Question 1151171
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Assuming there is no replacement of balls...

One color will be chosen twice, the other two colors will each be chosen exactly once, and we consider each color being drawn twice separately:

P(one of each color chosen on 4 draws) = 

(C(6,2)*C(4,1)*C(5,1) + C(6,1)*C(4,2)*C(5,1) + C(6,1)*C(4,1)*C(5,2)) / C(15,4)

where the 15 comes from the total number of balls, and C(n,r) = n!/((n-r)!r!)

This works out to about 52.7%