When someone buys a ticket for an airline flight there is a 0.0995 probability that the person will not show up for the flight (based on data from the IBM research paper by Lawrence, Hong, and Cherrier). An agent for Air America want to book 24 persons on an airplane that can seat only 22. If 24 persons are booked what is the probability that not enough seats will be available? Is this probability low enough so that overbooking is not a real concern?
To find the probability that not enough seats will be available when 24 persons are booked on an airplane that can only seat 22, we can use the concept of binomial probability.
In this scenario, each person booking a ticket can either show up for the flight (with a probability of 1 - 0.0995 = 0.9005) or not show up (with a probability of 0.0995). Since we want to calculate the probability that not enough seats will be available, we need to find the probability of having more than 22 people showing up.
To calculate this probability, we can use the binomial probability formula:
P(x > k) = 1 - P(x <= k)
Where:
- P(x > k) is the probability that x is greater than k (the number of people who actually show up)
- P(x <= k) is the cumulative probability that x is less than or equal to k
In this case, x follows a binomial distribution with n = 24 (total number of bookings) and p = 0.9005 (probability of each booking showing up). We need to find P(x > 22).
Using a binomial probability calculator or software, we can find:
P(x <= 22) ≈ 0.1918
Therefore, the probability that not enough seats will be available, P(x > 22), is:
P(x > 22) = 1 - P(x <= 22)
= 1 - 0.1918
≈ 0.8082
The calculated probability is approximately 0.8082, which means there is an 80.82% chance that not enough seats will be available if 24 persons are booked on the airplane.
Whether this probability is low enough to consider overbooking not a real concern is subjective and depends on the airline's policies, past experience, and risk tolerance.