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

 

[blue_raver22]

?

The solution to this problem involves using the concept of conditional probability. We can approach this problem by breaking it down into smaller steps.

First, we need to find the probability of drawing a white ball from the first urn. This is simply the number of white balls in the first urn divided by the total number of balls in the first urn (white and black). Let's say this probability is denoted as P(W1).

Next, we need to find the probability of drawing a white ball from the second urn, given that a white ball was transferred from the first urn. This is denoted as P(W2|W1). This can be calculated by dividing the number of white balls in the second urn (after the transfer) by the total number of balls in the second urn (after the transfer).

Similarly, we can find the probability of drawing a white ball from the third urn, given that a white ball was transferred from the second urn. This is denoted as P(W3|W2).

We continue this process until we reach the nth urn. The final probability we are looking for is the probability of drawing a white ball from the nth urn, given that white balls were transferred from all the previous urns. This can be written as P(Wn|W1,W2,...,Wn-1).

Using the concept of conditional probability, we can calculate this final probability by multiplying all the individual probabilities together. This can be written as:

P(Wn|W1,W2,...,Wn-1) = P(W1) * P(W2|W1) * P(W3|W2) * ... * P(Wn|Wn-1)

Substituting the individual probabilities we calculated in the first step, we get:

P(Wn|W1,W2,...,Wn-1) = (number of white balls in first urn / total number of balls in first urn) * (number of white balls in second urn / total number of balls in second urn) * ... * (number of white balls in nth urn / total number of balls in nth urn)

This final probability gives us the likelihood of drawing a white ball from the nth urn, given that white balls were transferred from all the previous urns. Therefore, it is the probability that we are looking for in this problem.