Question

The number of errors in a textbook follow a poisson distribution with a mean of 0.01 errors per page. what is the probability that there are 3 or less errors in 100 pages? round your answer to four decimal places (e.g. 98.7654).

Answer 1

Mean number of errors in each page = 0.01
Mean number of errors in 100 pages = 0.01*100=1

It is possible to use the cumulative distribution function (CMF), but the math is a little more complex, involving the gamma-function.  Tables and software are available for that purpose.

Thus it is easier to evaluate with a calculator for the individual cases of k=0,1,2 and 3.

The Poisson distribution has a PMF (probability mass function) 
$P(k):=\frac{\lambda^ke^{-\lambda}}{k!}$
with λ = 1
=>
$P(0):=\frac{1^0e^{-1}}{0!}=0.3678794$
$P(1):=\frac{1^1e^{-1}}{1!}=0.3678794$
$P(2):=\frac{1^2e^{-1}}{2!}=0.1839397$
$P(3):=\frac{1^3e^{-1}}{3!}=0.0613132$
=>
$P(k<=3)=P(0)+P(1)+P(2)+P(3)=0.9810118$
or
P(k<=3)=0.9810 (to four decimal places)