When finding the probability of at least $1$ thing occurring out of $X$ trials where order matters, why can't you do $P(1) + P(2) +\ldots + P(X)$? Basically, I have a homework problem here along with the solution:

However, I'm confused why I can't do it in the following way. Let's let C represent a crash and N represent not crashing. So this should be the set of probabilities where at least 1 car crashes: 


*

*$CCC = (.0423)^3$

*$CCN = (.0423)^2$

*$NCC = (.0423)^2$

*$CNC = (.0423)^2$

*$CNN = .0423$

*$NCN = .0423$

*$NNC = .0423$
So if I sum up all these probabilities, it should be the probability that at least $1$ car will crash. However, it ends up being $0.132343557$ which is incorrect. The correct answer should be $0.122$ as noted in the image above, but I'm confused why this method shouldn't work the same.
 A: You are forgetting that the terms CCN, NCC and CNC each have probability (0.0423)$^2$ (0.9577) and the terms CNN,NCN, and NNC each have probability (0.0423) (0.9577)$^2$.
A: Short answer is that you can of course find the solution your way, given that you list all possibilities and state the probabilities right, as Michael indicates.
Sometimes "excluding the unwanted event" is just much easier than "including all the wanted events". So in your case, as you listed, there are 23 possible scenarios, and it is much easier to calculate:
$1 - P(NNN)$
than calculating:
$P(CCC)+P(CCN)+P(NCC)+P(CNC)+P(CNN)+P(NCN)+P(NNC).$
Further, in the question, the three random variables of each car having an accident are independent. If they were somehow dependent, calculating each of the 7 elements in summation would not be as easy as multiplying their individual probabilities. You'd have to deal with conditional probabilities for each of those 7 cases. But if you choose to just exclude the unwanted, you'll have to do a one-line calculation with conditional probabilities.
If you had 4 cars instead of 3, and dependence; things would get incredibly messier, and your chances of making a mistake would exponentially grow. So better stick with the "1 minus unwanted event" approach.
