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URN problem

  1. Aug 20, 2010 #1
    Plz tell the solution to this problem

    Each of n urns contains a white balls and b black balls; the urns are numbered 1,2,
    . . . , n. One randomly selected ball is transferred from the firrst urn into the second,
    then another from the second into the third, and so on. Finally a ball is drawn at
    random from the nth urn. What is the probability that it is white
  2. jcsd
  3. Aug 21, 2010 #2
    Try solving the case n=2.
  4. Sep 5, 2010 #3


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    Let Xn denote the color of the ball selected from the nth urn.

    The probability that you are looking for is P[Xn=w]. We have P[Xn=w]=P[Xn=w|Xn-1=w]P[Xn-1=w] + P[Xn=w|Xn-1=b]P[Xn-1=b].

    We know P[Xn-1=b]=1-P[Xn-1=w]. So we have P[Xn=w]=P[Xn=w|Xn-1=w]P[Xn-1=w] + P[Xn=w|Xn-1=b](1-P[Xn-1=w]).

    Here P[Xn=w|Xn-1=w] denotes the probability of drawing white from the nth urn given that a white ball has been drawn from the previous urn. So it is equal to (a+1)/(a+b+1). And P[Xn=w|Xn-1=b] is the probability of choosing a white ball from the nth urn given that we had a black ball from the previous urn; so it is a/(a+b+1). If put these values into the equality for P[Xn=w], if I am not making a mistake, we obtain,

    P[Xn=w]=P[Xn-1=w]/(a+b+1) + a/(a+b+1),

    which is a recurrence relation with initial value P[X1=w]=a/(a+b), which can be solved to obtain the solution (which also makes me think there should be a much easier way to find that probability.)
  5. Sep 6, 2010 #4
    Is this true?

    All urns have the same amount of balls to start with: a+b
    Every time a ball is drawn there is probability a/(a+b) for the first urn and a/(a+b+1) or (a+1)/(a+b+1) for every next urn.
    Depending on which color the ball has a or b is increased by one.
    So it doesnt seem to matter how many urns there are!.
    It is always a+1 or b+1 to be divided by (a+b+1),
    so the probability for urn n = P(W) = (a+a/(a+b))/(a+b+1).
    Does that make sense?


    Last edited: Sep 6, 2010
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