Solutions - 5160-05

(a)

$X\sim \mathrm{Exp}(0.5)$, so the Poisson rate for a 2-mile stretch is $\alpha =0.5\cdot 2=1$, giving $Y\sim \mathrm{Poisson}(1)$.

The mean number of potholes in a 2-mile stretch is $E[Y]=\alpha =1$.

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(b)

$$\begin{array}{c}P[Y\ge 2]=1-P[Y=0]-P[Y=1]\\ \\ \gg \gg 1-e^{-1}\cdot \frac{1^0}{0!}-e^{-1}\cdot \frac{1^1}{1!}\gg \gg 0.2642\end{array}$$