Consider an urn with 12 beads, four of each color: red, yellow and blue. Pull these beads out in pairs. What is the probability that among these six pairs, every color combination is represented?
Consider an urn with 30 beads, 10 of each color. Pull these beads out in triplets. What is the probability that every color combination is represented?
I agree with Charlie's answer, 64/385. My laborious calculation, using conditional probabilities:
a) PR2 = The probability of two red pairs: Start with a
red. The probability that it pairs with another red =
3/11. Then pick a red not in that pair. The probability
that it pairs the last red = 1/9. So, the probability of two red
pairs = (3/11)*(1/9) = 1/33.
b) PR0 = The probability of no red pairs. Start with a red.
The probability that it does not pair with another red = 8/11.
Then pick a different red. The probability that it also does not
pair with a red = 7/9. Then pick a third red. The
probability that it also pairs a non-red, given that there are two
non-red pairs already, = 6/7. So PR0 = (8/11)*(7/9)*(6/6) = 16/33
c) PR1 = The probability of exactly one red pair = 1 - PR2 - PR0 = 16/33
d) Given that there is exactly one red pair, the probability that the
4th red is matched to a color different from the 3rd red = 4/7.
e) Given that there is a RR, a RY, and a RB, the probability that the other three are YY, BB, YB = 3/5.
So, the requested probability = (16/33)*(4/7)*(3/5) = 64/385
I don't recommend this approach for Part II, especially since there are more steps than what I wrote down. :-)
Part I (the long solution)
