Probability of draws at random with replacement of five tickets 
$400$ draws are made at random with replacement from $5$ tickets that are marked  $-2, -1, 0, 1,$ and $2$ respectively. Find the expected value of: the number of times positive numbers appear?

Expected value of $X$ number times positive number appear $= E(X)= (1\cdot(1/5))+ (2\cdot(1/5))= (1/5) + (2/5)=3/5=0.6$?
$400\cdot 3/5=240$
Expected Value of $X$ number times positive number appear $=E(X)=240/400=3/5=0.6$?

The idea is right, but note that $0$ is not positive. So the probability we get a positive on any draw is $\frac{2}{5}$. 
So if $X_i=1$ if we get a positive on the $i$-th draw, with $X_i=0$ otherwise, then $E(X_i)=\frac{2}{5}$. Now use the linearity of expectation to conclude that the expected number of positives in $400$ draws is $(400)\left(\frac{2}{5}  \right)$.
