A hire company has two lawnmowers which it hires out by the day. The number of demands per day may be modeled by a Poisson distribution with mean $1.5$. In a period of $100$ working days, how many times do you expect
- neither lawnmower to be used;
- some requests for a lawnmower to have to be refused?
Given: Mean $= 1.5$
$\therefore \;$ Parameter of Poisson's distribution $= \lambda = 1.5$
Number of lawnmowers $= 2$
- $P \left(\text{neither lawnmower is used}\right) = P \left(X = 0\right)$
$\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^{0}}{0!}\\\\ & = e^{- \lambda} \\\\ & = e^{- 1.5} = 0.2231 \end{aligned}$
$\therefore \;$ In $100$ working days, neither lawnmower will be used $= 0.2231 \times 100 = 22.31 \approx 22$ times
- $P \left(\text{some requests for a lawnmower to have to be refused}\right) = P \left(X > 2\right)$
Now, $P \left(X > 2\right) = 1 - P \left(X \leq 2\right)$
$\begin{aligned} P \left(X \leq 2\right) & = \sum \limits_{x = 0}^{2} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!}\right] \\\\ & = e^{- 1.5} \times \left[1 + \dfrac{1.5}{1} + \dfrac{1.5^2}{2} \right] \\\\ & = 0.2231 \times 3.625 = 0.8087 \end{aligned}$
$\therefore \;$ $P \left(X > 2\right) = 1 - 0.8087 = 0.1913$
$\therefore \;$ In $100$ working days, number of times a request for a lawnmower will be refused $= 0.1913 \times 100 = 19.13 \approx 19$ times