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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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    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

1 Answers 1

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When we draw the balls from urn 1 and urn 2 there are four things that can happen. Both balls can be black or both white, in which case $X_{n+1}=X_n$, or we can draw a white ball from bin 1 and a black ball from bin 2 in which case $X_{n+1} = X_n - 1$ or we can draw a black ball from bin 1 and a white ball from bin 2 so $X_{n+1} = X_n + 1$.

Now, at time $n$, as you have observed, bin 1 has $X_n$ white balls and $15-X_n$ black balls, and bin 2 has $10-X_n$ white balls and $15 - (10-X_n)=5+X_n$ black balls. So the probability of drawing a white ball from bin 1 and a black ball from bin 2 is $\frac{X_n}{15}\frac{5+X_n}{15}$ (we just multiply the probabilities because the two draws are simultaneous and independent.) Similarly, the probability that we draw a black ball from bin 1 and a white ball from bin 2 is $\frac{15-X_n}{15}\frac{10-X_n}{15}$. The probability that the two balls are the same is just 1 minus the sum or the two probabilities that they are different. So putting it all together, we get

$P(X_{n+1}=k | X_n=j)=\begin{cases}\frac{j^2+5j}{225}&\mbox{if }k=j-1\\\frac{-2j^2+20j+75}{225}&\mbox{if }k=j\\\frac{j^2-25j+150}{225}&\mbox{if }k=j+1\\0&\mbox{otherwise.}\end{cases}$

Check my algebra.

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    @Gerry Doh! I'll blame it on the superbowl.2012-02-06