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   Author  Topic: More Balls and Urns  (Read 1027 times)
ThudnBlunder
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More Balls and Urns  
« on: Dec 23rd, 2008, 7:57am »
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An urn contains w > 2 white balls and b black balls. If 3 balls are wihdrawn from the urn the probability that they are all white is p. Having another white ball in the urn would increase this probability to 4p/3.
 
What is the maximum possible value of b?
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towr
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Re: More Balls and Urns  
« Reply #1 on: Dec 23rd, 2008, 8:13am »
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    p = w/(w+b)(w-1)/(w+b-1)(w-2)/(w+b-2)
4/3p =(w+1)/(w+b+1)(w)/(w+b)(w-1)/(w+b-1)
 
divide the two to get  
4/3 = (w+1)/(w+b+1) / ((w-2)/(w+b-2))
4(w+b+1)(w-2) - 3 (w+1)(w+b-2) = 0
b = (w-2)(w+1) / (11-w)
 
b = 88 is the maximum (for w=10).
« Last Edit: Dec 23rd, 2008, 8:14am by towr » IP Logged

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SMQ
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Re: More Balls and Urns  
« Reply #2 on: Dec 23rd, 2008, 8:20am »
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Darn, towr beat me to it -- same calculation.  The other possible values are: w = 5, b = 3; w = 7, b = 10; w = 8, b = 18; and w = 9, b = 35.
 
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