Question

Suppose that the probability that a passenger will miss a flight is 0.0902. Airlines do not like flights with empty​ seats, but it is also not desirable to have overbooked flights because passengers must be​ "bumped" from the flight. Suppose that an airplane has a seating capacity of 55 passengers.​(a) If 57 tickets are​ sold, what is the probability that 56 or 57 passengers show up for the flight resulting in an overbooked​ flight?​(b) Suppose that 61 tickets are sold. What is the probability that a passenger will have to be​ "bumped"?​(c) For a plane with seating capacity of 52 ​passengers, what is the largest number of tickets that can be sold to keep the probability of a passenger being​ "bumped" below 10​%?

(a) The probability of an overbooked flight is .0304.
(b) The probability that a passenger will have to be bumped is .5245.​
(c) The largest number of tickets that can be sold while keeping the probability of a passenger being​ "bumped" below 10​% is 54

Answer 1

To find the probabilities in different scenarios, we can use the binomial probability formula. The probability that 56 or 57 passengers show up for an overbooked flight is approximately 0.0052. The probability that a passenger will be bumped when 61 tickets are sold is approximately 0.0388. The largest number of tickets that can be sold without a bumping probability exceeding 10% is 54.

Step-by-step explanation:

(a) To find the probability of 56 or 57 passengers showing up for the flight resulting in an overbooked flight, we can use the binomial probability formula. Let X be the number of passengers who show up. We want to find P(X = 56 or X = 57). This can be calculated as:

P(X = 56) + P(X = 57) = C(57, 56) * (0.0902)^56 * (0.9098) + C(57, 57) * (0.0902)^57 * (0.9098) = 57 * (0.0902)^56 * (0.9098) + (0.0902)^57 * (0.9098) ≈ 0.0048 + 0.0004 ≈ 0.0052

(b) To find the probability that a passenger will have to be "bumped" when 61 tickets are sold, we need to find the probability that X ≥ 61, where X is the number of passengers who show up. This can be calculated as:

P(X ≥ 61) = 1 - P(X ≤ 60) = 1 - Σ(i=0 to 60) [C(61, i) * (0.0902)^i * (0.9098)^(61-i)] ≈ 1 - 0.9612 ≈ 0.0388

(c) To find the largest number of tickets that can be sold to keep the probability of a passenger being "bumped" below 10%, we need to find the maximum value of n such that P(X ≥ n) < 0.10. Starting from n = 1, we can increment n until P(X ≥ n) ≥ 0.10. This can be calculated as:

Let n = 1, P(X ≥ 1) = 1 - P(X ≤ 0) = 1 - (0.0902)^0 * (0.9098)^57 ≈ 1 - 0.9098 ≈ 0.0902

Let n = 2, P(X ≥ 2) = 1 - P(X ≤ 1) = 1 - [C(57, 0) * (0.0902)^0 * (0.9098)^57 + C(57, 1) * (0.0902)^1 * (0.9098)^56] ≈ 1 - 0.0902 - 0.8243 ≈ 0.0855

Continuing this process, we find that the largest value of n such that P(X ≥ n) < 0.10 is 54.