Trouble understanding probability. A box contains six 40-W bulbs, five 60-W bulbs, and four 75-W bulbs. If bulbs are selected one by one in random order, what is the probability that at least two bulbs must be selected to obtain one that is rated 75-W?
The answer stated that $A$ is the event that two bulbs are selected and at least one is rated 75-W. The answer then goes on to say that $A'$ is the event that only one bulb must be selected to obtain a 75-W bulb.
I am not understanding how they made the jump from $A$ to $A'$. How is selecting one bulb the complement of selecting at least two? For example I understand that the complement of selecting all bulbs is selecting no bulbs. I am not really understanding the "at least" part of the problem. Also there is another problem which included "at most" and the complement of "at most" was "all of". Could someone explain this to me?
 A: I  draw bulbs one at a time until I get a $75$ watt bulb. Because I am easily irritated, I will be unhappy if I have to draw $2$ or more lightbulbs. The question asks for the probability I will be unhappy. 
The probability I will be unhappy is $1$ minus the probability I will be happy. And I will be happy only if I get a $75$ watt bulb immediately. The probability of this is $\frac{4}{15}$, so the probability I will be unhappy is $\frac{11}{15}$.
In terms of events, the number of bulbs I draw until we I a $75$ watt bulb is any of $1,2,3,4,\dots, 12$ (I assume I am drawing without replacement). The event $A$ that I will be unhappy is the set $\{2,3,4,\dots,12\}$. The event I will be happy is the complement $A'$ of $A$, it is the set $\{1\}$.
Or else we can think of the sample space as consisting of the following "words:" S, FS, FFS, FFFS, FFFFS, and so on where for example FFFS means I got a weak bulb three times in a row, and then got a $75$ watt bulb on the fourth try. Here F means failure and S means success. The event $A$ consists of FS, FFS, FFFS, and so on, everybody but just plain S. 
It is fairly often the case that to find the probability $\Pr(A)$ of an event $A$, it is easier to first find $\Pr(A')$, and then use the fact that $\Pr(A)=1-\Pr(A')$. 
Remark: If the explanation given is the one in your second paragraph, then the explanation is not good. The event $A$ is the event that at least two (two or more) bulbs are  selected.  
The experiment consists of selecting bulbs until we get a $75$ watt bulb, and then stopping. So selecting $0$ bulbs is not one of the possible outcomes of the experiment. 
As to the "at most" part, you have not described an explicit problem. But in the context of the lightbulb problem, "at most $3$" means $1$ or $2$ or $3$. 
A: Think are two alternative worlds, two endings to a story.
In the first story the hero picks a first bulb, but it's not a 75W bulb. He has to pick at least one more. We don't know how long the hero is there, maybe he's really unlucky, maybe he gets it on the second try. It doesn't matter, we know that the first bulb which he picked wasn't a 75W one, and he'll be picking at least two. This is the probability which you are after in the end.
In the second story, the first one he picks is a 75W bulb. No more picking required.
If you think about it, these are the only two possible stories. Either on the first go it's a 75W bulb (second story), or it's not and so he's picking more than one (at least two), (first story).
The crucial point is that the probability of the second story is easy to calculate. Four of the fifteen bulbs are 75W. You pretty much have the answer by writing down the question. And because the first and second story are alternatives and they cover all possibilities, the probability is one minus the other probability, 1-(4/15).
A: The problem is perhaps badly phrased. Think about it this way: If I select a single bulb, what is the probability that it is not a 75W bulb? If it isn't, then I must select at least one more. Maybe I'll get a 75W on that second pick, or maybe I'll have to pick again. In the worst case, I'll pick all the 40 and 60 Watt bulbs first (11 of them), which will guarantee a 75W bulb on the 12th pick.
But the question asks what the probability is that I have to pick at least 2, which means more than 1. So, I either get a 75W bulb on the first pick--the $A'$ case-- or I get one on a later pick--the $A$ case.
A: I think you have confused the idea of a set of bulbs with the idea of a set of possible events (which happen to involve the act of selecting bulbs from the box).
Indeed when you are speaking of subsets of the set of bulbs, the complement of one $75$-watt bulb is a set of fourteen bulbs consisting of all the bulbs except that one $75$-watt bulb. (I'm assuming here that the bulbs are individually identifiable; otherwise this makes no sense.) But these sets have little to do with the sets you need to consider.
If you're drawing without replacement, you will never encounter a set of fourteen bulbs in your set of events in this problem, because after drawing at most twelve bulbs you will have found one of the $75$-watt bulbs. (And if you are drawing with replacement, you can draw a lot more than fifteen times before stopping.)
The event of interest is, "I drew bulbs from the box; when I drew was a $75$-watt bulb, I stopped drawing bulbs, but at that point I had already drawn $2$ or more bulbs."
In other words, if $X$ was the number of bulbs drawn up to and including the first
$75$-watt bulb, then you observed that $X \ge 2.$
The complement of that event is the opposite of $X \ge 2,$
that is, $X < 2,$ which happens when you draw the first bulb and it is a $75$-watt bulb.
