5 odd-numbered taxis out of 9 to 3 airports 
A fleet of 9 taxis must be dispatched to 3 airports: three to airport A, five to B and one to C. If the cabs are numbered 1 to 9, what is the probability that all odd-numbered cabs are sent to airport B?

What I have have come up with so far:


*

*Probability the cabs are odd is $\frac59$

*Probability they are sent to only airport B is $\frac13$

*$\frac59 + \frac13 = \frac89$


Is this the right answer?

If I understand your question correctly, you can rephrase it as, "What's the probability of selecting all odd numbers when sampling uniformly from the collection $\{1,2, \ldots, 9\}$?"  Well, the probability of selecting an odd number out of these $9$ possible numbers is $P(1) = \frac{5}{9}$.  Having selected this first odd number, the probability of selecting another odd number out of the remaining $8$ numbers (for two odd numbers total) is $P(2 \mid 1) = \frac{4}{8}$.  Given we've selected two odd numbers, the probability of selected a third out of the remaining $7$ numbers is $P(3 \mid 1,2) = \frac{3}{7}$.  See where this is going?  The last two will be $P(4 \mid 1,2,3) = \frac{2}{6}$ and $P(5 \mid 1,2,3,4) = \frac{1}{5}$.  
To figure out how to combine these, use the law of total probability repeatedly:
\begin{align*}
P(5) & = P(5 \mid 1) \, P(1) \\
& = P(5 \mid 1, 2) \, P(2 \mid 1) \, P(1) \\
& = P(5 \mid 1,2,3) \, P(3 \mid 1, 2) \, P(2 \mid 1) \, P(1) \\
& = P(5 \mid 1,2,3,4) \, P(4 \mid 1,2,3) \, P(3 \mid 1, 2) \, P(2 \mid 1) \, P(1) \\
& = \frac{1}{5}\frac{2}{6}\frac{3}{7}\frac{4}{8}\frac{5}{9} \\
& = \frac{1}{126}
\end{align*}
