Solutions - 5160-04

(a)

$X\sim \mathrm{Poisson}(2/5)$

$$\begin{array}{c}P[X>2]=1-P[X\le 2]\\ \\ \gg \gg 1-(\frac{e^{-2/5}{\left(\frac{2}{5}\right)}^0}{0!}+\frac{e^{-2/5}{\left(\frac{2}{5}\right)}^1}{1!}+\frac{e^{-2/5}{\left(\frac{2}{5}\right)}^2}{2!})\\ \\ \gg \gg 1-0.9921\gg \gg 0.0079\end{array}$$

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(b)

$$\begin{array}{c}P[4\le T\le 5]=F_T(5)-F_T(4)\\ \\ \gg \gg (1-e^{-\frac{2}{5}(5)})-(1-e^{-\frac{2}{5}(4)})\\ \\ \gg \gg e^{-8/5}-e^{-2}\gg \gg 0.0666\end{array}$$