A book of $500$ pages has $500$ misprints. Using the Poisson distribution, estimate the probabilities that a given page contains
- exactly $3$ misprints;
- more than $3$ misprints.
Mean number of misprints per page $= \dfrac{500}{500} = 1$
$\therefore \;$ Parameter of Poisson's distribution $= \lambda = 1$
- $P \left(\text{exactly 3 misprints}\right) = P \left(X = 3\right)$
$\begin{aligned} P \left(X = 3\right) & = \dfrac{e^{- \lambda} \times \lambda^{3}}{3!} \\\\ & = \dfrac{e^{- 1} \times 1^3}{6} \\\\ & = \dfrac{0.3679}{6} = 0.06132 \end{aligned}$
- $P \left(\text{more than 3 misprints}\right) = P \left(X > 3\right) = 1 - P \left(X \leq 3\right)$
$\begin{aligned} P \left(X \leq 3\right) & = \sum \limits_{x = 0}^{3} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!}\right] \\\\ & = e^{- 1} \times \left[1 + \dfrac{1}{1} + \dfrac{1^2}{2} + \dfrac{1^3}{6}\right] \\\\ & = 0.3679 \times 2.6667 = 0.9811 \end{aligned}$
$\therefore \;$ $P \left(\text{more than 3 misprints}\right) = 1 - 0.9811 = 0.0189$