IIT JEE , AIEEE Forum - Mathematics , Algebra - Probablity question-1

Anupamagrawal13a

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.

Rates to all replies.

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dinesh_ddt

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

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

ayshwarya

hey i dont understand d quesn n d ansr explain plz

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