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!

In part a, there are exactly three ways for one out of three cars to pass, they are the three possibilities that you added. But there is only one way for all the cars to fail! The first car must fail, the second car must fail, and the third car must fail. Since there is only one way for this to happen you only consider this one probability. 
Btw there are also three ways for exactly two cars to pass and only one way for all three of them to pass. Hope this helps. 
