Question #344461

 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?


1
Expert's answer
2022-05-24T23:28:03-0400

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

p(x;Ξ»)=eβˆ’Ξ»Ξ»xx!p(x;\lambda)=\frac{e^{-\lambda}\lambda^x}{x!} for x=0, 1, 2...

(a) 4 or more errors

That is

P(Xβ‰₯4)=1βˆ’P(X≀3)=1βˆ’(p(0;2)+p(1;2)+p(2;2)+p(3;2))==1βˆ’(eβˆ’2200!+eβˆ’2211!+eβˆ’2222!+eβˆ’2233!)=1βˆ’eβˆ’2(1+2+2+4/3)β‰ˆ0.1429P(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)=eβˆ’2200!=eβˆ’2β‰ˆ0.1353P(X=0)=p(0;2)=\frac{e^{-2}2^0}{0!}=e^{-2}\approx0.1353


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