PMF given by Poisson

Question

I have a question for my probability class that I am struggling with. It asks to find F(2), meaning find the probability that $$f_Y(k\le2)$$ given:
$$f_Y(k) = \begin{cases} \frac{e^{-\lambda}\lambda^{k}}{k!} & k = 0,1,2,...\\ 0 & \text{otherwise} \end{cases}$$ I know that this is a Poisson distribution but I am not sure what values to use for λ and k. My best guess would be 2 and 1, however I am sure that is not correct.

Answer 1

$$F(2)=f(y \leq 2)=e^{-\lambda}[1+\lambda+\frac{\lambda^2}{2}]$$

to have a numerical value of this probability you need more information about $\lambda$, i.e. more information about the mean of your rv