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Homework Help: Probability (balls in a bucket)

  1. Dec 16, 2007 #1
    Q: 2 blue balls and 1 white ball are in a bucket. You draw one ball at a time, record the color of the ball and put it back to the bucket until you get at least 2 blue balls and 2 white balls. Let X be the number of blue balls you drew, Y be the number of white balls you drew, Z be the number of balls you drew. What is the probability mass functions of X,Y, and Z? (i.e. give the probabilities of all the possible outcomes for X,Y,Z, respectively)

    I understand the question, but don't understand the solution. Can someone please explain/help me out?

    Thanks a lot!
  2. jcsd
  3. Dec 16, 2007 #2


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    Are you saying you don't understand a given answer? Or are you saying you don't "see" the solution?

    What is the prob. z = 1? (It's zero. Why?)
    What's the prob z < 4? (Again 0.)
    Prob{z=4} = number of ways 2B+2W balls can be obtained/number of all color combinations in 4 draws
  4. Dec 16, 2007 #3


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    Just work on it systematically. First of all, you know what a pmf is:

    pX(x) = P(X = x)

    -- pY(y) and pZ(z) are defined similarly.

    Now, if X is the number of blue balls drawn, we know that X must be *at least* 2, because we're not going to *stop* drawing balls out until we get two blue ones. Therefore:

    pX(x) = 0 for x < 2
    likewise for pY(y)

    similarly, pZ(z) must be zero for z < 4, because that's the minimum number of draws required to get 2 blue and 2 white. So, that's a start...
  5. Dec 16, 2007 #4
    I don't understand a given answer...

    According to the solutions:

    P(X=2)=(3C1) (2/3)2 (1/3)2 + (2C1) (2/3) (1/3)2 + (1/3)2
    P(X=3)=(4C1) (2/3)2 (1/3)2

    P(Z=4)=(4C3) (2/3)2 (1/3)2

    But I don't get it...can someone please explain? :)
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