According to the Stirling's approximation:
n!≈2πn(en)n for large n
The mean of the binomial distribution μ=np⇒p=nμ
P(x=k)=Cnkpk(1−p)n−k==k!(n−k)!n!(nμ)k(1−nμ)n−k≈≈2π(n−k)((n−k)/e)n−kk!2πn(n/e)nnkμk(1−μ/n)k(1−μ/n)n
Let's use some approximations for large n:
(1−nμ)n≈e−μ(1−nμ)k≈1n−kn≈1((n−k)/e)n−k(n/e)n≈(1−k/n)neknk(1−k/n)k≈e−ke−knk=nk
P(k)≈nknkk!μke−μ=k!μke−μ - Poisson's distribution.
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