.7*0.3^65 + ... + 0.7^{12}*0.3^{54}$
What is wrong with this?
Q2: What is the probability that 2 nodes have exactly 1 common friend?
My approach: A and B connects to exactly C, and A and B cannot connect to the another node together. Any clues on how to approach this?
 A: NOTE : here I assume the question is to find, for two fixed vertices $A,B$ that are given, the probability that they have exactly one common neighbor.  The more general problem of giving the probability there there is at least one pair of vertices with exactly one neighbor (not necessarily $A,B$) is a tougher one and an answer to that would be welcome.
Let $E_C$ denote the event that a fixed vertex $C$ is the only common neighbor of $A$ and $B$.
If we assume that $A,B$ both are neighbors of $C$, 
$A$ can have $0, 1, \ldots, 9$ neighbors other than $B$ and $C$, and if $A$ has say $k$ neighbors, then $B$ has probability $0.3^k$ of not having any of these as neighbors.
Then 
$P(E_C) = .7^2\sum\limits_{k = 0}^{9}{9 \choose k}0.7^k0.3^{9 - k}0.3^k $
Informally, this means $P(E_C)$ = $P(A,B$ neighbors with $C$) $\sum\limits_{k = 0}^9 P(A$ has $k$ neighbors$)P(B$ has none of them as neighbors$)$. 
Now, $C$ can take on 9 different values.  And for some other vertex $D$, the event that $C$  is the only common neighbor of $A,B$ is mutually exclusive with the event that $D$ is.  Therefore, we can just sum this probability over every vertex, and the probability that $A,B$ have exactly one common neighbor is $9 * P(E_C) \approx 0.01$
Here is a very lazy way to get this.
