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)

$P(0,0) =  \frac{3}{4} * \frac{3+c}{4+c} = \frac{9+3c}{16+4c}$

$P(0,1) =  \frac{3}{4} * \frac{1}{4+c} = \frac{3}{16+4c}$

$P(1,0) =  \frac{1}{4} * \frac{3}{4+c} = \frac{3}{16+4c}$

$P(1,1) =  \frac{1}{4} * \frac{1+c}{4+c} = \frac{1+c}{16+4c}$

b)

$E[X_{1}] = E[(X_{1})^2] =  1 * \frac{1}{4} = \frac{1}{4}$ 

$E[X_{2}] = E[(X_{1})^2] =  1 * \frac{1+c}{4+c} + 1 * \frac{1}{4+c} = \frac{2+c}{4+c}$ (all other sum terms are 0)

$E[X_{1}X_{2}] = 1*1*\frac{1+c}{4c+16}$ (all other sum terms are 0)

$Var(X_{1}) = \frac{1}{4} - \frac{1}{16} = \frac{3}{16}$

$Var(X_{2}) = \frac{2+c}{4+c} - (\frac{2+c}{4+c})^2 = \frac{4+2c}{(4+c)^2}$

$Cov(X_{1},X_{2}) = \frac{1+c}{4c+16} - \frac{1}{4} * \frac{2+c}{4+c} = -1$

$Cor(X_{1},X_{2}) = \frac{-1}{\sqrt{\frac{3}{16} * \frac{4+2c}{(4+c)^2}}} = \frac{-c-16}{\sqrt{12+6c}}$

And then when c goes to inf, the correlation goes to -inf, which doesn't seem to make sense...

Did I go wrong somewhere?