Question

Suppose that the probability that a passenger will miss a flight is

0.0925

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

59

passengers.

​(a) If

61

tickets are​ sold, what is the probability that

60

or

61

passengers show up for the flight resulting in an overbooked​ flight?

​(b) Suppose that

65

tickets are sold. What is the probability that a passenger will have to be​ "bumped"?

​(c) For a plane with seating capacity of

61

​passengers, how many tickets may be sold to keep the probability of a passenger being​ "bumped" below

55​%?

Answer 1

Here we have given that the probability that a passenger will miss a fight is 0.0925.

The probability that the passengers will show up for the flight is

P(show) = 1 – 0.0925 = 0.9075

1. If 61 tickets are sold ten we want to find the probability that 60 or 61 passengers show up for the flight resulting an overbooked flight.

P(60 or 61 show) = P(60 show) + P(61 show)

Here n = 61, P(show) = 0.9075 x = 60 or 61

P(60 show) = ⁶¹C₆₀ (0.9075)⁶⁰ (0.0925)¹ = 0.0167…….. using excel function (=binom.dist) function

P(61 show) = ⁶¹C₆₁ (0.9075)⁶¹ (0.0925)⁰ = 0.0027

P(60 or 61 show) = 0.0167 + 0.0027 = 0.0194

Probability that 60 or 61 passengers show up for the flight resulting an overbooked flight is 0.0194.

2. 65 tickets are sold and flight has seating capacity is 59 passengers.

We want to find the probability of a passenger will have to be bumped.

Here n = 65, x = 59, p = 0.9075

P(bump) = P( more than 59 passengers) = ⁶⁵C₅₉ (0.9075)⁵⁹ (0.0925)⁶ = 0.4358

Suppose 65 tickets are sold. The probability that a passenger will have to be​ "bumped" is 0.4358.

3. Plane with seating capacity of 61 passengers to find the number of tickets may be sold to keep the probability of passenger of being “bumped” below 5%.

Let consider the number of tickets n = 62, P( more than 61 show up) = ⁶²C₆₁ (0.9075)⁶² (0.0925)¹ = 0.0024

Suppose n = 63, P( more than 61 show up) = ⁶³C₆₁ (0.9075)⁶³ (0.0925)² = 0.0164

Suppose n = 64, P( more than 64 show up) = ⁶⁴C₆₁ (0.9075)⁶⁴ (0.0925)³ = 0.0571

Here if 64 tickets sold out then the probability of passengers of being “bumped” is above 5%

Therefore, for a plane with seating capacity of 61 passengers, 63 tickets may be sold to keep the probability of passenger of being “bumped” below 5%.