Question

The number of typing errors made by a typist has a Poisson distribution with an average of two errors per page. If more than two errors appear on a given page, the typist must retype the whole page. What is the probability that a randomly selected page does not need to be retyped? (Round your answer to three decimal places.)

Answer 1

Answer: 0.6767

Step-by-step explanation:

Given : Mean =$\lambda=2$ errors  per page

Let X be the number of errors in a particular page.

The formula to calculate the Poisson distribution is given by :_

$P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}$

Now, the probability that a randomly selected page does not need to be retyped is given by :-

$P(X\leq2)=P(0)+P(1)+P(2)\\\\=(\dfrac{e^{-2}2^0}{0!}+\dfrac{e^{-2}2^1}{1!}+\dfrac{e^{-2}2^2}{2!})\\\\=0.135335283237+0.270670566473+0.270670566473\\\\=0.676676416183\approx0.6767$

Hence, the required probability :- 0.6767