Probability of an 8-bit binary string that begins with 111 or ends with 111 
Consider the uniform probability space where the set of outcomes consists of all $8$-bit binary strings. What is the probability of the following event?
A string that begins with $111$ or ends with $111$

So since each bit is independent of the next, I took the space of all three bit strings and find the probability that a given one is $111$. I calculated this to be $1/8$. Now for the end being $111$, I also calculated that to be $1/8$.
Finally I said $Pr[\text{Begins with "111" or ends with "111"}] = 1/8 + 1/8 = 1/4.$
Am I doing this problem correctly?
 A: You're almost right.
Don't forget the strings that both begin and end with $111$. By only counting either, you overcount by the number of strings that have both. It's basically the inclusion-exclusion principle at work. 
That said, this is on the assumption that "or" here is not the exclusive "or". If it is the exclusive "or", in that you count those that begin or end with the string, but not those with both. I feel like it's likely to be the former though - but at the same time it is ambiguous, so it's hard to give a conclusive answer on the matter.
Rectifying the matter is simple, though:


*

*In the "inclusive or" case, you need to account for those that have both kinds and the double-counting that resulted. Note that there are $2^2$ out of $2^8$ total strings that meet both conditions (ergo, you'd need to subtract $2^2/2^8 = 1/2^6 = 1/64$.

*In the "exclusive or" case, you cannot count the strings that have both whatsoever. Those have been included twice over from the counting of strings that meet each condition. $1/64$ of the strings meet both conditions, and they were added in twice - once from the counting of the strings that start in $111$ and once from counting the strings that end in it. Therefore, you need to subtract $1/64$ twice to remove their probability (i.e. subtract $1/32$ from the probability).

Note: in the comments of this post OP clarified that it was probably the inclusive case, but I'm including both for posterity because it's still useful knowledge in general. I made some edits after Henry noted that I hadn't done the calculation quite correctly (I had forgotten that the "exclusive or" case meanted we needed to drop even more strings).
