Answer to Question #344461 in Statistics and Probability for Mine

Question

 On average, a textbook author makes two wordprocessing errors per page on the first draft of her textbook. What is the probability that on the next page

she will make

(a) 4 or more errors?

(b) no errors?

Answer 1

Let 𝑋 be the number of errors made in one page. Then 𝑋 has a Poisson distribution with 𝜆 = 2 per page.

$p(x;\lambda)=\frac{e^{-\lambda}\lambda^x}{x!}$ for x=0, 1, 2...

(a) 4 or more errors

That is

$P(X\ge4)=1-P(X\le3)=1-(p(0;2)+p(1;2)+p(2;2)+p(3;2))=\\ =1-(\frac{e^{-2}2^0}{0!}+\frac{e^{-2}2^1}{1!}+\frac{e^{-2}2^2}{2!}+\frac{e^{-2}2^3}{3!})=1-e^{-2}(1+2+2+4/3)\approx0.1429$

(b) no errors

That is

$P(X=0)=p(0;2)=\frac{e^{-2}2^0}{0!}=e^{-2}\approx0.1353$