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?

Here is another way to arrive at the answer, without doing the whole "double count and then correct for it" dance:
Of all possible octets (8-bit strings), half of them will begin with $1$.  Of the other half (i.e. those that begin with $0$), a quarter will end with $01$.  Since there are $2^8$ possible octets, we have:
$$
2^8 \times \frac{1}{2} + 2^8 \times \frac{1}{2} \times \frac{1}{4} \\
2^7 + 2^5
$$
While this may not look identical to the other answers, note that:
$$
2^5 = 2^6 - 2^5
$$
because
$$
2^6 - 2^5 = 2 \times 2^5 - 2^5 = 2^5 + 2^5 - 2^5 = 2^5
$$
