At a telephone enquiry system the number of phone cells regarding relevant enquiry follow Poisson distribution with an average of 5 phone calls during 10 minute time intervals. The probability that there is at the most one phone call during a 10 - minute time period is

Solution :
$$P\left( {X = r} \right) = \frac{{{e^{ - m}}{m^r}}}{{r!}}$$
$$P$$ (at most 1 phone call)
$$\eqalign{ & = P\left( {X \leqslant 1} \right) = P\left( {X = 0} \right) + P\left( {X = 1} \right) \cr & = {e^{ - 5}} + 5 \times {e^{ - 5}} \cr & = \frac{6}{{{e^5}}} \cr} $$