Water bus

The only situations in which the bus is overbooked are (1) when nobody is absent, and (2) when only one is absent. The probabilities are $P(\text{overbooked}) = \left(\frac12\right)^{11} + 11\cdot \left(\frac12\right)^{11} = \frac{12}{2^{11}} = \frac{3}{2^9}.$ Thus the desired probability is $P(\text{not overbooked}) = 1 - \frac{3}{2^9} = \frac{2^9 - 3}{2^9},$ and the answer is $2^9 - (2^9 - 3) = \boxed{3}$ .