It appears to me that 1/2 of all possible strings start with 0, and 1/4 end with 00. So the issue then is how many bit strings of eight digits can be made.
check my thinking.
How many bit strings of length eight either start with a 0 bit or end with the two bits 00?
3 answers
If the first bit (left most bit) is a 0, then it can be filled in only each of the bit string can be filled in one way and when bit strings of length eight end with the two bits 00. Each of the remaining five position is represent in the bit string can be filled in 2 ways (i.e., either by 0 or 1).
Hence, there are 1×2×2×2×2×2×1×1 = 25 = 32 bit strings of length eight either start with a 0 bit or end with the two bits 00.
Hence, there are 1×2×2×2×2×2×1×1 = 25 = 32 bit strings of length eight either start with a 0 bit or end with the two bits 00.
if it begins with 0 we have two to the power of 7 ways of choosing the rest, and if it ends in 00 then we have two to the power of 6 ways of choosing the rest. As the question uses the word OR we add these results to get our answer!!
3 answers