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 20120116 19:26:15 