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. how do i solve this

Question

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. how do i solve this

Answer 1

THE ANSWER IS .15 USE BINOMIAL

Answer 2

To determine the percent of time that on any given flight, at least one passenger holding a reservation will not have a seat, you can use the concept of complementary probability.

First, calculate the probability that all passengers show up for the flight. Since each reservation is independent, the probability that a passenger shows up is the complement of the percentage of people who do not show up. In this case, the probability of a passenger showing up is 1 - 0.20 = 0.80.

Next, you want to find the probability that all 120 passengers have a seat. This can be calculated using the binomial distribution formula:

P(X = k) = (n choose k) * p^k * (1 - p)^(n - k),

where n is the total number of passengers booked (120), k is the number of passengers who have a seat (120), and p is the probability of a passenger showing up (0.80).

P(X = 120) = (120 choose 120) * 0.80^120 * (1 - 0.80)^(120 - 120)

Simplifying this expression:

P(X = 120) = 1 * 0.80^120 * 0.20^0

Since any number raised to the power of 0 is equal to 1, the expression becomes:

P(X = 120) = 0.80^120

Now, the probability that at least one passenger does not have a seat is equal to 1 minus the probability that all 120 passengers have a seat:

P(at least one passenger does not have a seat) = 1 - P(X = 120)

P(at least one passenger does not have a seat) = 1 - 0.80^120

To convert this to a percentage, simply multiply by 100:

P(at least one passenger does not have a seat) = (1 - 0.80^120) * 100

Thus, you can solve this problem by calculating:

P(at least one passenger does not have a seat) = (1 - 0.80^120) * 100.

Using a calculator or statistical software, you can find the answer.