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?
 A: 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*}
A: 
What I have have come up with so far: The probability the cabs are odd: 5/9

Well, that is the probability that a cab will be odd.

The probability they are sent to only airport 2: 1/3

Why do you think this?  There isn't an equal number of cabs sent to each airport.

5/9 + 1/3 = 8/9

Why are you adding?  If there were only two airports would the probability be $5/9+1/2= 19/18$?  That makes no sense.

There are nine cabs, five have odd numbers, and there are to be five sent to airport $\#2$.
There is only $1$ way to select five from five odd cabs, and 
(how many?) ways to select any five from all nine cabs.
Divide and calculate.

The probability that cab#1 is one of the five sent to airport#2 is $5/9$.
Given that, the probability cab#3 is one of the four other cabs sent to that airport is $4/8$, and so forth.
The probability that all five odd cabs are sent to airport two is then 

 $$\dfrac 59\dfrac 48 \dfrac 37\dfrac 26\dfrac 15$$

A: There is only $1$ way to send all $5$ odd numbered cabs to $B$,
against an unrestricted distribution of $\binom95 = 126$
Thus $Pr = \dfrac1{126}$
