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In the polya urn scheme what is the probability that the 3rd ball is red given exactly one of the first four balls is red?

so P(R3|exactly one of the first four balls is red) =

P(R3 and exactly one of the first four balls is red)/(exactly one of the first four balls is red)

P(R3) = P(R1)P(R2|R1)P(R3|R2 andR1) + P(B1)P(R2|B1)P(R3|B1 and R2) + P(B1)P(B2|B1)+P(R3|B1 and B2) + P(R1)P(B2|R1)P(R3|R1 andB2).

This is equal to (r/b+r)(r+c/b+r+c)(r+2c/b+r+2c) + (b/b+r)(b+c/b+r+c)(r/b+r+2c) + (b/b+r)(r/b+r+c)(r+c/b+r+2c)+(r/b+r)(b/b+r+c)(r/b+r+2c) = r/b+r I think.

P(exactly 1 is red out of 4) = (4 choose 1)(r/b+r)(b/b+r)^3

So do you divide P(R3)= r/b+r by P(exactly 1 is red out of 4)?

I am stuck here.

Also how would you find the probability that the 2nd ball is red and the nth ball is red? The probability second ball is red is still r/b+r. The probability that the nth ball is red is 1. So would it just be r/b+r?

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I might be able to help you if I knew what the Polya scheme was... – mixedmath Sep 13 '11 at 21:46
Can anyone help me? – lord12 Sep 13 '11 at 22:03
up vote 2 down vote accepted

There are four possibilities of having exactly one of the first four balls red, and one of these will have the third ball red. You need to calculate the probabilities:

  • $\Pr(R_1,B_2,B_3,B_4)$
  • $\Pr(B_1,R_2,B_3,B_4)$
  • $\Pr(B_1,B_2,R_3,B_4)$
  • $\Pr(B_1,B_2,B_3,R_4)$

Your answer will then be the third divided by the sum of all four.

Hints: $$\Pr(B_1,B_2,R_3,B_4) =\Pr(B_4 | B_1,B_2,R_3)\Pr(R_3 | B_1,B_2) \Pr(B_2 | B_1) \Pr(B_1)$$ and $$\Pr(B_4 | B_1,B_2,R_3)= \frac{b+2c}{r+b+3c}.$$

You should find the denominators keep repeating and useful repetition in the numerators too. The final answer is wonderfully simple.

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Thanks for the help. But isn't the probability of exactly one red in 4 balls just a binomial probability (4 choose 1)(r/b+r)(b/b+r)^3? – lord12 Sep 13 '11 at 22:38
No - the point about a Polya urn model is that the probability of each event does depend on what happened before. So if you draw a red ball then you put it back with $1$ more red ball (or, if I read your link correctly, $c$ more red balls) so increasing the probability that the next ball drawn is also red. – Henry Sep 13 '11 at 23:10

The final answer is wonderfully simple because the problem exhibits a wonderful form of symmetry called exchangeability. There is a explanation of how to use exchangeability in probability calculations for Polya's urn scheme in Section 11.2 (page 135-136) of Problems and Snapshots from the World of Probability by Blom, Holst, and Sandell.

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