This answer is pretty good. But only accounts for one of the combinations that have 3 different colours of the 27 possible outcomes.
There are 6 occasions where the balls are different colours therefore multiply your answer by 6.
An urn contains n red balls, n white balls and n blue balls. Three balls are drawn at random from the urn, one at a time, without replacement.
What is the probability that three balls are all different colours.
Answer: n/3n x n/3n-1 x n / 3n-2
Let me baby-step through an explanation.
Sorry? There are six different occasions?
If it's different colours, then its red, blue and white isnt it?
What other possible combinations can there be?
What is the probability that the balls are drawn in the order Red-White-Blue?
What is the probability that the balls are drawn in the order Blue-Red-White?
What is the probability that the balls are drawn in the order White-Red-Blue?
. . . etc. . . . etc.
Ok. The answer is actually infact (nC3)^3/3nC3. But i was confused here, since i thought it they picked one and didnt replace it, then it cant be computed using combinatorics.
Namely, if the question stated, if it was picked and replaced, is the solution still (nC3)^3/3nC3?