λ=0.0001×100=0.01P(X≥2)=1−P(X<2)=1−[P(X=0)+P(X=1)]=1−[e−0.01(0.01)00!+e−0.01(0.01)11!]=1−0.99995=0.00005\lambda=0.0001\times 100=0.01\\ P(X\geq 2)=1-P(X<2)\\ =1-[P(X=0)+P(X=1)]\\ =1-[\frac{e^{-0.01}(0.01)^{0}}{0!}+\frac{e^{-0.01}(0.01)^{1}}{1!}]\\ =1-0.99995=0.00005λ=0.0001×100=0.01P(X≥2)=1−P(X<2)=1−[P(X=0)+P(X=1)]=1−[0!e−0.01(0.01)0+1!e−0.01(0.01)1]=1−0.99995=0.00005
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