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

Answer:

a) 0.2135 = 21.35% probability that overbooking occurs.

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

Step-by-step 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.

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

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.