Probability of an event if the sample space has identical elements Suppose we have a box, with only one small hole. Suppose 10 distinct black balls and 20 distinct white balls are put in the box. Now, in a random draw of 1 ball, the probability that the ball drawn is black is  $$\frac{\text{number of ways in which the favourable event occurs}}{\text{total number of outcomes}}$$
where the event is the drawing of a black ball, which comes out to be $\frac{10}{30}$. So far, so good. 
Now, suppose that we have another box, again with a small hole. Suppose also that 10 identical black balls and 20 identical white balls are put into the box. What is the probability now, that a randomly drawn ball will be black?
According to my understanding, (which I'm not too sure about) there are only 2 distinct outcomes - drawing a white ball and drawing a black ball, because all the balls are identical. From that point of view the probability should be $\frac{1}{2}$.
On the other hand, it just doesn't seem right. This logic says that whatever be the number of items of each type, only the number of types can affect the probability. But that doesn't seem realistic. If I hade a huge number of white balls, and only 1 black ball the probability of getting a black ball would be very very small. But this result says otherwise. 
I'm in a dilemma here, can any one sort out this problem?
 A: The answer is again $\frac{1}{3}$, as can be seen from the fact that in your answer to the first part you did not [need to] rely on the distinctness of the white (black) balls from others of the same colour.
Suppose some outside agent has labelled the balls from $1,\dots,30$ in some way that is undetectable to the touch. Now the balls are again distinct, and the probability of drawing a black ball is again $\frac{10}{30}$. The mere act of labeling the balls did not change their probability of selection (since it was undetectable to you), so this probability must be the same as for unlabeled balls.

Probability of choosing 1 black ball and 0 white balls from 10 non-identical black balls and 20 non-identical white balls is:
$\Pr[X=1] = \dfrac{{10 \choose 1}{20 \choose 0}}{{30 \choose 1}} = \dfrac{10 \times 1}{30} = \dfrac{10}{30}$ 
If on the other hand the black balls are identical and we wish to rely on this fact, the answer is obtained by simple ratio of number of black balls to the total:
$\Pr[X=1] = \dfrac{10}{30}$ 
