An executive in a company makes on an average 5 telephone calls per hour at a cost of Rs.2 per cell. The probability that in any hour the cost of the calls exceeds a sum of Rs. 4 is

Average of 5 telephone calls per hour at a cost of 2 Rs Per cell.

    $\mathrm{\lambda }=5$

Now $\mathrm{P}(\mathrm{X}=9)=\frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^{\mathrm{r}}}{\mathrm{r}!}$

In any hour the cost of the calls exceeds a sum of 4 Rs.

Calls exceed by $\frac{4}{2}=2$

Required Probability $=1-\mathrm{P}(X=0)-\mathrm{P}(X=1)-\mathrm{P}(X=2)$

     $$\begin{gathered} =1-\frac{5 {\mathrm{e}}^{-5}}{0!}-\frac{5^1 {\mathrm{e}}^{-5}}{1!}-\frac{5^2 {\mathrm{e}}^{-5}}{2!} \\ =1-{\mathrm{e}}^{-5} \left(1+5+\frac{25}{2}\right) \\ =1-{\mathrm{e}}^{-5} \left(\frac{37}{2}\right) \\ =1-\frac{37}{2{\mathrm{e}}^5} \\ =\frac{2{\mathrm{e}}^5-37}{2{\mathrm{e}}^5} \end{gathered}$$