Different sample space for essentially the same problem

As an exercise in elementary probability theory, we had to determine the probabilistic models of extracting a ball from a box with $b$ black and $w$ white balls (this was exactly how the exercise was formulated).

Now I solved the exercise, by saying that $\Omega =\{x_1,\ldots,x_b,y_1,\ldots,y_w \}$ and with $p(t)=\frac{1}{b+w},\ t\in \Omega$

But in the solution to the exercise, it was indicated that $\Omega=\{b,w \}$ with $p(b)=\frac{b}{b+w},p(w)=\frac{w}{b+w}$.

Now my question is: Who was right ? Or - are we both right ?

My guess is that I modeled the case where we can distinguish the balls whereas in the solution the case is modeled, where one can't distinguish the balls (this case being a special case of mine, since $p(\text{"white"})=w\cdot \frac{1}{b+w}=\frac{w}{b+w}$. Am I correct ?

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You are both right. Because the two-element model is so simple, and in such widespread use, there is a case to be made that the answer given in the solution to the exercise is "better." But for example when we are tossing two dice, and are only interested in the total, the analogue of your answer would be better. –  André Nicolas Mar 11 '12 at 20:17
If your professor is repeating the experiment of drawing a ball (with replacement), your answer is that of a front-row student who sees the numbers marked on each ball and thus can, after a large number of trials, be fairly sure that the urn contains $w$ white balls and $b$ black balls (all numbers have been seen repeatedly). The book's answer is that of a back-row student who sees black and white balls being drawn from the urn but cannot see the numbers but can estimate $P(b)=\frac{b}{w+b}$. So the outcomes are only $2$ for the back-row student but $w+b$ for the front-row student. –  Dilip Sarwate Mar 11 '12 at 21:46