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Each unordered triple of vertices is a triangle in a graph with probability $p^3$. Consequently the triangle count in a binomial random graph satisfies $\mathbb{E}X=\left(\genfrac{}{}{0ex}{}{n}{3}\right)p^3\le (np{)}^3/6$. If $np\to 0$, the Markov inequality gives $\mathbb{P}(X>0)\le \mathbb{E}X\to 0$. Hencethe sparse direction of the triangle-existence threshold in a binomial random graph.

Two distinct triangles in a graph with at most one common vertex use disjoint edges, so their indicator random variables are independent and have zero covariance. A pair sharing an edge has covariance $p^5-p^6$. The count from part (i) givesWith $\mu =\left(\genfrac{}{}{0ex}{}{n}{3}\right)p^3$, the triangle variance in a binomial random graph therefore satisfiesBoth terms tend to zero when $np\to \infty$, because $n^{2}p=n(np)\to \infty$. Applying part (ii) provescompleting the triangle-existence threshold in a binomial random graph.