I see a flaw with your second theory....
If the white ball is picked first, I agree that we have:
1/6 * 4/5 * 3/4 = 12/120
But, if the white ball is picked second, we should have:
4/6 * 1/5 * 3/4 = (still) 12/120
And, if the white ball is picked third, we should have:
4/6 * 3/4 * 1/5 = (still) 12/120
So, I would agree with your first post....
P = 12/120 * C(3,1) = 3/10
I'm still learning with these wonderful probability exercises... if there's a mistake in my logic above, please let me know! (But pls go easy on me!)
I racked my brain on this one, and I have a new solution. The logic is as follows
The white ball can be drawn in the first, second or the third try.
Probability of the white ball being drawn in the first try, with the black ball NOT being drawn in the second and the third tries is 1/6*4/5*3/4=12/120
Probability of the white being drawn in second try is 5/6*1/5*3/4=15/120
Probability of the white being drawn in third is 5/6*4/5*1/4 = 20/120
The sum is 47/120
This assumes that the balls are drawn without replacement. If they are drawn with replacement, then the answer will be 3C1*1/6*5/6*5/6 = 25/72
Note: What do you mean when you say the balls are drawn without repetition?