The number of dropped connections per call follows a Poisson distribution. From four calls, the number of dropped connections is $2,\,0,\,3,\,1$.
Obtain the maximum likelihood estimate that the next two calls will be completed without any accidental drops.
I know the maximum likelihood estimate of $\lambda$ is $1.5$. I think I am supposed to use this formula: $$f(x|\lambda) = \frac{\lambda^x e^{-\lambda}}{x!}$$
However, I do not know what $x$ should be. The answer should be $0.0498$.