Question

1. A small regional carrier accepted 21 reservations for a particular flight with 18 seats. 16 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 54% chance, independently of each other.
a. Find the probability that overbooking occurs.
b. Find the probability that the flight has empty seats.

Answer 1

Say X be the event that the remaining passengers arrive for the flight. X follows a binomial distribution with number of trials = 5 and p = 0.54

X ~ b(0.54, 5)

a) If overbooking occurs, three or more passengers arrive. X = 3, 4, 5

Probability = P(X=3) + P(X=4) + P(X=5)

= 5C3.(0.54)³(0.36)² + 5C4.(0.54)⁴(0.36)¹ + 5C5.(0.54)⁵(0.36)⁰ = 0.575

Hence, the probability that overbooking occurs is 0.575

b) If the flight has empty seats, X = 0 or 1

P(X = 0) + P(X = 1) = 5C0.(0.54)⁰(0.36)⁵ + 5C1.(0.54)¹(0.36)⁴ = 0.141

Hence, the probability that the flight has empty seats is 0.141