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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?

http://docs.google.com/viewer?a=v&q=cache:nJ28V6_L-RcJ:www.stat.berkeley.edu/~sourav/exercise.pdf+polya+urn+scheme&hl=en&gl=us&pid=bl&srcid=ADGEESgyeAHWaSTiuBbYmBBOnsMpmfG2EX7UdfmbZ9UHO0VSCpUNuvP8ognAso-7pWryOfcht5ZpRMGgU70w4OXNi3OH6p5otHaBEZqML0WaCMjysT15VpxPpfQDzW80u1vLkRDtQXm8&sig=AHIEtbTTtlJD5yTkqPJ_gVNViBDbcVa_5A

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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    Can anyone help me?2011-09-13

2 Answers 2

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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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    No - the point about a [Polya urn model](http://en.wikipedia.org/wiki/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.2011-09-13
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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.