Solution

Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 36 / 2 / b / iii / Solution

{title2=$\Pr(\operatorname{diam}G(n,p)\leq2)\to1$}

For fixed $0<p<1$, the <missing common-neighbour count in a binomial random graph> has <expectation> tending to zero. The <Markov inequality> therefore shows that every pair has a <common neighbour> with probability tending to one. On that event every <graph distance> is at most two, so the <graph> is <connected> and its <graph diameter> is at most two.

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 Incoming links: Diameter bound for a fixed-density binomial random graph