If the probability of a defective fuse from a manufacturing unit is $2\%$ in a box of $200$ fuses, find the probability that
- exactly $4$ fuses are defective;
- more than $3$ fuses are defective. $\;\;\;$ Given: $e^{- 4} = 0.0183$
Probability that a fuse is defective $= p = 2 \% = 0.02$
Total number of fuses $= n = 200$
$\therefore \;$ Mean of Poisson distribution $= \lambda = n \times p = 200 \times 0.02 = 4$
- $P \left(\text{exactly 4 fuses are defective }\right) = P \left(X = 4\right)$
$\begin{aligned} P \left(X = 4\right) & = \dfrac{e^{- \lambda} \times \lambda^{4}}{4!} \\\\ & = \dfrac{e^{- 4} \times 4^4}{24} \\\\ & = \dfrac{0.0183 \times 256}{24} = 0.1952 \end{aligned}$
- $P \left(\text{more than 3 fuses are defective}\right) = P \left(X > 3\right) = 1 - P \left(X \leq 3\right)$ where
$P \left(X \leq 3\right) = P \left(\text{at most 3 fuses are defective}\right)$
$\begin{aligned} P \left(X \leq 3\right) & = \sum \limits_{x = 0}^{3} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!}\right] \\\\ & = e^{- 4} \times \left[1 + \dfrac{4}{1} + \dfrac{4^2}{2} + \dfrac{4^3}{6}\right] \\\\ & = 0.0183 \times 23.67 = 0.4331 \end{aligned}$
$\therefore \;$ $P \left(\text{more than 3 fuses are defective}\right) = 1 - 0.4331 = 0.5669$