So I write this out as,
$$E(X\mid X>2) = \frac 1 {P(X>2)}\sum_{k=2}^\infty k\frac{3^k e^{-3}}{k!}$$
For $P(X>2)$ I get .80085 and when I multiply by the sum I get 3.55947. My book, however, says that the answer is 4.16525.
PS, I computed $P(X>2)$ by subtracting $P(X=0) + P(X=1)$ from $1.$ I hope I didn't mess this up somehow so I assume I made a mistake in computing the summation.
Thank you in advance. Pardon any mistakes in my math. Hopefully it's not something too silly :P
You are asked to require $X>2$, so $2$ itself is not included. Your computation correctly gives $E[X \mid X \geq 2]$.