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29 Dec 2007 10:16:35 IST

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Probablity question-1

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Suppose a sample space consist of the integers 1,2,3,,,,,,,,,,,,,,,,,,,2n.The probablity of choosing integer k is proportional to log k.Show that the conditional probablity of choosing the integer 2,given that an even integer is choosen, is log2/nlog2+log(n!) .

Please answer it with giving total solution.

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Dinesh Babu

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Joined: 18 Dec 2006

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29 Dec 2007 10:31:12 IST

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see this,

p(2/even) = p(2 n even) / p(even) rite??
now,
p(even) = p(2) + p(4) + .... p(2n)
p(even) = k.(log2 + log4 + .... log 2n) * k is some real constant
p(even) = k.(log(2.4...2n)
p(even) = k.(log(2^n(1.2.3..n)) %%important man%%
so,
p(even) = k.log(2^n.n!) rite??
p(even) = k(nlog2 + log(n!))
and p(2 n even) = k.log2 ( choosing 2 and even is equivalent to choosing 2)

so p(2/even) = k.log2 / k.(nlog2 + log n!)

cancelling k gives ur ans
got it?????

puneet

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Joined: 19 Oct 2006

Posts: 1966

29 Dec 2007 13:51:32 IST

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hii

well the answer is here .. i must point out that it is important here thta u take

P(n) as k.log(n) ..

gud work .

cheers

dont u know me

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Joined: 5 Dec 2007

Posts: 704

29 Dec 2007 23:49:05 IST

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hey i dont understand d quesn n d ansr explain plz

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