My revised solution is below:

Edwin

First we calculate the number of possible ways to
choose 8 seats for students to sit in.  Then we'll divide
by the number of ways the 8 students can sit in any seat. 

Suppose the 12 seats are represented by these 12 lines:
_ _ _ _ _ _ _ _ _ _ _ _


Case 1.  A student will sit in the first seat

There must be a student in every other seat, let
S represent a student.  So we must have students
in these 6 positions

S _ S _ S _ S _ S _ S _

But we have 8 students; 2 more to seat. In this case 
we can pick two of the 6 vacant seats to seat two 
other 2 students. The number of ways to 
choose these two seats is C(6,2) = 15 ways. 


Case 2.  No student sits in the first seat.

As before, there must be a student in every other seat, So we
must have students in these seats.  

_ S _ S _ S _ S _ S _ S

However in this case there are only 5 choices of seats for 
the two remaining students, since in this case the first
seat must be vacant.  So the number of ways to choose the
2 seats for two other students to sit in is C(5,2) = 10 ways. 

That makes 15+10 or 25 ways to choose 8 seats for students to 
sit in:


Now we must divide that by the number of ways that any of the 8 students
can sit anywhere which is:

We can choose the 8 seats any of C(12,8)

So the probability is 25/495 = 5/99

Edwin