Probability of taking different colored balls out of a jar

1. Jul 25, 2009

bantalon

1. The problem statement, all variables and given/known data
You are given a jar with 9 balls in it. 4 of which are red, 3 are green, and 2 are blue. The 9 balls are taken out of the jar one by one.
1. What are the odds that the first 3 balls have 3 different colors?
2. What are the odds that the last 3 ball have 3 different colors?
3. What are the odds the the first, the third, and the seventh balls have 3 different colors?

2. Relevant equations
Will be shown in the solution attempt.

3. The attempt at a solution
I think I managed to solve the first question, but I would appreciate your remarks, if you have any. I am looking for an elegant way to solve the second and the third questions.
1. [(4*3*2)*3!*6!]/9!
Explanation: We have 4 options of choosing red ball, 3 of green, and 2 of blue, we multiply by 3! as the order matters. We then multiply by 6! options of taking out the next 6 available balls. We finish by dividing by 9! which is total number of options of taking out the 9 balls.

2. Jul 25, 2009

HallsofIvy

Staff Emeritus
I would have done (1) in quite a different way. Initially, there are 9 balls, 4 of which are red. The probability that the first ball is red is 4/9. Now there are 8 balls, 3 of which are green. The probability that the second ball is green is 3/8. Finally, there are 7 balls left, 2 of which are blue. The probability the third ball is blue is 2/7. The probability that the first three balls drawn are "red, green, blue" in that order, is (4/9)(3/8)(2/7). Since there are 3!= 6 ways to order those three colors, the probability of three different colors (red, green, and blue) in any order is 6(4/9)(3/8)(2/7). An important observation in concluding that we can just multiply by 6 is that while the individual fractions will be different (for example, "blue", "green", "red" would be (2/9)(3/8)(4/7)) the numerators and denominators are the same, just with the numerators in different orders. Is that the same as what you got?

3. Jul 25, 2009

bantalon

Yes, this gives the exact same solution.