I'm reviewing a question I did for the Poisson random variable and I can't remember how I got the the answer to part b). The problem goes as follows: "Suppose that the number of typographical erros on a page has a Poisson probability distribution with a parameter $\lambda=.25$. a)What is the probability that there is no error on that page? b)What is the probability that there are at least two errors?"
Here's what I've done a) $P\{X=0\}=.779$
$P\{X \ge2\}=1-P\{X=1\}= 1-\sum_{i=0}^{1}\frac{e^{-.25}.25^1}{1!}=1-(.779-0.1947)=.416$
I can't seem to figure out why this procedure it taken because now I take the sum from zero to one and subtract it from $1$ to get $.8$. Any help is appreciated
$$\Pr[X = k] = e^{-\lambda} \frac{\lambda^k}{k!}, \quad k = 0, 1, 2, \ldots.$$
Therefore, $$\Pr[X = 0] = e^{-\lambda} = e^{-1/4} = 0.778801.$$ And we also have $$\Pr[X = 1] = e^{-\lambda}\lambda = \frac{e^{-1/4}}{4} = 0.1947.$$ Therefore, $$\Pr[X \ge 2] = 1 - \Pr[X \le 1] = 1 - (\Pr[X = 0] + \Pr[X = 1]) = 0.026499.$$