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.)

Required:
a. Find the probability that overbooking occurs.
b. Find the probability that the flight has empty seats.

Answer 1

a) 0.2135 = 21.35% probability that overbooking occurs.

b) 0.4625 = 46.25% probability that the flight has empty seats.

Explanation:

For each booked passengers, there are only two possible outcomes. Either they arrive, or they do not. The probability of a passenger arriving is independent of other passengers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

We are interest in the 19-14 = 5 remaining reservations, each of whom arrive with 52% probability

This means that
`n = 5, p = 0.52`

a. Find the probability that overbooking occurs.

There are 17-14 = 3 seats remaining. So overbooking occurs if more than 3 arrive.

`P(X > 3) = P(X = 4) + P(X = 5)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 4) = C_(5,4).(0.52)^(4).(0.48)^(1) = 0.1755`

`P(X = 5) = C_(5,5).(0.52)^(5).(0.48)^(0) = 0.0380`

`P(X > 3) = P(X = 4) + P(X = 5) = 0.1755 + 0.0380 = 0.2135`

21.35% probability that overbooking occurs.

b. Find the probability that the flight has empty seats.

3 seats remaining, so this is the probability that less than 3 passengers arrive.

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)`

`P(X = 0) = C_(5,0).(0.52)^(0).(0.48)^(5) = 0.0255`

`P(X = 1) = C_(5,1).(0.52)^(1).(0.48)^(4) = 0.1380`

`P(X = 2) = C_(5,2).(0.52)^(2).(0.48)^(3) = 0.2990`

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0255 + 0.1380 + 0.2990 = 0.4625`

0.4625 = 46.25% probability that the flight has empty seats.