Dilution calculations
I haven't checked your numbers but you're assuming the $\ce{PbI2}$ to be perfectly insoluble. That's a reasonable assumption but in reality its solubility product is $K_{sp}=4.41\times 10^{-9}$ (Wikipedia), so it does have some solubility.

To take this into account it becomes an equilibrium problem. Assume the initial concentrations (just after mixing)  to be $\ce{C(Pb^2+)}$ and $\ce{C(I^-)}$.

After precipitation the lead concentration has been reduced by$X$ and the iodide concentration by $2X$.

With the solubility product after precipation:

$$K_{sp}=[\ce{Pb^2+}]\times [\ce{I^-}]^2$$

So we have:

$$K_{sp}=\Big(\ce{C(Pb^2+)}-X\Big)\times \Big(\ce{C(I^-)}-2X\Big)^2$$

Note that this is a third degree polynomial equation in $X$. Solve for $X$ to find the real equilibrium concentrations $[\ce{Pb^2+}]$ and $[\ce{I^-}]$.

We can also use the equation for the case where $K_{sp}=0$. Then:

$$0=\Big(\ce{C(Pb^2+)}-X\Big)\times \Big(\ce{C(I^-)}-2X\Big)^2$$

First case: $\ce{C(Pb^2+)}-X=0 \implies \ce{C(Pb^2+)}=X \implies [\ce{Pb^2+}]=0$

Also: $[\ce{I^-}]=\ce{C(I^-)}-2\times\ce{C(Pb^2+)}$

In the second case we assume $\Big(\ce{C(I^-)}-2X\Big)^2=0$ and proceed as above. 

Which case applies depends on which reagent is the limiting one.