Question

A small regional carrier accepted 19 reservations for a particular flight with 17 seats. 14 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 52% chance, independently of each other. (Report answers accurate to 4 decimal places.)

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

Answer 1

(a) The probability of overbooking is 0.2135.

(b) The probability that the flight has empty seats is 0.4625.

Explanation:

Let the random variable X represent the number of passengers showing up for the flight.

It is provided that a small regional carrier accepted 19 reservations for a particular flight with 17 seats.

Of the 17 seats, 14 reservations went to regular customers who will arrive for the flight.

Number of reservations = 19

Regular customers = 14

Seats available = 17 - 14 = 3

Remaining reservations, n = 19 - 14 = 5

P (A remaining passenger will arrive), p = 0.52

The random variable X thus follows a Binomial distribution with parameters n = 5 and p = 0.52.

(1)

Compute the probability of overbooking as follows:

P (Overbooking occurs) = P(More than 3 shows up for the flight)

`=P(X>3)\\\\={5\choose 4}(0.52)^(4)(1-0.52)^(5-4)+{5\choose 5}(0.52)^(5)(1-0.52)^(5-5)\\\\=0.175478784+0.0380204032\\\\=0.2134991872\\\\\approx 0.2135`

Thus, the probability of overbooking is 0.2135.

(2)

Compute the probability that the flight has empty seats as follows:

P (The flight has empty seats) = P (Less than 3 shows up for the flight)

`=P(X<3)\\\\1-P(X\geq 3)\\\\=1-[{5\choose 3}(0.52)^(3)(1-0.52)^(5-3)+{5\choose 4}(0.52)^(4)(1-0.52)^(5-4)+{5\choose 5}(0.52)^(5)(1-0.52)^(5-5)]\\\\=1-[0.323960832+0.175478784+0.0380204032]\\\\=0.4625399808\\\\\approx 0.4625`

Thus, the probability that the flight has empty seats is 0.4625.