Solving the URN Problem: Finding the Probability of Drawing a White Ball

[stuttgart311]

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

 

[awkward]

Try solving the case n=2.

 

[fi1]

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.)

 

[Stellaferox]

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 doesn't 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?

grtz,

Marc