How many bit strings of length 8 start with "1" or end with "01"? A bit string is a finite sequence of the numbers $0$ and $1$. Suppose we have a bit string of length $8$ that starts with a $1$ or ends with an $01$, how many total possible bit strings do we have?
I am thinking for the strings that start with a 1, we would have $8 - 1 = 7$ bits to choose, so $2^7$ possible bit strings of length $8$ that starts with a $1$? 
Can I go about the second condition the same way and just add the total's together? That is, if my logic is even correct in the first place?

The strategy you seem to be proposing is to note that there are $2^7$ bit strings starting with $1$ and $2^6$ ending with $01$, since one may make $7$ choices in the first case and $6$ choices in the second. If we add these up to get $2^6+2^7$, this doesn't quite work to count the number of strings satisfying either condition. In particular, consider a string like
$$10000001$$
it both starts with $1$ and ends with $01$, so the above method would have counted it twice. In particular, the remedy for this is to subtract out the number of strings that satisfy both conditions from the sum $2^6+2^7$ to compensate for counting those strings twice.
This is the inclusion-exclusion principle.
