Question

A small regional carrier accepted 12 reservations for a particular flight with 10 seats. 5 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. (Report answers accurate to 4 decimal places.)​

Answer 1

Answer: We remove the 16 regular customers and their seats from consideration.

Thus, we consider the 22 - 16 remaining customers = 6 and the 18-16 = 2 seats

overbooking means 3 or more of the 6 remaining passengers; each has a 58% chance remaining. We can either use the binomial distribution formula P(x) = C(6, x) .58^x .42^(6-x) or use packages that provide the result, including cumulative probabilities.

Overbooking = 1 - P(X <= 2) In Excel, we use =1 - binom.dist(2,6,.58,1) ; the 1 is for cumulative.

The solution is 0.7920

Probability of empty seats = P(x <= 1) = binom.dist(1,6,.58,1) = 0.0510