Let π be the number of errors made in one page. Then π has a Poisson distribution with π = 2 per page.
p(x;Ξ»)=x!eβλλxβ for x=0, 1, 2...
(a) 4 or more errors
That is
P(Xβ₯4)=1βP(Xβ€3)=1β(p(0;2)+p(1;2)+p(2;2)+p(3;2))==1β(0!eβ220β+1!eβ221β+2!eβ222β+3!eβ223β)=1βeβ2(1+2+2+4/3)β0.1429
(b) no errors
That is
P(X=0)=p(0;2)=0!eβ220β=eβ2β0.1353
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