Rearrangement of Students (flaw in my solution) 
There are 11 students in a class including A, B and C. The 11 students have to form a straight line. Provided that A cannot be the first person in the line, what is the probability that in any random rearrangement of line, A comes before B and C.

For eg, this is a valid rearrangement (1-8 are other students)
1 2 3 A 4 5 C 6 7 B 8
Here's my solution -
Probability that A goes before B and C without any restrictions is $\frac13$. (Notice the symmetry. The answer will be same for B goes first and C goes first.)
Probability that A goes first without any restrictions is $\frac{1}{11}$
Hence the answer is
$$\frac13 - \frac{1}{11} = \frac{8}{33}$$
But the answer is $\frac{4}{15}$ according to my textbook. Please help me to find flaw in my solution.

Short way:
As OP has remarked, by symmetry, the probability that $A$ comes before $B$ and $C$ is $\dfrac13$.
The only object that need concern us is the one immediately preceding $A$.
(Others won't affect the probability computation)
There are $8$ ways with the constraints, as against $10$ unconstrained ways for this object.
Thus $Pr = \dfrac13\cdot\dfrac8{10} = \dfrac4{15}$

Added explanation:
For constrained arrangements, we know that there are $8$ ways to choose the ball *immediately preceding $A$, both lumped together as a $\huge\bullet$
$B$ (say) can be put next to it in one way: ${\huge\bullet}\Large\uparrow\bullet\uparrow$, but C can now be introduced in $2$ ways.
The next object can now be introduced in four ways   $\Large\uparrow{\huge\bullet}\Large\uparrow\bullet\uparrow\bullet\uparrow$, including before $A$, and for each  succeeding object, the number of ways will increase by one.
Similarly for unconstrained ways, except that it will start with $10$ choices for the object immediately preceding $A$, and succeeding objects can be placed on either side.
We thus get $\dfrac {8\times1\times2\times4\times 5\times ...\times 10}{10\times2\times3\times4\times 5\times ...\times 10}$
Apart from the initial $\dfrac8{10}$, the rest of it simplifies to $\dfrac13$,
the symmetric probability that $A$ comes before $B$ and $C$
So we can clearly see the rationale of the short way
