Answer: 0.6767
Step-by-step explanation:
Given : Mean =[tex]\lambda=2[/tex] errors per page
Let X be the number of errors in a particular page.
The formula to calculate the Poisson distribution is given by :_
[tex]P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}[/tex]
Now, the probability that a randomly selected page does not need to be retyped is given by :-
[tex]P(X\leq2)=P(0)+P(1)+P(2)\\\\=(\dfrac{e^{-2}2^0}{0!}+\dfrac{e^{-2}2^1}{1!}+\dfrac{e^{-2}2^2}{2!})\\\\=0.135335283237+0.270670566473+0.270670566473\\\\=0.676676416183\approx0.6767[/tex]
Hence, the required probability :- 0.6767
