
Poisson Distribution
The number of errors per page made by a typist has a Poisson distribution with mean 0.2 Find the Probabilities
a) makes no errors on the first two pages
b) makes exactly one error in the first three pages
c) makes her first error on the third page
d) makes no error on exactly two of the first five pages.
a) Po(0.4)
P(X< or = 0) = 0.6703 is this correct?
b) Po(0.6)
P(X< or = 1)  P(X< or = 0)
0.87810.5488
=0.3292 is this correct?
how do i do part c) and d)
thanks.

Assume that the errors in different pages are independent to each other.
Then, for (a), no errors on the first two pages = $\displaystyle P(X=0).P(X=0) = 0.818^2 = 0.669$
For (b), exactly one error in the first three pages = 3.P[exactly 1 error in 1 page and no error in other 2 pages] (3 is multiplied because you can choose 1 page (which contains error) form 3 pages in 3 ways) $\displaystyle = 3.P(X=1).P(X=0).P(X=0) = 0.329$
For (c), first error on the third page= $\displaystyle P(X=0).P(X=0).P(X=1)$
And for (d), no error on exactly two of the first five pages= $\displaystyle {{5}\choose{2}}.P(X=0).P(X=0)$
Quote:
P(X< or = 0)......
It has no meaning since Poisson distribution takes values 1,2,3,.... and so on (not 0).
Also, always stick to the original parameter (here 0.2) while calculating the probabilities.