Number the red and yellow balls from $1$ to $n$, where $n=r+y$. Don't bother to number the blue balls.
Let $X_i=1$ if the $i$-th numbered ball is drawn before any blue ball, and let $X_i=0$ otherwise. Then the total number of draws is $1+(X_1+\cdots+X_n)$. It follows that the expected number of draws is
$$1+E(X_1+\cdots+X_n).$$
By the linearity of expectation, this is
$$1+E(X_1)+\cdots +E(X_n).$$
All the $X_i$ have the same expectation. so let us find $E(X_1)$.
The probability that ball with label $1$ comes before any blue ball is $\frac{1}{b+1}$. This is because all orderings of ball $1$ and the blue balls are equally likely. So $E(X_1)=\frac{1}{b+1}$.
It follows that the number of draws has mean
$$1+\frac{n}{b+1}.$$
Remark: We used the powerful tool of indicator random variables. One can alternately find the probability distribution of the number of draws, and then find the mean by using the standard formula, and a manipulation of binomial coefficients. That is substantially more messy. It is useful to know that in many cases the expectation of a random variable $Y$ can be computed without explicit knowledge of the probability distribution function of $Y$.
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.
Best Answer
You are on the right path in solving the problem, however, there is a nice technique for evading the calculations, which is finding the expected value using indicator functions.
First, number the red balls from $1$ to $r$ and define $X_i$ to be the function indicating whether ball number $i$ is drawn before the first blue ball or not. So $X_i$ is $1$ if the i'th ball is drawn before the first blue ball and zero otherwise.
Now, it is easy to see that $X$ is the sum of indicators, i.e. $$X = 1 + \sum_{i=1}^r X_i $$
so the expected value of both sides are equal, and using the expected of the sum property and the symmetry between $\mathbb{E}[X_i]'s$, we get:
$$\mathbb{E}[X] = 1 + \sum_{i=1}^r \mathbb{E} [X_i] = 1 + r\mathbb{E}[X_1] = 1 + r\mathbb{P}[X_1] $$
the problem now boils down to finding the probability that the first red ball is drawn before the first blue ball, which is just $\frac{1}{b+1}$, since out of b+1 different ways of putting 1 red ball and b blue balls next to each other, only the one which the first red ball comes first is valid. you can then arrange other red balls where ever you want.
Finally, the expected value is given by:
$$\mathbb{E}[X] = 1 + \frac{r}{b+1} $$