Edit: The indicator random variables have been radically changed, so that one gets a very quick computation of the mean and variance.
Let $Y$ be the number of reds drawn before the first blue. Suppose that the red balls have labels $1, 2, 3,\dots,r$. Let $X_i=1$ if red ball with label $i$ is drawn before the first blue is drawn, and let $X_i=0$ otherwise.
Then $Y=X_1+\cdots+X_r$. Note that the number of draws up to an including the first blue is $Y+1$. But $Y+1$ and $Y$ have the same variance.
To calculate the variance of $X_i$, we first calculate the mean. By linearity of expectation we have
$$E(Y)=E(X_1)+E(X_2)+\cdots+E(X_r).$$
By symmetry, all the $E(X_i)$ are the same. The probability red with label $i$ comes before any of the $b$ blue is $\frac{1}{b+1}$. It follows that $E(Y)=\frac{r}{b+1}$.
To calculate the variance of $Y$, calculate $E(X_1+\cdots +X_{r})^2$ and subtract the square of $E(Y)$, which we know.
To find $E(X_1+\cdots+X_r)^2$, expand the square and use the linearity of expectation. We know the expectation of $\sum X_i^2$, since $X_i^2=X_i$. So we need the expectations of the cross terms.
For $i\ne j$, $X_iX_j=1$ if both red ball $i$ and red ball $j$ come before any blue. This has probability $\frac{2}{(b+2)(b+1)}$. Multiply by $2\binom{r}{2}$ to get the sum of the cross terms.
Randomly select an urn, draw a ball without replacement, then again randomly select an urn and draw a second ball. Let $D_n$ be the colour of the $n^{th}$ draw, and $U_n$ be the urn from which it is drawn.
Using k for black, the urns are: $a = \{(w,4), (r,2)\}, b = \{(r,3), (k,3)\}$
Now, if we know a black ball is going to be removed in the second draw, then only one of the other two black balls, or one of the three red balls, could be removed from the second urn during the first draw.
So the probability of drawing a red ball in the first draw give that knowledge is:
$$P(D_1=r \mid D_2=k) \\ = P(D_1=r \cap U_1=a \mid D_2=k)+P(D_1=r \cap U_1=b \mid D_2=k) \\ = P(U_1=a)P(D_1=r \mid U_1=a \cap D_2=k)+P(U_1=b)P(D_1=r \mid U_1=b \cap D_2=k) \\ = \frac{1}{2}\frac{2}{6}+\frac{1}{2}\frac{3}{5} \\ = \frac{7}{15}$$
Best Answer
Let $R_1$ and $R_2$ be the event of picking a red ball on the first or second draw, respectively. Then: $$P(R_1 \mid R_2) = \frac{2}{8} = \frac{1}{4}$$
Reason: If you've picked 1 red ball then $2$ of the remaining $8$ balls are red. The order of doing this does not matter, so if you know the second ball is red then there is a $\frac14$ chance of the first ball was also red.
Or take the long way, using Bayes theorem and total probability:
We know: $\mathrm{P}(R_1)=\frac{3}{9}, \mathrm{P}(\overline{R_1})=\frac{6}{9}, \mathrm{P}(R_2\mid R_1)=\frac{2}{8}, \mathrm{P}(R_2\mid \overline{R_1})=\frac{3}{8}$
So:$$\begin{array}{l}\mathrm{P}(R_1 \mid R_2) \\ = \dfrac{\mathrm{P}(R_1\cap R_2)}{\mathrm{P}(R_2)} \\ = \dfrac{\mathrm{P}(R_1)\cdot\mathrm{P}(R_2\mid R_1)}{\mathrm{P}(R_1)\cdot\mathrm{P}(R_2\mid R_1)+\mathrm{P}(\overline{R_1})\cdot\mathrm{P}(R_2\mid\overline{R_1})} \\ = \dfrac{\frac{3}{9}\cdot\frac{2}{8}}{\frac{3}{9}\cdot\frac{2}{8}+\frac{6}{9}\cdot\frac{3}{8}} \\ = \dfrac{1}{4}\end{array}$$