$E[\min(X,2)]$ for$ X\sim Po(3)$ 
Let $X\sim Po(3), W=\min(X,2)$
Find $E[W]$.

My attempt:
$W=\begin{cases}
X & X \leq 2 \\
2 & X \gt 2
\end{cases}$
So $$P(W=k)=P(\min(X,2)=k)=P(\{\min(X,2)=k\} \cap \{X\le2\})+P(\{\min(X,2)=k\} \cap \{X\gt2\}) $$ (because the sets are disjoint)
$$=\begin{cases}
P(X=k) & k=0,1,2 \\
0 & \text{else}
\end{cases} + \begin{cases}
1 & k=2 \\
0 & \text{else}
\end{cases} $$
The problem with this is that, among others, $P(W=2) > 1$ so clearly I have done some things wrong.
My questions are:


*

*Where did I go wrong?

*How can I solve this problem?
Thanks for your help!
 A: Second question
To efficiently find the expectation $E[W]$, I suggest you to divide it into two cases $\{X \le 2\}$ and $\{ X > 2 \}$.
\begin{align}
E[W] &= E[W1_{\{X \le 2\}}] + E[W1_{\{ X > 2 \}}] \\
&= E[X1_{\{X \le 2\}}] + E[21_{\{ X > 2 \}}] \\
&= \sum_{k = 0}^2 k\frac{3^k}{k!} e^{-3} + 2E[1_{\{ X > 2 \}}] \\
&= \left( 3 + 9 \right) e^{-3} + 2P(X > 2) \\
&= 12 e^{-3} + 2 \left[ 1 - \left( 1 + 3 + \frac92 \right) e^{-3} \right] \\
&= 2 - 5e^{-3}
\end{align}
First question
First term in the sum is correct
The transition from $P(\{\min(X,2)=k\} \cap \{X\le2\})$ to
$\begin{cases}
P(X=k) & k=0,1,2 \\
0 & \text{else}
\end{cases}$
is OK, since in the event $\{X\le2\}$, $\{\min(X,2)=k\} = \{ X = k\}$.  When $k \in \{0,1,2\}$, $\{ X = k\}$ is a subset of $\{X\le2\}$, so $\{X\le2\}$ can be omitted in the calculation of the probability $P(\{\min(X,2)=k\} \cap \{X\le2\})$.  Otherwise, these two events in the probability calculation has no intersection, one gets zero.
Second term in the sum is incorrect
The transition from $P(\{\min(X,2)=k\} \cap \{X>2\})$ to
$\begin{cases}
\color{red}{1} & \color{red}{k=2} \\
0 & \text{else}
\end{cases}$
is incorrect.  In fact, $$\{\min(X,2) = k\} = \{ X \le k\} \cap \{ k \ge 2\}.$$  When $k=2$, $\{X>2\} \cap \{ X \le k\} = \varnothing$, so $$P(\{\min(X,2)=k\} \cap \{X>2\}) = P(\varnothing) = 0.$$
