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Topic: Molina's urns (Read 2867 times) |
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karcoms
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Molina's urns
« on: Oct 12th, 2005, 11:38pm » |
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Two urns contain the same total numbers of balls, some blacks and some whites in each. From each urn are drawn n balls with replacement, where n >= 3. Find the number of drawings and the composition of the two urns so that the probability that all white balls are drawn from the first urn is equal to the probability that the drawing from the second is either all whites or all blacks.
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karcoms
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Re: Molina's urns
« Reply #1 on: Oct 12th, 2005, 11:45pm » |
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This is a very tricky question that converts itself to one of the fundamental number theory equations - Fermat's theorem
Let x be the number of white balls in the first urn (obviously it has to be all whitwe since we want all white balls to be there)
Let y be the number of white balls in the 2nd urn and z be the number of black balls in the 2nd urn. The 2nd urn has both white and black and the total = y + z which also happens to be the same number of balls in the first urn
So we have the probabilities as follows:
first urn -> x / y + z in each turn hence for n turns we have
p1 = (x / y + z)^n
Similarly for the 2nd urn we have probabilities as follows:
For all whites: (y / y + z)^n
For all Blacks: (z / y + z)^ n
Hence for all whites OR all blacks add the 2 probs
= (y / y + z)^n + (z / y + z)^ n
Now the questions says the 2 probs from the 2 urns are equal hence the equation is
(x / y + z)^n = (y / y + z)^n + (z / y + z)^ n
Hence x^n = y^n + z^n
On an observation it is said that for n < 2000 this is impossible but maybe for some n > 2000 it is possible. Interesting problem.
MNK
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Barukh
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Re: Molina's urns
« Reply #2 on: Oct 12th, 2005, 11:59pm » |
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There is an older thread discussing thins problem.
on Oct 12th, 2005, 11:45pm, karcoms wrote:| On an observation it is said that for n < 2000 this is impossible but maybe for some n > 2000 it is possible. Interesting problem. |
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What do you mean? The non-existence of a solution for any n was proved 10 years ago.
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karcoms
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Re: Molina's urns
« Reply #3 on: Oct 13th, 2005, 7:25am » |
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I tried searching the forum and could not find it, hence thought I would start the thread with this information. Thanks for the update, and yes ,there is no solution for n >= 3
MNK
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