Question

It is known that in a glass items production in the average one in every 1000 items contain bubbles. Assuming Poisson’s distribution, find the probability that in 6000 items: (a) Exactly 5 items contain bubbles. (b) Less than 4 items contain bubbles

Answer 1

Poisson distribution with mean 6.

(a) $P(=5)$

$p(=5) = \frac{e^{-6}6^5}{5!}=0.160623141$

(b) $P(\le3)$

$p(\le3) = \sum_0^3\frac{e^{-6}6^x}{x!}$

$=\frac{e^{-6}}{0!}+\frac{e^{-6}6^1}{1!}+\frac{e^{-6}6^2}{2!}+\frac{e^{-6}6^4}{3!}$

$=0.002478752+0.014872513+0.044617539+0.089235078$

$=0.151203883$