SOLUTION: I have 27 balls. 14 are red and 13 blue. I deceded to group the balls into a group of 3s ? the balls will be choosen randomly. What is the probablity that the group will have 2 r

Question 167214: I have 27 balls. 14 are red and 13 blue. I deceded to group the balls into a group of 3s ?
the balls will be choosen randomly.
What is the probablity that the group will have 2 red and 1 blue.
2)-the probability that the group will have all reds or all blue
Found 2 solutions by scott8148, Edwin McCravy:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
possible groups of 3 out of 27 __ 27C3 __ 27*26*25/(3*2*1) __ 2925

possible groups of 2 red out of 14 __ 14C2 __ 14*13/(2*1) __ 91

possible groups of 1 blue out of 13 __ 13

probability of 2 red and 1 blue __ 91*13/2925 __ .404 (approx)

possible groups of 3 red out of 14 __ 14C3 __ 14*13*12/(3*2*1) __ 364

probability of 3 red __ 364/2925 __ .124 (approx)

possible groups of 3 blue out of 13 __ 13C3 __ 13*12*11/(3*2*1) __ 286

probability of 3 blue __ 286/2925 __ .098 (approx)
Answer by Edwin McCravy(20086) (Show Source): You can put this solution on YOUR website!
Edwin's solution:
I have 27 balls. 14 are red and 13 blue. I decided to group the balls into a group of 3s the balls will be choosen randomly.
Then one of the groups of three was chosen at random.
What is the probablity that the group chosen will have 2 red and 1 blue.

We can pick the two reds 14C2 ways and the one blue 13C1 ways. So the numerator of the probability is (14C2)(13C1) The denominator is 27C3. So the probability is 2)-the probability that the group will have all reds or all blue P(all reds or all blue) = P(all reds) + P(all blue) P(all reds) = P(all blues) = P(all reds or all blue) = P(all reds) + P(all blue) = = = . Edwin

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