Airline overbooking is a common practice. Many people make reservations on several flights due to uncertain plans and then cancel at the last minute or simply fail to show up. Jet Green airlines’ planes hold 100 passengers. Past records indicate that 20% of the people making a reservation do not show up for the flight. We will assume that all reservations are independent, That is, each reservation is for one person and these reservations are made independent of one another. Suppose that Jet Green airlines has decided to book 120 people for each flight. Determine the percent of time that on any given flight, at least one passenger holding a reservation will not have a seat.
To determine the percent of time that on any given flight, at least one passenger holding a reservation will not have a seat, we can use the concept of probability.
First, let's find the probability that a specific passenger does not show up. We are given that 20% of the people making a reservation do not show up, so the probability of a specific passenger not showing up is 0.20 or 20% (in decimal form, 0.20).
Next, we need to find the probability that all 120 passengers show up for the flight. Since each reservation is made independently, we can multiply the probability of each passenger showing up together. Therefore, the probability that all passengers show up is (1 - 0.20)^120, which is approximately 0.026.
The percent of time that on any given flight, at least one passenger holding a reservation will not have a seat can be found by subtracting the probability that all passengers show up from 1 (since we want the complementary event). Therefore, the percent is approximately 1 - 0.026 = 0.974 or 97.4%.
So, on any given flight, there is approximately a 97.4% chance that at least one passenger holding a reservation will not have a seat.
To determine the percent of time that on any given flight, at least one passenger holding a reservation will not have a seat, we can use the concept of complementary probability.
First, let's find the probability that a specific passenger with a reservation will not show up for the flight. Given that 20% of the people making a reservation do not show up, the probability of a passenger not showing up is 0.20.
Now, let's find the probability that all passengers will show up for the flight. Since all reservations are independent, the probability that a specific passenger will show up is 1 - 0.20 = 0.80.
We want to determine the percent of time that at least one passenger holding a reservation will not have a seat. In other words, we want to find the complementary probability that all passengers will have a seat.
Since there are 120 people booked for each flight, and the planes hold 100 passengers, we can subtract the probability that all passengers will have a seat from 1.
The probability that all passengers will have a seat can be calculated as:
(0.80)^120 ≈ 1.5507 x 10^-7
Therefore, the percent of time that on any given flight, at least one passenger holding a reservation will not have a seat is approximately:
1 - 1.5507 x 10^-7 = 0.99999984493
This is equivalent to approximately 99.999984493%, or almost 100%.