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 44% chance, independently of each other.
(Report answers accurate to 4 decimal places.)

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

Answer 1

Solution :

Total reservation accepted = 16

Total number of seats = 12

Number of regular customers who will surely arrive = 8

Remaining (16 - 8) = 8 customers has probability of arriving 44/100 = 0.44.

a) The overbooking will occur if out of 8 irregular customers who have 44% chance of arriving, more than 4 will arrive.

Let X represents the number of customers who arrive out of 8 irregular customers.

Let us consider "an irregular customer who will arrive" as success.

Probability of success (p) = 0.44

Number of trials (n) = 8

X can be considered as a binomial distributed random variable with parameters n = 8 and P = 0.44.

According to binomial probability law, probability of occurrence of exactly x successes in n trials is given by,

We have to obtain P(X > 4).

P(X > 4) = 1 - P(X ≤ 4)

P(X > 4) = 1 - P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

Using binomial probability law we get,

Hence, the probability that overbooking occurs is 0.2416.

b) The flight will have empty seats if out of 8 irregular customers, less than 4 will arrive.

Hence, we have to obtain P(X < 4).

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Using the previous part we get,

P(X < 4) = 0.0097 + 0.0608 + 0.1672 + 0.2627

P(X < 4) = 0.5004

Hence, the probability that the flight has empty seats is 0.5004.