Suppose we have 2 sets of billiard/pool balls, what is the probability of drawing 4 unique balls? The Cue ball is not included in either set, so we just have 1-15 in each set (30 total balls).
 A: Starting with the second draw, the probability of drawing a different ball is $28/29$ (there is only 1 ball out of 29 left that would be not unique).  On the third draw, if the second was unique, the probability is $26/28$.  The probability of the fourth being unique, given that the first 3 are distinct, is $24/27$.  So the answer is 
$$\frac{28}{29}\times\frac{26}{28}\times\frac{24}{27}$$
A: Hint:
Draw the balls in sequence rather than simultaneously.
The first ball cannot match any previous balls drawn (since there were no previous balls to match with).
Once the first ball has been drawn and when going to draw the second ball, only $28$ of the remaining $29$ balls are a number different than what was drawn in the first round.
Once the first and second ball have been drawn, given that they were both different, how many balls remain that don't match either of the first or second ball?
Continue in this fashion and apply what you know about conditional probability, in particular that $Pr(A\cap B\cap C\cap D) = Pr(A)Pr(B\mid A)Pr(C\mid A\cap B)Pr(D\mid A\cap B\cap C)$
