Elementary probability problem involving multiple events 
A company minibus has $9$ passenger seats and on a routine run it is estimated that the probability of any passenger seat being occupied is $0.58$.  Calculate, correct to 3 decimal places, the probability that on a routine run there will be:
a) no passengers
b) one passenger
c) at least $2$ passengers

This is how I approached this problem:
$Let\quad O_n=n\space seats\space are\space occupied\\
P(O_n)=0.58^n$
So for question a:
$1-P(O_9)\approx 0.003$
For question b:
$P(O_1)=0.580$
For question c:
${\sum_{i=2}^9 0.58^i} - {\Pi_{n=2}^9 0.58^n}\approx 0.783$
I feel like I've gone in a whole different direction.
If I've done it wrong, can you please post explanations as to where I've gone wrong. I need it for my exam and this is the only section I struggle with.
Thank you :).

For exactly $n$ seats to be occupied, you would need exactly $9-n$ seats to be unoccupied.  There are $9\choose n$ ways for this to happen.  Thus, the probability that exactly $n$ seats are occupied is
$$P(O_n)={9\choose n}0.58^n(1-0.58)^{9-n}$$
Can you do the rest?
