Answer to Question #330178 in Statistics and Probability for Robera

Question

A secretary makes 2 errors per page on the average. What is the probability that on the next page she makes

a) 4 or more errors?

b) No errors?

Answer 1

According to the Poisson's distribution $P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!},$

where $\lambda$ - average value

$a)~P(x\ge4)=1-P(x\le3)=\\ 1-(P(x=0)+P(x=1)+\\ +P(x=2)+P(x=3))=\\ =1-(\frac{2^0}{0!}+\frac{2^1}{1!}+\frac{2^2}{2!}+\frac{2^3}{3!})/e^2=\\ =1-(1+2+2+\frac{4}{3})/e^2\approx0.143=14.3\%$

$b)~P(x=0)=\frac{2^0}{e^20!}=1/e^2\approx0.135=13.5\%$