![=1-[P(X=0)+P(X=1)+P(X=2)]](/tpl/images/1009/5329/cbcf3.png)
![=1-(\dfrac{e^{-2.3}2.3^0}{0!}+\dfrac{e^{-2.3}2.3^1}{1!}+\dfrac{e^{-2.3}2.3^2}{2!})\ \ \ [\text{Poisson distribution formula: }P(X=x)=\dfrac{e^{-\mu}\mu^x}{x!}]\\\\=1-0.596038825932=0.403961174068\approx0.40396](/tpl/images/1009/5329/3f915.png)
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Mathematics, 30.12.2020 14:00 emily200705
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Solution needed with a fair explanation
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