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Probability of taking different colored balls out of a jar

  1. Jul 25, 2009 #1
    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. jcsd
  3. Jul 25, 2009 #2

    HallsofIvy

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    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?
     
  4. Jul 25, 2009 #3
    Yes, this gives the exact same solution.
     
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