IIT JEE , AIEEE Forum - Mathematics , Algebra

rishipratimm

if the sum of two +ve integers is 2n,then find the probability that their product is greater than 3/4 th of their maximum possible product.

i)2/n

ii)n/n-1

iii)n-1/2n-1

iv)none of these

sboosy

a+b = n
Max. product is obtained when nos. are equal
Thus max. product is n^2

Let a=k,b=2n-k
k(2n-k) >3/4 n^2
which gives
4k^2 - 8nk +3n^2< 0
(2k-3n)(2k-n)< 0
n/2 < k <3n/2
Thus there are n-1 possibilities for k

In general a and b can take values from 1,2,3...2n-1

Thus required probability is (n-1)/(2n-1)

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