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Topic: More Balls and Urns (Read 1027 times) |
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ThudnBlunder
wu::riddles Moderator Uberpuzzler
The dewdrop slides into the shining Sea
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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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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
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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).
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« Last Edit: Dec 23rd, 2008, 8:14am by towr » |
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SMQ
wu::riddles Moderator Uberpuzzler
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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. --SMQ
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--SMQ
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