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Consider a polyurn scheme of more than two colors. Let us draw a ball from the urn and replace it with another ball of the color we picked from the urn. We assume that $w$ is the number of white balls in the total of $N$ balls.

If $S_n$ is the number of white balls in a random sample of $n$ random draws, we need to obtain a recursive equation for $\Pr(S_n=x)$, where we have $$P(S_1=x)= \left(\frac{W}{N} \right)^X \left(\frac{N-W}{N} \right)^{1-X} ,$$ and we have $X=0,1$.

My approach1:

Given: Sn= Number of white balls in a random sample of n draws. To start with , we have P(Sn=x) where x =1 , which means we picked a white ball, the probability of picking a white ball can be written as: W/N.

Next we have P(Sn=x) where x=0, which means we did not pick a white ball. Then the probability of NOT picking a hite ball can be written as 1-W/N.

If we combine the first 2 equations i just obtained, we get the probability p(SN=X) = (W/N)^X * (1-W/N)^(1-X) => lets call this equation 3. where u put in values of X as 0 and 1. Lets put X=1 in equation 3, we get W/N. Lets put X=0, we get (1-W/N)

My approach 2:

P(S2 = 1) which is the probability of getting one white ball in a sample of (n=2) draws. There can be two ways of drawing the said combination. White- Non White or Non white - Whit. therefore the probability can be written as:

= 2 * (W/N)*(N-W/N+1)

P(S2=0) which is the probability of getting ZERO white balls in a sample of (n=2) draws.

= 2 * (N-W/N+1) * (N-W+1/N)

P(S3 = 1) which is the probability of getting one white ball in a sample of (n=3) draws. The various forms the draws can assume are => W-NW-NW , NW-W-NW, NW-NW-W. (Where NW=> Non white and W=> White) When we write down the probabilities we obtain:

= [3*(W/N)(N-W/N+1)(N-W+1/N+2)]

P(S3=0) which is the probability of getting ZERO white balls in a sample of (n=3) draws. The various combinations possible are: NW-NW-NW during all the three draws since we get zero white balls. Writing the probabilities: = [3*(N-W)(N-W+1)(N-W+2)] / [N*(N+1)*(N+2)]

P(S4 = 1) which is the probability of getting one white ball in a sample of (n=4) draws. The various combinations possible are: W-NW-NW-NW , NW-W-NW-NW, NW-NW-W-NW, NW-NW-NW-W (Where NW => NON WHITE & W => WHITE) Writing the probabilities: =4*(W/N) * ((N-W) / (N+1)) * (N-W+1/ N+2) * ((N-W+2) / (N+3))

P(S4=0) which is the probability of getting ZERO white balls in a sample of (n=4) draws. The various combinations possible are: (NW-NW-NW-NW ) *4

=4*(N-W/N) * (N-W+1)/(N+1) * (N-W+2 / N+1) * (N-W+3 / N+3)

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Given: Sn= Number of white balls in a random sample of n draws. To start with , we have P(Sn=x) where x =1 , which means we picked a white ball, the probability of picking a white ball can be written as: W/N. Next we have P(Sn=x) where x=0, which means we did not pick a white ball. Then the probability of NOT picking a hite ball can be written as 1-W/N. If we combine the first 2 equations i just obtained, we get the probability p(SN=X) = (W/N)^X * (1-W/N)^(1-X) =>eqn 3. where u put in values of X as 0 and 1. Lets put X=1 in equation 3, we get W/N. Lets put X=0, we get (1-W/N) –  user19974 Nov 27 '11 at 1:48
    
@srivatsan could u tell me if my approach to the problem is correct? –  user19974 Nov 27 '11 at 4:04
    
I had a look at your computations. Your P(S2=1) (which I indicated) is correct but explained in a way and written in an order such that I doubt you know why. Your P(S2=0) is wrong.You do not mention P(S2=2) and you do not check that the sum of the nonzeroes P(S2=k) is 1. I did not read further. I suggest you go back to this n=2 case and try to solve it completely and accurately. // Please avoid fallacies like my god this is only a simple case, how am I going to do the general n case, the opposite is true! If you solve the n=2 case and understand it, helping you through the rest will be easy. –  Did Nov 29 '11 at 7:18
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1 Answer

could u tell me if my approach to the problem is correct?

No it is not.

For example $P(S_n=0)$ is not $1-P(S_n=1)$ unless $n=1$ because $S_n$ is any integer from $0$ to $n$ with positive probability.

And $P(S_n=1)$ is not $W/N$ unless $n=1$ because one wants the probability that, amongst the $n$ first draws, exactly one produced a white ball. What you write is the probability that the first draw produced a white ball. And note that the total number of balls is not the same before every draw.

Take $n=2$. Then the draw white-nonwhite has probability $(W/N)(1-(W+1)/(N+1))$ (why?) and the draw nonwhite-white has probability $(1-W/N)(W/(N+1))$ (why?) hence $P(S_2=1)$ is their sum $2W(N-W)/(N(N+1))$.

You might want to expand on this and, for starters, to write down the full distribution of $S_2$.

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thanks for the correction and Whats the right way to solve this problem? –  user19974 Nov 27 '11 at 15:46
    
My post is full of leads. Did you try to follow any of them? For starters, you could complete the distribution of $S_2$. –  Did Nov 27 '11 at 16:12
    
I could not exactly follow what u wrote here. could you be able to elaborate a bit more? –  user19974 Nov 27 '11 at 21:53
    
I certainly could if I would but, since you seem to refuse to follow ANY of the suggestions I made in my post, I fail to see the point. Please do not tell me you cannot complete the distribution of $S_2$, because, as they say, this would insult my intelligence... :-) –  Did Nov 27 '11 at 22:20
    
I do understand what u hv posted and follow it completely. My question is will the expansion continue as the next two combinations being white-white and then non white-non white? –  user19974 Nov 27 '11 at 23:30
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