Question

6. a) Find the probability that at most 5 defective fuses will be found in a box of 200, if

experience shows that 20% of such fuses are defective.

Answer 1

The Poisson's Distribution is

$$P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!},\ x=0,1,2,3,..$$

Let X denote the defective fuse

$p=\frac{20}{100}\\n=200\\mean=\lambda=np=200\times\dfrac{20}{100}=40\\\Rightarrow P(atmost \ 5\ defective\ fuses)=\sum_{x=0}^5\dfrac{e^{-40}40^x}{x!}$

$= e^{-40}\times[\frac{40^0}{0!}+\frac{40^1}{1!}+\frac{40^2}{2!}+\frac{40^3}{3!}+\frac{40^4}{4!}+\frac{40^5}{5!}]\\=4.125\times 10^{-12}$