Let $G(n, \lambda/n)$ be an Erdös-Rényi random graph. I need to show that $$\lim_{n\to\infty}\mathbb{P}(\text{vertex $1$ is part of a triangle}) = 0.$$
(This post also asks a similar question as mine)
Below is my attempt.
Let $[n]$ denote the set of vertices of $G$ and $x , y \in [n]\backslash\{1\}$. Let $\Delta(1, x,y)$ be the sets of all possible triangles in $G$ such that vertex $1, x$ and $y$ form a triangle. Then, $$ \mathbb{P}(\text{vertex $1$ is part of a triangle}) \leq \mathbb{P}\left(\bigcup_{x, y \in [n]\backslash\{1\}}|\Delta(1, x, y)| \geq 1\right)\\ \leq \sum_{x, y \in [n]\backslash\{1\}}\mathbb{E}[|\Delta(1, x, y)|] $$ The last inequality is obtained used the bound for union and Markov's inequality. This leaves me with computing the expected number of triangles in $G(n, \lambda/n)$. I have seen on the internet that the expected number of triangles in Erdös-Rényi graph is $${n \choose 3}\frac{\lambda^3}{n^3}.$$ But shouldn't in my case $$\mathbb{E}[|\Delta(1, x, y)|] = {n-1 \choose 2}\frac{\lambda^2}{n^2}?$$ If so, it means that $$\mathbb{P}(\text{vertex $1$ is part of a triangle}) \leq (n-2)(n-3){n-1 \choose 2}\frac{\lambda^2}{n^2}$$ which does not go to $0$ if $n \to \infty$. So I am not sure what to do now.
Denote by $ |\Delta(1, *, \tilde\star)|$ the number of triangles that contain node 1. For each triple $1,x,y$, the chance that all three edges connecting them are selected in $G(n.\lambda/n)$ is $(\lambda/n)^3$. Thus $$\mathbb{E}[|\Delta(1, *, \tilde\star)|] = {n-1 \choose 2}\frac{\lambda^3}{n^3} \to 0 \quad \text{as} \quad n \to \infty \,. $$