Calculating Seat Availability for an Overbooked Flight Using Probability and Statistics

Imagine a flight with 100 seats, where a hotel chain decides to sell 105 tickets to increase their chances of a full house. Given that the probability of a person who booked the flight actually showing up is 0.90, what is the probability that everyone who shows up will have a seat?

Modeling the Problem Using the Binomial Distribution

This scenario can be modeled using the binomial distribution, where the number of tickets sold is the number of trials, and the probability of a person showing up is a constant probability.

Let:

n 105 (number of tickets sold) p 0.90 (probability that a person shows up) Number of seats available 100

The random variable X represents the number of passengers that show up. X follows a binomial distribution: X ~ Binomial(n105, p0.90).

Calculating the Probability

We need to find the probability that the number of passengers who show up is less than or equal to the number of seats available, i.e., P(X ≤ 100).

Using the Normal Approximation to the Binomial Distribution

Since n is large, we can use the normal approximation to the binomial distribution:

Mean (μ): Variance (σ^2): Standard Deviation (σ):

Applying the Continuity Correction

Using the continuity correction, we calculate:

P(X ≤ 100) ≈ P(Z ≤ (100.5 - 94.5) / 3.07) P(Z ≤ 6 / 3.07) P(Z ≤ 1.95)

Z is a standard normal variable.

Looking Up the Z-Score

Using a standard normal distribution table or calculator:

P(Z ≤ 1.95) ≈ 0.9744

Therefore, the probability that everyone who shows up will have a seat is approximately 97.44%.

Alternative Methods and Results

There are alternative methods of calculating this directly, but the binomial distribution for n 105 and p 0.90 may also be approximated with a normal distribution with mean 94.5 and standard deviation 3.07.

Using the normal approximation:

P(X ≤ 100) ≈ 0.9833 P(Z ≤ 1.95) ≈ 0.9745

However, the normal approximation method is more straightforward and often sufficient for practical purposes.

Conclusion

Understanding the probability of seat availability based on overbooking can help airlines and other transportation services make informed decisions. Using the binomial distribution and its normal approximation provides a practical and accurate method to assess such scenarios.