Conditional Expected Value and distribution question 
The distribution of loss due to fire damage to a warehouse is:
$$
\begin{array}{r|l}
\text{Amount of Loss (X)} & \text{Probability}\\
\hline
0 & 0.900 \\
500 & 0.060  \\
1,000 & 0.030\\
10,000 & 0.008 \\
50,000 & 0.001\\
100,000 & 0.001 \\
\end{array}
$$
Given that a loss is greater than zero, calculate the expected amount of the loss.

My approach is to apply the definition of expected value:
$$E[X \mid X>0]=\sum\limits_{x_i}x_i \cdot p(x_i)=500 \cdot 0.060 + 1,000 \cdot 0.030 + \cdots + 100,000 \cdot 0.001=290$$
I am off by a factor of 10--The answer is 2,900.  I am following the definition of expected value, does anyone know why I am off by a factor of $1/10$?
Should I be doing this instead???
$E[X \mid X>0] = \sum\limits_{x_i} (x_i \mid x_i > 0) \cdot \cfrac{\Pr[x_i \cap x_i>0]}{\Pr(x_i > 0)}$
Thanks.

You completely missed the word "given".  That means you want a conditional probability given the event cited.  In other words, your second option is right.
For example the conditional probability that $X=500$ given that $X>0$ is
$$
\Pr(X=500\mid X>0) = \frac{\Pr(X=500\ \&\ X>0)}{\Pr(X>0)} = \frac{\Pr(X=500}{\Pr(X>0)} = \frac{0.060}{0.1} = 0.6.
$$
