SUppose an urn contains $r$ red balls and $b$ blue balls. Each time a ball is drawn, it is returened along with $m$ additional balls of the same colour. Let $X_k$ be the number of red balls in $k$ draws from the urn. Find $E(X_2)$
My attempt : I think conditional expectation will be the best approach so I started with $$E(X_k) = E(E(X_k|Y))$$ where $X_k=$ # red colour in $k$ draws and $Y=1$ if at $k-1$ we get a red color and $Y=0$ if its blue. Bu I dont know how to proceed further.
On
Here's a probability tree Carl Heckman is talking about. Each node corresponds to the resulting marble for that draw, where $r$ means we draw a red marble in the draw, for example. The probabilities for each draw are annotated along the arrows leading to that outcome. The probabilities are computed such that, starting from the base of the arrow, the probability along that arrow is the probability of the pointed-to outcome, starting from the base. For example, the probability of drawin $r$ both times is $$ \frac{r}{r+b} \cdot\frac{r+m}{r+m+b}. $$
We can compute the expected number of red marbles after two draws as $$ (r+2m)\left(\frac{r}{r+b}\right)\left(\frac{r+m}{r+m+b}\right) + (r+m)\left(\frac{r}{r+b}\right)\left(\frac{b}{r+m+b}\right) + (r+m)\left(\frac{b}{r+b}\right)\left(\frac{r}{r+b+m}\right) + (r)\left(\frac{b}{r+b}\right)\left(\frac{b+m}{r+b+m}\right), $$ which I'm bound to make a mistake simplifying.
By definition $$E[X_2]=0\cdot P(X_2=0)+1\cdot P(X_2=1)+2P(X_2=2).$$ The probability that $X_2=1$ can be split up in the probability of first drawing blue and then red, and vice versa. Then we obtain: $$E[X_2]=1\cdot\left(\frac{r}{b+r}\frac{b}{r+m+b}+\frac{b}{b+r}\frac{r}{b+m+r}\right)+2\cdot\frac{r}{b+r}\frac{r+m}{r+m+b}$$ that is, $$E[X_2]=\frac{2br+2r^2+2rm}{(b+r)(r+m+b)}=\frac{2r}{b+r}$$