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.
2.36
To determine the percentage of time that on any given flight, at least one passenger holding a reservation will not have a seat, we can use probability.
First, let's calculate the probability that a person does not show up for the flight. Given that past records indicate that 20% of the people making a reservation do not show up, the probability of a person not showing up is 0.20.
Next, let's calculate the probability that all 120 passengers will show up for the flight. Since each reservation is independent of one another, we can calculate this probability by multiplying the probability of each individual showing up for the flight: (1 - 0.20)^120. Simplifying this expression, we get (0.80)^120.
Now, the probability that at least one passenger holding a reservation will not have a seat is equal to 1 minus the probability that all passengers show up. So the result would be 1 - (0.80)^120.
To calculate this percentage, we can multiply the result by 100. Therefore, the percent of time that on any given flight, at least one passenger holding a reservation will not have a seat is:
(1 - (0.80)^120) * 100 ≈ 88.863%
So, approximately 88.863% of the time, at least one passenger holding a reservation will not have a seat on any given flight.