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!
 A: The case that the first car passes and the next two fail is different from the case that the first car fails, the second passes, and the third fails. But there is only one case in which all cars fail: namely, all cars fail. 
Think of it with coin tosses: if you flip a penny, a nickel, and a dime (different coins, in case you are not in the U.S. or Canada), then there is only one way for all three coins to come out tails,  namely, the penny comes out tails, the nickel comes out tails, and the dime comes out tails. But there are three ways in which exactly one coin comes out heads: you can have the penny come out heads and the nickel and dime come out tails; you can have the nickel come out heads and the penny and dime come out tails; or you can have the dime come out heads and the penny and nickel come out tails. These are three different outcomes, but you do not have three different ways in which the three can come out tails.
A: 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. 
A: 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 :)
