What is wrong with my reasoning? The Questions

$70\%$ of all vehicles pass inspection. Assuming vehicles pass or fail independently. What is the probability:
a) exactly one of the next $3$ vehicles  passes
b) at most $1$ of the next $3$ vehicles  passes

The answer to a) is $.189.$ The way I calculated it was:
$P(\text{success}) \cdot P(\text{fail}) \cdot P(\text{fail})\\
 + P(\text{fail}) \cdot P(\text{success}) \cdot P(\text{fail})\\
+ P(\text{fail}) \cdot P(\text{fail}) \cdot P(\text{success})\\
=.7\cdot.3\cdot.3 + .3\cdot.7\cdot.3 + .3\cdot.3\cdot.7\\
 = .189$
I summed the $3$ possible permutations of $1$ success and $2$ failures.
For b) the answer is $.216.$ To get that answer you take your answer to a) and add the probability of exactly $0$ successes which is $P(\text{fail}) \cdot P(\text{fail}) \cdot P(\text{fail}) = .189 + .3\cdot.3\cdot.3 = .216$
What I don't understand is why the probability of exactly $0$ successes doesn't follow the pattern of exactly $1$ success. Why doesn't the "formula" work:
$P(\text{fail}) \cdot P(\text{fail}) \cdot P(\text{fail})\\
 + P(\text{fail}) \cdot P(\text{fail}) \cdot P(\text{fail})\\
 + P(\text{fail}) \cdot P(\text{fail}) \cdot P(\text{fail})\\
 = .3\cdot.3\cdot.3+.3\cdot.3\cdot.3+.3\cdot.3\cdot.3\\
 = .081$
$\Rightarrow .189 + .081 = .27$ (not $.216$)
Now I'm wondering if I calculated the answer to a) the wrong way, and it was merely a coincidence that I got the right answer!

I'm not too confident with my probability, but I'll try my hand at an explanation.
In the first part, you are considering the next 3 trials as successive events so you can have a success in either the first spot, second spot, or third spot.
In part b, you require all 3 events to be failures so there is only 1 way that the 3 failures can occur in the next 3 events.
Does that make any sense? I never felt probability was my strong point :)
