Probability of Getting a Seat on an Overbooked Flight: A Statistical Analysis

 As airlines strive to maximize their profits, overbooking flights has become a common practice. However, this can lead to passengers being bumped off the flight if there are no available seats. In this blog post, we will analyze the probability of getting a seat on an overbooked flight based on a certain airline policy.

An airline knows that 5 percent of the people making reservations on a certain flight will not show up. Consequently, their policy is to sell 52 tickets for a flight that can hold only 50 passengers. What is the probability that there will be a seat available for every passenger who shows up? 

To solve this problem, we will use the binomial distribution, which is a discrete probability distribution that describes the number of successes in a fixed number of independent trials. In this case, the "success" is defined as a passenger showing up for the flight and the "failure" is defined as a passenger not showing up for the flight.

Let's denote the probability of success as p = 0.95 (since 5 percent of the people will not show up), the number of trials as n = 50 (since the flight can hold only 50 passengers), and the number of tickets sold as k = 52 (since the airline sells 52 tickets). Then, the probability of getting a seat on the flight can be calculated as follows:

P(X <= n) = Σ (n choose x) * p^x * (1-p)^(n-x) for x = 0 to k-n

Here, X is a binomial random variable that represents the number of passengers who show up for the flight. The expression "n choose x" represents the number of ways to choose x items from a set of n items, and can be calculated as follows:

(n choose x) = n! / (x! * (n-x)!)

Using the above formula, we can calculate the probability of getting a seat on the flight as:

P(X <= n) = Σ (50 choose x) * 0.95^x * 0.05^(50-x) for x = 0 to 2

Here, we have used the fact that k-n = 2, since the airline has sold two more tickets than the maximum capacity of the flight.

Calculating the above sum using a calculator or a statistical software package gives us:

P(X <= n) = P(X = 0) + P(X = 1) + P(X = 2) ≈ 0.9177

Therefore, the probability of there being a seat available for every passenger who shows up is approximately 0.9177 or 91.77%.

In this blog post, we analyzed the probability of getting a seat on an overbooked flight based on a certain airline policy. Using the binomial distribution, we calculated the probability of there being a seat available for every passenger who shows up as approximately 91.77%. This suggests that the airline policy of selling two more tickets than the maximum capacity of the flight is a reasonable one, as there is a high probability that all passengers who show up will get a seat.

However, it is important to note that the above calculation assumes that the passengers who show up are independent of each other, which may not always be the case in practice. Additionally, there may be other factors that affect the probability of getting a seat on an overbooked flight, such as the time of day, the day of the week, and the popularity of the destination.