A calculus problem by Mark Recio

Let's say, an encoder makes an average of three errors per page. Assume the number of errors per page has the Poisson distribution, the probability that at least one error will be found is:

$p (x = X) = \cfrac{e^{-\lambda} \times \lambda ^{X}}{X!} \\ \lambda = 3 \\ \implies \sum_{x = 1}^{\infty} \cfrac{e^{-3} \times 3^x}{x!} = 1 - p (x = 0) \\ \text{(Note: the complement of the probability getting no error is the probability of getting at least one error.)} \\ = 1 - \cfrac{e^{-3} \times 3^0}{0!} \\ = 1 - 0.04979 \\ = \boxed{0.95021} \ .$

$\begin{aligned} S & = \sum_{x=1}^\infty \frac {e^{-3}3^x}{x!} \\ & = \frac 1{e^3} \left(\sum_{\color{#D61F06}x=0}^\infty \frac {3^x}{x!} - \frac {3^0}{0!} \right) \\ & = \frac 1{e^3} \left(e^3 - 1 \right) \\ & = 1 - \frac 1{e^3} \\ & \approx \boxed{0.95021} \end{aligned}$