Problem: Let $G ∼ G(n, n^{−1/3})$. Prove that G has diameter 2 almost surely.
That is, prove that almost surely, every pair of distinct vertices u, v has some common neighbor in G.
Idea: Here is what I am thinking... we will let $X_{uv}$ be the indicator for the event $A_{uv}$ that no other vertex in G is adjacent to u or v.
Let X = # of "bad" things happening (not diameter 2) = $\sum_{(u,v), u,v \in V(G)} X_{uv}$
Then we'd like to show that $E[X] \rightarrow 0$ as $n \rightarrow \infty$, (i.e. it has diameter 2 almost surely).
$$E[X] = E[\sum_{(u,v), u,v \in V(G)} X_{uv}] = \sum_{(u,v), u,v \in V(G)} E[X_{uv}] = \sum_{(u,v), u,v \in V(G)} P[A_{uv}] = {n\choose2}(1-p)(1-p^2)^{n-2}$$
And in this situation $p = n^{-1/3}$. How can I show that this goes to zero?
\begin{eqnarray} \binom n2(1-p)\left(1-p^2\right)^{n-2} &=& \binom n2\left(1-n^{-\frac13}\right)\left(1-n^{-\frac23}\right)^{n-2} \\ &\sim_{n\to\infty}&\binom n2\left(1-n^{-\frac13}\right)\exp\left(-n\cdot n^{-\frac23}\right) \\ &=& \binom n2\left(1-n^{-\frac13}\right)\exp\left(-n^{\frac13}\right) \\ &\to_{n\to\infty}&0\;. \end{eqnarray}