Mathematics - Probability Question with Solution | TestHub

Since, $\mathrm{P}(\mathrm{X}=\mathrm{r})=\frac{e^{-\lambda }{\lambda }^r}{r!}$ (where $\lambda =$ mean) $\begin{array}{rl}\therefore & \mathrm{P}(\mathrm{X}=\mathrm{r}>1.5)=\mathrm{P}(2)+\mathrm{P}(3)+\dots \infty \\ & =1-\mathrm{P}(\mathrm{X}=\mathrm{r}\le 1)=1-\mathrm{P}(0)-\mathrm{P}(1)\\ & =1-\left(e^{-2}+\frac{e^{-2}\times 2}{1!}\right)=1-\frac{3}{e^2}\end{array}$