What is the smallest and largest possible values for the variance? Suppose
$P( X \in \{1,2,3\}) = 1$ and  $E(X) =2.5.$

What is the smallest and largest possible values for the variance?

My understand: So what I understand is variance finds the distance between each element and the mean. So the closer 2 is to E(X) the farther 1 and 3 can be from E(X). But I have no clue how to obtain this..

Let $P(X=i)=p_i$ where $i=1,2,3$ and $p_1+p_2+p_3=1$.
$E(X)=2.5$ gives $p_1+2p_2+3p_3=2.5$. 
Further, $Var(X)=2.25p_1+0.25p_2+0.25p_3=2.25p_1+0.25(1-p_1)=2p_1+0.25$
Now notice that $p_2+2p_3=1.5$ from subtracting the first two equations hence $p_2=1.5-2p_3$. This shows $p_1=1-p_2-p_3=1-1.5+2p_3-p_3=-0.5+p_3$
So the feasible region is $(p_1,p_2,p_3)=(p_3-0.5,1.5-2p_3,p_3)$
Now since $p_i\in(0,1)$ for all $i$, $p_3-0.5>0$ implies $p_3>0.5$ further $1.5-2p_3>0$ implies $p_3<0.75$ and $1.5-2p_3<1$ implies $p_3>0.25$. So the range of $p_3$ is $(0.5,0.75)$.
Now writing $p_1$ in terms of $p_3$ in the variance form, $2p_1+0.25=2p_3-1+0.25=2p_3-0.75$.
This is therefore maximum for $p_3=0.75$ i.e. $Var_{max}=2\times 0.75-0.75=0.75$ and this is minimum for $p_3=0.5$ i.e. $Var_{min}=2\times0.5-0.75=0.25$.
There you go!
