Question

A small regional carrier accepted 16 reservations for a particular flight with 12 seats. 6 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 50% chance, independently of each other.

Find the probability that overbooking occurs.    
Find the probability that the flight has empty seats.  

Answer 1

Answer: A small regional carrier accepted 16 reservations for a particular flight with 12 seats. 6 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 50% chance, independently of each other.

Solution:

Total reservation accepted = 16

Total confirm passenger = 6

Seats left = 12 - 6 = 6

Unconfirmed passenger = 16 - 6 = 10

p = 0.50

q = 1 - p =1 - 0.50 = 0.50

Therefore,

X: No. of extra passenger boarding the flight.

X ~ Bin(10, 0.50)

* the probability that overbooking occurs:  

P(overbooking) = P(X > 6)

P(X > 6) = P(X=7) + P(X=8) + P(X=9) + P(X=1

= 10C7(0.5)⁷(0.5)³+10C8(0.5)⁸(0.5)²+10C9(0.5)⁹(0.5)¹

+10C10(0.5)¹⁰(0.5)⁰

= 0.117195 + 0.043943 + 0.009765 + 0.000977

= 0.17188

Therefore, the probability that overbooking occurs is 0.17188.

* the probability that the flight has empty seats:

P(empty seats) = 1 - P(X 6)

= 1 - P(X=6) - P(X>6)

= 1 - [10C6(0.5)⁶(0.5)⁴] - 0.17188

= 1 - 0.205078 - 0.17188

= 0.623042

Therefore, the probability that the flight has empty seats is 0.623042.