Question

A small regional carrier accepted 13 reservations for a particular flight with 11 seats. 9 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 58% chance, independently of each other.
Find the probability that overbooking occurs.   
Find the probability that the flight has empty seats.   

Answer 1

There are 11 seats in the flight and 9 passengers will arrive for the flight for sure

So there are 2 empty seats that can be filled

Number of seats available = 2

Number of seats booked = 4   

Probability of showing up, q = 0.58  

Probability of not showing up, p = 1 - 0.58 = 0.42
  
a) Probability that over booking occurs mean that more than 2 passenger showed up, P(X > 2) =    P(3) + P(4)

= 4!/(3!* 1!) * 0.58^3 * 0.42^1 + 4!/(4!* 0!) * 0.58^4 * 0.42^0  

= 0.3278 + 0.1132  

= 0.4410

b) Probability that the flight has empty seats means less than less than 2 people showed up = P(0) +P(1)

= 4!/(0!* 4!) * 0.58^0 * 0.42^4 + 4!/(1!* 3!) * 0.58^1 * 0.42^3

= 0.0311 + 0.1719

= 0.203