As a slight tweak to the problem, suppose the question was "what is the probability that at least one black ball is drawn in three b***s?"
Prob that 1st ball is non-black = 6/9
Prob that next ball is non-black = 5/8
Prob that final ball is non-black = 4/7
So probability of at least 1 black ball = 1 - 6*5*4/(9*8*7) = 16/21
Using DavidQD and Chris's values above: Probability of at least 1 black ball = 64/84 = 16/21
Using my answer to the original problem: Probability of at least 1 black ball = 384/504 = 16/21
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