Probability of a random graph being triangle-free 
Let $S$ be the set of all graphs with vertex set $\{1,2,\ldots,n\}$. A random graph $G\in S$ has probability $2^{-{n \choose 2}}$. Show that a random graph almost surely contains a triangle.

My attempt so far: we want to show that as $n$ goes to infinity, the probability of a triangle-free graph goes to zero. I tried to find a general formula for the probability of a triangle-free graph with $n$ vertices, but it seems impossible. I know that for any three vertices, the probability of not being a triangle is $\frac{7}{8}$, but I can't calculate the probability of the union of all such three vertices. Is there any other way to approach this problem?

Consider a random graph on $3n$ vertices. Divide the vertices into $n$ triplets. The probability that a given triplet is not a triangle is $\frac78$, so the probability that no triplet is a triangle is $\left(\frac78\right)^n$. The probability that the graph is triangle-free is thus at most $\left(\frac78\right)^n$, which tends to $0$ as $n$ increases without bound.
