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You have 30 balls 20 black and 10 white let $X_{n}$ be the current number of white balls in urn 1 and each urn holds 15 balls, find
$P_{r}$ ($X_{n+1}$)=k |$X_{n}$=j).

You are taking a ball from urn 1 and urn 2 and placing the ball from urn 1 into 2 and the ball from urn 2 and putting it in urn 1. I should have clarified that earlier.

So I have this representation Urn 1: \begin{equation} \frac{X_{n}\:\:\:15-X_{n}}{15}; \end{equation} with $15-X_{n}$ representing black balls in urn 1

Urn 2: \begin{equation} \frac{10-X_{n}\:\:\:15-(10-X_{n})}{15}; \end{equation}

with $\frac{15-(10-X_{n})}{15}$ being the black balls and $\frac{10-X_{n}}{15}$ being the white balls in urn 2.

So would my answer be $P_{r}$ ($X_{n+1}$)=k |$X_{n}$=j)=$ \frac{15-(10-X_{n})}{15}; $

I realize that my variables might not match up at this point; but I think I'm on the right track but I am just missing something.

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    I think you forgot to write part of the question: what kind of ''2 urns and 30 balls'' problem is it? What does the $n$ in $X_n$ mean?2012-02-05
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    Each urn holds 15 balls each and $X_{n}$ is the current number of balls in urn 1.2012-02-05
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    What are you doing to the balls to transition from one state to another? You have to be doing something to get from one state to the next, taking one ball out of each urn and putting it into the other, taking a ball out of a random urn and painting it black and putting it back, etc. Without knowing the process, finding the probabilities is difficult.2012-02-06
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    You are taking a ball from urn 1 and urn 2 and placing the ball from urn 1 into 2 and the ball from urn 2 and putting it in urn 1. I should have clarified that earlier.2012-02-06
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    Your final formula doesn't depend on $k$; it gives the same answer for 10 white balls as for 1; can that possibly be right?2012-02-06

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