Poisson LawWarningThis post is more than a year old. Information may be outdated.b(n, p)→dP(λ) if n≫1 & p≪1 & λ=npb(n,~p) \xrightarrow{d} P(\lambda) ~ \text{if} ~ n \gg 1 ~ \& ~ p \ll 1 ~ \& ~ \lambda = npb(n, p)d Backlinks1230202Comments