In the refectory of a college both tea and coffee are sold.
The number of cups of coffee and tea sold per five minute interval may be considered to be independent Poisson distributions with means $2.7$ and $1.5$ respectively.
Calculate the probabilities that, in a given five minute interval,
- exactly one cup of coffee and one cup of tea are sold,
- exactly two drinks are sold,
- more than five drinks are sold.
Number of cups of coffee sold per five minute interval is a Poisson distribution with mean $= 2.7$
$\therefore \;$ Parameter of Poisson distribution (for coffee) $= \lambda_c = 2.7$
Number of cups of tea sold per five minute interval is a Poisson distribution with mean $= 1.5$
$\therefore \;$ Parameter of Poisson distribution (for tea) $= \lambda_t = 1.5$
- $P \left(\text{exactly one cup of coffee and one cup of tea are sold}\right)$
$\hspace{3cm}$ $= P \left(\text{1 cup of coffee sold}\right) \times P \left(\text{1 cup of tea sold}\right)$
$\begin{aligned} P \left(1 \text{ cup of coffee sold}\right) & = P \left(X = 1\right) \\\\ & = \dfrac{e^{- \lambda_c} \times \left(\lambda_c\right)^{1}}{1!} \\\\ & = e^{- 2.7} \times 2.7 \\\\ & = 0.0672 \times 2.7 = 0.1814 \end{aligned}$
$\begin{aligned} P \left(1 \text{ cup of tea sold}\right) & = P \left(X = 1\right) \\\\ & = \dfrac{e^{- \lambda_t} \times \left(\lambda_t\right)^{1}}{1!} \\\\ & = e^{- 1.5} \times 1.5 \\\\ & = 0.2231 \times 1.5 = 0.3347 \end{aligned}$
$\therefore \;$ $P \left(\text{exactly one cup of coffee and one cup of tea are sold}\right)$
$\hspace{3cm}$ $= 0.1814 \times 0.3347 = 0.0607$
- Parameter of Poisson distribution for both number of cups of coffee and tea sold per five minute interval $= \lambda = \lambda_c + \lambda_t = 2.7 + 1.5 = 4.2$
$P \left(\text{exactly 2 drinks are sold}\right) = P \left(X = 2\right)$
$\begin{aligned} P \left(X = 2\right) & = \dfrac{e^{- \lambda} \times \lambda^2}{2!} \\\\ & = \dfrac{e^{- 4.2} \times \left(4.2\right)^2}{2} \\\\ & = \dfrac{0.0150 \times 17.64}{2} = 0.1323 \end{aligned}$
- $P \left(\text{more than 5 drinks are sold}\right) = P \left(X > 5\right)$
$P \left(X > 5\right) = 1 - P \left(X \leq 5\right)$
$\begin{aligned} P \left(X \leq 5\right) & = \sum \limits_{x = 0}^{5} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} + \dfrac{\lambda^{5}}{5!} \right] \\\\ & = e^{- 4.2} \times \left[1 + \dfrac{4.2}{1} + \dfrac{4.2^2}{2} + \dfrac{4.2^3}{6} + \dfrac{4.2^4}{24} + \dfrac{4.2^5}{120} \right] \\\\ & = 0.0150 \times 50.2243 = 0.7534 \end{aligned}$
$\therefore \;$ $P \left(X > 5\right) = 1 - 0.7534 = 0.2466$