Each time you draw a ball, the probability that it's red is $n/X$. (The conditional probability that the second one is read, given that the color of the first one, depends on the color of the first one. But the probability that the second ball is red, given the composition of the population and no information about the color of the first one, is still $n/X$. Lack of replacement doesn't change that number, but it does alter the conditional probabilities.)
Here's a probability mass function: \begin{align} f(w) & = \Pr(\text{first red is on $w$th trial}) \\[10pt] & = \Pr(\text{exactly 2 red in first $w-1$ trials})\cdot\Pr(\text{red on $w$th trial} \mid \text{exactly 2 red in first $w-1$ trials}) \\[10pt] & = \frac{\text{number of ways to choose 2 red out of $n$ and $w-3$ others out of $X-n$}}{\text{number of ways to choose $w-1$ out of $X$}} \cdot \frac{n-2}{X-w+1} \\[10pt] & = \frac{\dbinom{n}{2}\dbinom{X-n}{w-3}}{\dbinom{X}{w-1}} \cdot \frac{n-2}{X-w+1}. \end{align}
Now one could say the expected number of trials is $\displaystyle\sum_{w=3}^X w f(w)$, and then seek a closed form.
I am not altogether happy with this answer for two reasons: (1) I haven't given the closed form; and (2) I think there ought to be a slick way that doesn't require finding $f(w)$. Often it's easier to find the expected value of a random variable than the whole probability mass function.
More later, perhaps . . . . . .