Answer to Question #346738 in Statistics and Probability for Madiha

Question

A secretary makes 2 errors per page on the average. What is the probability that on the next 

page she makes; 4 or more errors and no error?

Answer 1

Average error per page $\lambda=2$

Let X denote the number of Errors: $X\sim Po(\lambda)$

(a) 

$$P(X\ge4)=1-P(X=0)-P(X=1)$$

$$-P(X=2)-P(X=3)$$

$$=1-\dfrac{e^{-2}(2)^0}{0!}-\dfrac{e^{-2}(2)^1}{1!}-\dfrac{e^{-2}(2)^2}{2!}$$

$$-\dfrac{e^{-2}(2)^3}{3!}=0.14287$$

(b) 

$$P(X=0)=\dfrac{e^{-2}(2)^0}{0!}=0.13534$$