Question

A small regional carrier accepted 16 reservations for a particular flight with 12 seats. 8 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 48% chance, independently of each other.
A) Find the probability that overbooking occurs.
B) Find the probability that the flight has empty seats.

Answer 1

Answer:

a) 32.04% probability that overbooking occurs.

b) 40.79% probability that the flight has empty seats.

Step-by-step explanation:

For each booked passenger, there are only two possible outcomes. Either they arrive for the flight, or they do not arrive. 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Our variable of interest are the 8 reservations that went for the passengers with a 48% probability of arriving.

This means that $n = 8, p = 0.48$

A) Find the probability that overbooking occurs.

12 seats, 8 of which are already occupied. So overbooking occurs if more than 4 of the reservated arrive.

$P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{8,5}.(0.48)^{5}.(0.52)^{3} = 0.2006$

$P(X = 6) = C_{8,6}.(0.48)^{6}.(0.52)^{2} = 0.0926$

$P(X = 7) = C_{8,7}.(0.48)^{7}.(0.52)^{7} = 0.0244$

$P(X = 8) = C_{8,5}.(0.48)^{8}.(0.52)^{0} = 0.0028$

$P(X > 4) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.2006 + 0.0926 + 0.0244 + 0.0028 = 0.3204$

32.04% probability that overbooking occurs.

B) Find the probability that the flight has empty seats.

Less than 4 of the booked passengers arrive.

To make it easier, i will use

$P(X < 4) = 1 - (P(X = 4) + P(X > 4))$

From a), P(X > 4) = 0.3204

$P(X = 4) = C_{8,4}.(0.48)^{4}.(0.52)^{4} = 0.2717$

$P(X < 4) = 1 - (P(X = 4) + P(X > 4)) = 1 - (0.2717 + 0.3204) = 1 - 0.5921 = 0.4079$

40.79% probability that the flight has empty seats.