The Poisson is given by
\(\displaystyle \frac{{\lambda}^{x}\cdot e^{-\lambda}}{x!}\)
Now, since it says "at least 3", the best thing to do is find the probabilities from 0 to 2 and subtract from 1.
\(\displaystyle 1-\sum_{x=0}^{2}\frac{(4.6)^{x}\cdot e^{-4.6}}{x!}\)
Intuitively, we know that the probability is going to be fairly high, since the average is 4.6 and they are asking for 3 and up.
