Answer: (a) 0.3679
(b) 0.0803
Explanation:
Given : A book of 600 pages contains 600 such errors.
Then , the average number of errors per page = [tex]\dfrac{600}{600}=1[/tex]
[tex]\text{i.e. }\lambda=1[/tex]
The Poisson distribution function is given by :-
[tex]P(x)=\dfrac{e^{-\lambda}\lambda^x}{x!}[/tex]
Then , the probability that that page 1 contains no errors ( Put [tex]x=0[/tex] and [tex]\lambda=1[/tex]) :-
[tex]P(x=0)=\dfrac{e^{-1}1^0}{0!}=0.3678794411\approx0.3679[/tex]
Now, the probability that page 1 contains at least three errors :-
[tex]P(x\geq3)=1-(P(0)+P(1)+P(2))\\\\=1-(\dfrac{e^{-1}1^0}{0!}+\dfrac{e^{-1}1^1}{1!}+\dfrac{e^{-1}1^2}{2!})=0.0803013970714\approx0.0803[/tex]
