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If $E(X) \sim \mathsf{Poisson}(\lambda),$ what is

a) $P(X=2)$

b) $P(X>2)$

c) $P(X^2>2)$

I don't understand what this is trying to say and I don't see how I calculate the Poisson when I'm only given that information. Please help!

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    As a hint for b, because the distribution is discrete $P(X>2)=P(X\geq 3) = 1-P(X\leq 2) = 1 - F_X(2)$ where $F_X(\cdot)$ is the cdf of $X$. For c, it is similar, but note that $X^2>2$ if, and only if, $X\geq 2$. – 2017-02-15

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By definition, a random variable $X \sim \text{Poisson}(\lambda)$ has pmf \begin{equation*} f_X(x)=P(X=x)=e^{-\lambda}\frac{\lambda^x}{x!}, \text{ where } x \in \{0,1,\dots\} \end{equation*} If you are not given a specific value for $\lambda$, you answers will just be in terms of $\lambda$. Can you answer your questions now?

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    Yes that helped thanks ! – 2017-02-16