f(n) represents the sum of the digits of the binary representation of n, where n is a positive integer with 1 ≤ n ≤ 2012.

Determine the probability that f(n) ≥ 3. What is the probabiliy that f(n) ≥ 4?

The probability of f(n)
>=  1 is 2012/2012 = 100.00%
>=  2 is 2001/2012 ~= 99.45%
>=  3 is 1946/2012 ~= 96.72%
>=  4 is 1781/2012 ~= 88.52%
>=  5 is 1451/2012 ~= 72.12%
>=  6 is  989/2012 ~= 49.16%
>=  7 is  528/2012 ~= 26.24%
>=  8 is  203/2012 ~= 10.09%
>=  9 is   49/2012 ~=  2.44%
>= 10 is    6/2012 ~=  0.30%
>= 11 is    0/2012  =  0.00%

Edited on January 16, 2012, 7:27 pm

Posted by Dej Mar on 2012-01-16 19:26:15