Drawing balls out of pocket

First calculate the probabilities for each value of $X$ , given that X can be either 0, 1 or 2:

$\begin{array}{l} P(X = 0) = \frac{n}{{n + 2}} \times \frac{{n - 1}}{{n + 1}} = \frac{{n(n - 1)}}{{(n + 2)(n + 1)}}\\ P(X = 1) = \frac{n}{{n + 2}} \times \frac{2}{{n + 1}} + \frac{2}{{n + 2}} \times \frac{n}{{n + 1}} = \frac{{4n}}{{(n + 2)(n + 1)}}\\ P(X = 2) = \frac{2}{{n + 2}} \times \frac{1}{{n + 1}} = \frac{2}{{(n + 2)(n + 1)}} \end{array}$

Now calculate the expected values of $X$ and $X^2$ : $\begin{array}{l} E(X) = 1*\left( {\frac{{4n}}{{(n + 2)(n + 1)}}}\right) + 2*\left( {\frac{2}{{(n + 2)(n + 1)}}} \right) = \frac{{4n + 4}}{{(n + 2)(n + 1)}} = \frac{4}{{n + 2}}\\ E({X^2}) = 1^2*\left( {\frac{{4n}}{{(n + 2)(n + 1)}}}\right) + 2^2*\left( {\frac{2}{{(n + 2)(n + 1)}}} \right) = \frac{{4n + 8}}{{(n + 2)(n + 1)}} = \frac{4}{{n + 1}} \end{array}$

Now calculate the variance using the standard formula: $\begin{array}{c} {\rm{Var}}(X) = E({X^2}) - {\left[ {E(X)} \right]^2}\\ = \frac{4}{{n + 1}} - {\left( {\frac{4}{{n + 2}}} \right)^2}\\ = \frac{4}{{n + 1}} - \frac{{16}}{{{{(n + 2)}^2}}}\\ = \frac{{4{{(n + 2)}^2} - 16(n + 1)}}{{(n + 1){{(n + 2)}^2}}}\\ = \frac{{4{n^2} + 16n + 16 - 16n - 16}}{{(n + 1){{(n + 2)}^2}}}\\ = \frac{{4{n^2}}}{{(n + 1){{(n + 2)}^2}}} \end{array}$

We know the variance must be $\frac{1}{3}$ , therefore: ${\rm{Var}}X = \frac{1}{3} \Rightarrow \frac{{4{n^2}}}{{(n + 1){{(n + 2)}^2}}} = \frac{1}{3}$ $\begin{array}{c} (n + 1){(n + 2)^2} = 12{n^2}\\ (n + 1)({n^2} + 4n + 4) = 12{n^2}\\ {n^3} + 4{n^2} + 4n + {n^2} + 4n + 4 = 12{n^2}\\ {n^3} - 7{n^2} + 8n + 4 = 0\\ (n - 2)({n^2} - 5n - 2) = 0 \end{array}$

$\therefore n = 2,{\rm{ }}n = \frac{{5 \pm \sqrt {33} }}{2}$

But $n$ must be an integer, so $n = 2$ .