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# Intro Probabil MATH 425

UM

GPA 3.87

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This 1 page Class Notes was uploaded by Mrs. Preston Lehner on Thursday October 29, 2015. The Class Notes belongs to MATH 425 at University of Michigan taught by Staff in Fall. Since its upload, it has received 6 views. For similar materials see /class/231506/math-425-university-of-michigan in Mathematics (M) at University of Michigan.

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Date Created: 10/29/15

Luann 440 becnon 3 ball AUUO ART S CHIPS Problem We are given 71 chips which we are to place into two piles say h of them in a heads7 pile and t n 7 h of them in a tails7 pile We have a coin that comes up head with probability 1 We ip the coin and if it comes up head then we remove a chip from the heads pile if there are any remaining in it lf there are no chips left in the heads pile then we do nothing lf the coin comes up tails we remove a chip from the tails pile if there are any remaining lf there is no chip left in the tails pile then we do nothing We repeat this until all chips are removed Let X denote the total number of ips of the coin Then is a function of n p and h We have no control over 71 and p but we are free to choose h How should h be chosen so as to minimize Solution We treat this as an exercise in expectation computed by conditioning Let Y denote the number of head in the rst 71 tosses Thus Y is binomial with parameters np Given the value of Y it is easy to determine the expected number of additional tosses required to remove the remaining chips Suppose that Y k lt h Then there are no more chips in the tails pile and indeed there have been h 7 k occasions after that pile was exhausted when the coin came up tails and nothing was done However there are h 7 k chips left in the heads pile The number of further trials required can be modeled as a sum of h 7 k independent geometric variables and hence the expected number of further ips is h 7 lf Y k h then all chips have been eliminated with no waste and no further ips are required lf Y k gt h then all chips in the heads pile have been removed with a waste of k 7 h cycles and there are k 7 h chips remaining in the tails pile These can be eliminated in k 7 h1 7 p ips on average Let X denote the number of additional ips that are required after the rst 71 Thus X n X and hence Ele 7 n EiXi n Emmin n faxw kpy k 190 h TL 7 n Z ltpk17p k7h 7 k Z ltpk17p kilih 190 p khl 1 As a function of h this rst decreases and then increases To see this let fh denote the above Then after some simpli cations we nd that h7l n M 7 m 71gt7 Zpklt17pgtnk 7 Zpklt17pgt 190 19 7 YlthiPYZh T p 17p 39 lf PY lt h lt p then the rst term is lt 1 and the second term is gt 1 so the above is negative But if PY lt h gt p then the rst term is gt 1 and the second term is lt 1 so 1

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