If the probability of chossing an interger "k" out of 2m integers `1,2,3,….,2m` is inversely proportional to `k^(4)(1leklem).Ifx_(1)` is the probability that chosen number is odd and `x^(2)` is the probability that chosen number is even, then

Options

A. `x_(1)gt1//2`
B. `x_(1)gt2//3`
C. `x_(2)lt1//2`
D. `x_(2)lt2//3`

Answer

A, C

Explanation

Let one probabilitt of choosing one integer k be P (K) `=lamda//k^(4).` (`lamda` is one constant of proportionality). Then
`underset(k=1)overset(2m)sum(lamda)/(k^(4))=1`
`or underset(k=1)overset(2m)sum(1)/(k^(4))=1`
Let `x_(1)` be the probability of choosing the odd numer. Then
`x_(1)underset(k=1)overset(m)sumP(2k-1)=lamdaunderset(k=1)overset(m)sum(1)/((2k-1)^(4))`
Also, `1-x_(1)=underset(k=1)overset(m)sumP(2k)`
`=lamdaunderset(k=1)overset(m)sum (1)/((2k)^(4))`
`ltlamdaunderset(k=1)overset(m)sum(1)/((2k-1)^(4))`
`implies1-x_(1)ltx_(1)`
`impliesx_(1)gt1//2`
`impliesx_(2)lt1//2`