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.

There are two possibilities for the places for A, B, C:
1) If they do not include the first place, then there are $\dbinom{10}{3}$ choices for the combined places for A, B, C, and only $\frac{1}{3}$ of the arrangements will have A before B and C.
2) If they do include the first place, then there are $\dbinom{10}{2}$ choices for the combined positions for A, B, C, and only $\frac{2}{3}$ of the arrangements will not have A in the first place.
Therefore the probability is given by $\displaystyle\frac{\frac{1}{3}\binom{10}{3}}{\binom{10}{3}+\frac{2}{3}\binom{10}{2}}=\frac{40}{120+30}=\frac{4}{15}$.
