The water in a tank is contaminated with bacteria. The bacteria are located randomly and independently. The mean number per milliliter of liquid is known to be $1.1$.
Find the probability that a sample of $1$ ml of liquid contains more than $2$ bacteria.
Five samples, each of $1$ ml of liquid, are taken.
- Find the mean number of bacteria for the five samples, taken together.
- Find the probability that there are in total
- $0$ bacteria,
- less than $3$ bacteria
Mean number of bacteria per milliliter of liquid $= 1.1$
$\because \;$ The bacteria are located randomly and independently, Poisson distribution can be applied to determine the number of bacteria.
parameter of Poisson distribution $= \lambda = 1.1$
$P \left(\text{Sample contains more than 2 bacteria}\right) = P \left(X > 2\right)$
$P \left(X > 2\right) = 1 - P \left(X \leq 2\right)$
$\begin{aligned}
P \left(X \leq 2\right) & = \sum \limits_{x = 0}^{2} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\
& = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!}\right] \\\\
& = e^{- 1.1} \times \left[1 + \dfrac{1.1}{1} + \dfrac{1.1^2}{2} \right] \\\\
& = 0.3329 \times 2.705 = 0.9005
\end{aligned}$
$\therefore \;$ $P \left(X > 2\right) = 1 - 0.9005 = 0.0995$
- $\because \;$ Poisson distribution can be used to determine the number of bacteria present,
$\therefore \;$ mean number of bacteria present in $5$ samples $= 1.1 \times 5 = 5.5$
- For $5$ samples, parameter of Poisson distribution $= \lambda = 5.5$
- $P\left(\text{0 bacteria present in 5 samples}\right) = P \left(X = 0\right)$
$\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^0}{0!} \\\\ & = e^{- \lambda} \\\\ & = e^{- 5.5} = 0.0041 \end{aligned}$
- $P \left(\text{less than 3 bacteria present in 5 samples}\right) = P \left(X < 3\right)$
$\begin{aligned} P \left(X < 3\right) & = \sum \limits_{0}^{2} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!}\right] \\\\ & = e^{- 5.5} \times \left[1 + \dfrac{5.5}{1} + \dfrac{5.5^2}{2} \right] \\\\ & = 0.0041 \times 21.625 = 0.0887 \end{aligned}$
- $P\left(\text{0 bacteria present in 5 samples}\right) = P \left(X = 0\right)$