Consider polya\'s urn model: an urn contains b black balls and rred balls. A bal

Question

Consider polya's urn model: an urn contains b black balls and rred balls. A ball is drawn at random. It is replaced, and , inaddition, c (>0)balls of the same color are added. A new ball isdrawn from the urn, and this procedure is repeated. The samplingcan continue for any number of draws from the urn. (Polya's urnmodel was introduced to model the spread of contagion in apopulation.)

Q1. Find the probability that the second ball drawn isblack.

Q2. Given that the second ball drawn was black, what is theprobability that the first was black?

Explanation / Answer

Q1. Case 1: ball drawn in the first draw isred         c morered balls are added to the urn.          forthe second draw, we have             no.of red balls = r + c            no. of black balls = b            P(the second ball drawn is black given the first balldrawn was red) = b/(r+b+c)        Case 2: balldrawn in the first draw is black        c more blackballs are added to the urn.         for thesecond draw, we have            no.of red balls = r           no. of black balls = b + c            P(the second ball drawn is black given that the first balldrawn was black ) = (b+c)/(r+b+c)     Hence P(the second ball drawn isblack) = b/(r+b+c) + (b+c)/(r+b+c)                                                                  =(2b+c)/(r+b+c)      Q2. We have to note here that the first draw happens beforethe second draw and it is independent of the outcomeof the second draw. P(first ball drawn was black given that the second balldrawn is black) = P(first ball drawn is black)                                                                                                              = b/(r+b)        c more blackballs are added to the urn.         for thesecond draw, we have            no.of red balls = r           no. of black balls = b + c            P(the second ball drawn is black given that the first balldrawn was black ) = (b+c)/(r+b+c)     Hence P(the second ball drawn isblack) = b/(r+b+c) + (b+c)/(r+b+c)                                                                  =(2b+c)/(r+b+c)      Q2. We have to note here that the first draw happens beforethe second draw and it is independent of the outcomeof the second draw. P(first ball drawn was black given that the second balldrawn is black) = P(first ball drawn is black)                                                                                                              = b/(r+b)

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