Answer to Question #186885 in Statistics and Probability for MiyamiZu

Question

A secretary makes 2 errors per page, on average. What is the probability that on the next page she will make (a) 3 or more errors; (b) no errors?

Answer 1

Average error per page $\lambda=2$

Let X denote the number of Errors.

(a) $P(X\ge3)=1-P(X<3)$

$=1-[P(X=0)+P(X=1)+P(X=2)]\\$

$=1-[\dfrac{e^{-\lambda}.\lambda^0}{0!}+\dfrac{e^{-\lambda}.\lambda^1}{1!}+\dfrac{e^{-\lambda}.\lambda^2}{2!}]$

$=1-[\dfrac{e^{-2}.2^0}{0!}+\dfrac{e^{-2}.2^1}{1!}+\dfrac{e^{-2}.2^2}{2!}]$

$=1-(0.135+0.2706+0.2706)\\=1-0.6764\\=0.3236$

(b) $P(X=0)=\dfrac{e^{-\lambda}.\lambda^0}{0!}$

$=\dfrac{e^{-2}.2^0}{0!} =0.135$