Correlation when adding balls to urn It is a pretty simple question, but I wanted to know if I got it right (and I had a weird result in the end):
There are 3 White balls and 1 Green ball in a urn. We take one ball from it, write down the colour then put it back with +c balls of the same colour. Now we take a second ball. $X_{i} = 1$ if the ith ball is green and $X_{i} = 0$ if it is white, for $i = 1,2$.
a) What is the joint probability distribution of $X_{1}$ and $X_{2}$
b) What is the correlation between $X_{1}$ and $X_{2}$. What whappens when c goes to inf?
Here's what I did:
a)
$P(0,0) =  \frac{3}{4} \cdot \frac{3+c}{4+c} = \frac{9+3c}{16+4c}$
$P(0,1) =  \frac{3}{4} \cdot \frac{1}{4+c} = \frac{3}{16+4c}$
$P(1,0) =  \frac{1}{4} \cdot \frac{3}{4+c} = \frac{3}{16+4c}$
$P(1,1) =  \frac{1}{4} \cdot \frac{1+c}{4+c} = \frac{1+c}{16+4c}$
b)
$E[X_{1}] = E[(X_{1})^2] =  1 \cdot \frac{1}{4} = \frac{1}{4}$ 
$E[X_{2}] = E[(X_{1})^2] =  1 \cdot \frac{1+c}{4+c} + 1 \cdot \frac{1}{4+c} = \frac{2+c}{4+c}$ (all other sum terms are 0)
$E[X_{1}X_{2}] = 1 \cdot 1 \cdot \frac{1+c}{4c+16}$ (all other sum terms are 0)
$Var(X_{1}) = \frac{1}{4} - \frac{1}{16} = \frac{3}{16}$
$Var(X_{2}) = \frac{2+c}{4+c} - (\frac{2+c}{4+c})^2 = \frac{4+2c}{(4+c)^2}$
$Cov(X_{1},X_{2}) = \frac{1+c}{4c+16} - \frac{1}{4} \cdot \frac{2+c}{4+c} = -1$
$Cor(X_{1},X_{2}) = \frac{-1}{\sqrt{\frac{3}{16} \cdot \frac{4+2c}{(4+c)^2}}} = \frac{-c-16}{\sqrt{12+6c}}$
And then when c goes to inf, the correlation goes to -inf, which doesn't seem to make sense...
Did I go wrong somewhere?
 A: I'm having trouble following your logic, but yes, you've made some mistakes (a correlation cannot exceed one in absolute value, for example).  $\text{E}(X_1)$ is easy enough to find so let's start by calculating $\text{E}(X_2)$.  The key is to condition on $X_1$ and then calculate the expectation in pieces.
\begin{align}
\text{E}(X_2) &= \text{E} [ \text{E} (X_2 \mid X_1) ] \\
&= P(X_1 = 1) \text{E} (X_2 \mid X_1 = 1) + P(X_1 = 0) \text{E}(X_2 \mid X_1 = 0) \\
&= \frac{1}{4} \cdot \frac{1 + c}{4 + c} + \frac{3}{4} \cdot \frac{1}{4 + c} \\
&= \frac{4 + c}{4 (4 + c)} \\
&= \frac{1}{4} .
\end{align}
This is interesting as it says that on average $X_2$ behaves just like $X_1$.  Now since these are Bernoulli random variables with the same expectation the variances are easy:
\begin{align}
\text{Var}(X_i) &=  \text{E}(X_i^2) - \text{E}(X_i)^2 \\
&= \frac{1}{4} - \frac{1}{16} \\
&= \frac{3}{16} .
\end{align}
The only thing left to calculate is the covariance and we can use the identity $\text{Cov}(X_1, X_2) = \text{E}(X_1 X_2) - \text{E}(X_1) \text{E}(X_2)$.  We already know the rightmost term so for the other we have
\begin{align}
\text{E}(X_2 X_2) &= P(X_1 = 1 \cap X_2 = 1) \\
&= P(X_1 = 1) P(X_2 = 1 \mid X_1 = 1) \\
&= \frac{1 + c}{4 (4 + c)}
\end{align}
yielding
\begin{align}
\text{Cov}(X_1, X_2) &= \frac{1 + c}{4 (4 + c)} - \frac{1}{16} \\
&= \frac{3c}{16 (4 + c)} .
\end{align}
If we now divide by this $\sqrt{\text{Var}(X_1) \text{Var}(X_2)}$ we get
\begin{align}
\text{Corr}(X_1, X_2) &= \frac{c}{4 + c} .
\end{align}
This makes sense since as $c \to \infty$ we have $X_1 = X_2$ with high probability so the correlation should approach one.
