A problem on my discrete mathematics homework.
My thought process is as follows: For there to be at least twice as many 0's as 1's, there can only be at most, 2 1's.
Total possible strings: $2^8 = 256$
Exactly 2 1's $= C(8, 2) = 28$
Exactly 1 1's $= 8$
Does that mean that for there to be at least twice as many 0's as 1's, I would have $8+28$ possible strings, leaving $36$ total?