I have given a random graph G(n, p) with n = 5000 vertices and an edge probability of p = 0.004.
I calculated the expected number of edges which is (0.004 * maximum number of possible edges) $pE = 0.004 * n(n-1)/2 = 0.004 * 5000 * 4999 * \frac{1}{2} = 49990$
So the expected degree of a vertex should be
$\frac{pE}{n} = 9.998 \approx 10$
right?
I wonder what would be the probability to find m = 100 edges in the graph? And what is the probability, that a vertex in the graph has a degree of one?
I am quite new to probability-theory it would be great if someone could help me.
Thank you so much.
A vertex can have degree at most $4999$. The probability of having degree 0 is $(1-p)^{4999}$. In general the probability of having degree $k$ is $\binom{4999}{k}p^k(1-k)^{4999-k}$ because we have to choose $k$ edges that are included in which case $4999-k$ edges are not included. The expected degree of a vertex is therefore $$\sum_{k=0}^{4999}{k\binom{4999}{k}p^k(1-p)^{4999-k}}=4999p\sum_{k=1}^{4999}{\frac{4998!}{(k-1)!(4998-(k-1))!}p^{k-1}(1-p)^{4998-(k-1)}}$$ and this is also equal to $4999p=19.996$.
The probability that the graph has exactly $100$ edges is $$\binom{\binom{5000}{2}}{100}p^{100}(1-p)^{\binom{5000}{2}-100}$$ which is extremely small (about $9.292984327\times 10^{-21203}$).
The probability that a given vertex has degree exactly $1$ is $4999p(1-p)^{4998}$.