Exercise about probability on plane seats If there are 30 open seats on an airplane (12 aisle, ten window and eight middle), what is the probability that the next passenger will get an aisle seat? What is the probability that the next three passengers will get two aisle and one middle seat (in any order)?
 A: That is just a rephrasing of the standard example of the drawing balls from a box problem. 
Assuming each seat category has same probability:


*

*Q: What is the probability that the next passenger will get an aisle seat?

*A: 12/30 

*Q: What is the probability that the next three passengers will get two aisle and one middle seat (in any order)?

*A: 


*

*Total number of placing 3 people in 30 seats is $$\binom{30}{3} = 4060$$ 

*Possibility of placing 2 people on the aisle: $$\binom{12}{2} = 66$$

*Possibility of placing 1 person in a middle seat: $$\binom{8}{1}=8$$

*Total probability: $$\frac{66*8}{4060} = 0.13$$
Btw: Those are examples of independent events. You tagged your question with "conditional-probability", which has nothing to do with this question in the first place, but for independent events always holds: conditional-probability = unconditional-probability
A: Next Passenger Aisle Seat: $12/30$
For the next three passengers:
There are 3 combinations for 2 aisle(A) and 1 middle(M) seat:
$AAM |AMA |MAA$
$12/30*11/29*8/28 + 12/30*8/29*11/28 + 8/30*12/29*11/28 ≈ 0.13$ 
or
$$
\dfrac{{12 \choose 2} * {8 \choose 1}}{{30 \choose 3}} ≈ 0.13
$$
or
$$
{3 \choose 2}* \dfrac{12}{30}*\dfrac{11}{29}* {1 \choose 1} *\dfrac{8}{28} ≈ 0.13
$$
