Probability of getting a red or white ball on first draw:$\frac{1}{2}$
probability of getting the remaining ball that is either red or white $\frac{1}{3}$
probability both of these happen: $\frac{1}{2}\cdot \frac{1}{3}\approx0.167$
Think of all the balls as distinguishable, and imagine taking out the balls one at a time until they are all gone. Then all sequences of balls are equally likely. Temporarily, let us call Red Ball $i$ and all the black balls special. Each of the $m+1$ special balls is equally likely to be the first special ball drawn. It follows that the probability Red Ball $i$ is drawn before any black is $\frac{1}{m+1}$.
The Bernoulli random variable $X_i$ takes on value $1$ with probability $\frac{1}{m+1}$, and therefore $E(X_i)=\frac{1}{m+1}$.
By the linearity of expectation we then have $E(X)=E(X_1)+\cdots+E(X_n)=\frac{n}{m+1}$.
Remark: In the second part of the post, you were going after the distribution of the random variable $X$. So let us do that. There are $\binom{n+m}{m}$ equally likely ways to choose the positions of the $m$ black balls. Now we count the number of ways to choose positions of the black balls if the first $k$ positions are to be black and the next one red. That leaves $n+m-k-1$ positions, and $m-k$ black. Positions for them can be chosen in $\binom{n+m-k-1}{m-k}$ ways, or equivalently $\binom{n+m-k-1}{n-1}$ ways. It follows that
$$\Pr(X=k)=\frac{\binom{n+m-k-1}{n-1}}{\binom{m+n}{m}}\tag{1}$$
for $k=0,1,\dots, m$.
Now we could use (1) to write down an expression for $E(X)$. With some fooling around with binomial coefficients, we could after a while simplify this to $\frac{n}{m+1}$. However, that is a painful way to find $E(X)$. The method of Indicator Random Variables described in the first part of the OP is far easier.
Best Answer
Hint: what is the probability that ball #$j$ has a label larger than that of ball #$1$?