A business man receives an average of $2$ e-mails per hour related to his business and $1.5$ e-mails per hour on personal matters.
Find the probability that in any randomly chosen hour
- he receives no e-mails,
- he receives more than 5 e-mails.
Applying Poisson distribution to determine the number of e-mails received per hour (both business and personal), we have
parameter of Poisson distribution for total e-mails $= \lambda = 2 + 1.5 = 3.5$
- $P \left(\text{receives no e-mail}\right) = P \left(X = 0\right)$
$\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^0}{0!} \\\\ & = e^{- \lambda} \\\\ & = e^{- 3.5} = 0.0302 \end{aligned}$
- $P \left(\text{receives more than 5 e-mails}\right) = P \left(X > 5\right)$
$P \left(X > 5\right) = 1 - P \left(X \leq 5\right)$
$\begin{aligned} P \left(X \leq 5\right) & = \sum \limits_{x = 0}^{5} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} + \dfrac{\lambda^{5}}{5!} \right] \\\\ & = e^{- 3.5} \times \left[1 + \dfrac{3.5}{1} + \dfrac{3.5^2}{2} + \dfrac{3.5^3}{6} + \dfrac{3.5^4}{24} + \dfrac{3.5^5}{120} \right] \\\\ & = 0.0302 \times 28.4003 = 0.8577 \end{aligned}$
$\therefore \;$ $P \left(X > 5\right) = 1 - 0.8577 = 0.1423$