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In a city it is found that 10 accidents took place in a span of 50 days. Assuming that the number of accidents follow the Poisson distribution, the probability that there will be 3 or more accidents in a day in that city is
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$1-(1.2) \mathrm{e}^{0.2}$
$X \sim$ poisson $(\lambda)$
$\operatorname{Mean}(\lambda)=\frac{10}{50}=\frac{1}{5}$
$P(X \geq 3)=1-P(X=0)-P(X=1)-P(X=2)$
$\begin{aligned} & =1-\frac{e^{-\lambda} \cdot \lambda^0}{0 !}-\frac{e^{-\lambda} \cdot \lambda^1}{1 !}-\frac{e^{-\lambda} \cdot \lambda^2}{2 !} \\ & =1-e^{-\frac{1}{5}}\left[\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)+\frac{1}{2}\left(\frac{1}{25}\right)\right] \\ & =1-e^{-\frac{1}{5}}\left[1+\frac{1}{5}+\frac{1}{50}\right] \\ & =1-e^{-0.2}\left[\frac{61}{50}\right]=1-e^{-0.2} \times 1.22 \\ & \Rightarrow P(X \geq 3)=1-1.22 e^{-0.2}\end{aligned}$
$\operatorname{Mean}(\lambda)=\frac{10}{50}=\frac{1}{5}$
$P(X \geq 3)=1-P(X=0)-P(X=1)-P(X=2)$
$\begin{aligned} & =1-\frac{e^{-\lambda} \cdot \lambda^0}{0 !}-\frac{e^{-\lambda} \cdot \lambda^1}{1 !}-\frac{e^{-\lambda} \cdot \lambda^2}{2 !} \\ & =1-e^{-\frac{1}{5}}\left[\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)+\frac{1}{2}\left(\frac{1}{25}\right)\right] \\ & =1-e^{-\frac{1}{5}}\left[1+\frac{1}{5}+\frac{1}{50}\right] \\ & =1-e^{-0.2}\left[\frac{61}{50}\right]=1-e^{-0.2} \times 1.22 \\ & \Rightarrow P(X \geq 3)=1-1.22 e^{-0.2}\end{aligned}$
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