Question

A small regional carrier accelted 19 reservations for a particular flight with 15 seats. 10 reservations went to regular customers who will arrive for the flight . Each of the remaining passengers will arrive for the flight with a 44% chance, independently of each other. find the probability that over booking occurs

find the probabilty that flight has empty seats

Answer 1

Number of seats = 15

Number of bookings = 19

Number of sure reservations = 10

Number of remaining seats = 15 - 10 = 5

Number of bookings remaining, n = 9

Binomial distribution: P(X) = nCx p^x q^n-x

P(a passenger who is not regular, actually arriving for the flight), p = 0.44

q = 1 - 0.44 = 0.56

P(over booking occurs) = P(X > 5) = P(6) + P(7) + P(8) + P(9)

= 9C6x0.44⁶x0.56³ + 9C7x0.44⁷x0.56² + 9C8x0.44⁸x0.56 + 0.44⁹

= 0.1070 + 0.0360 + 0.0071 + 0.0006

= 0.1507

P(the flight has empty seats) = P(X < 5) = P(0) + P(1) + P(2) + P(3) + P(4)

= 0.56⁹ + 9x0.44x0.56⁸ + 9C2x0.44²x0.56⁷ + 9C3x0.44³x0.56⁶ + 9C4x0.44⁴x0.56⁵

= 0.0054 + 0.0383 + 0.1204 + 0.2207 + 0.2601

= 0.6449