Question

Suppose that the probability that a passenger will miss a flight is 0.0928. 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 54 passengers. ​(a) If 56 tickets are​ sold, what is the probability that 55 or 56 passengers show up for the flight resulting in an overbooked​ flight? ​(b) Suppose that 60 tickets are sold. What is the probability that a passenger will have to be​ "bumped"? ​(c) For a plane with seating capacity of 59 ​passengers, how many tickets may be sold to keep the probability of a passenger being​ "bumped" below 5​%?

Answer 1

(a)

Probability of a passenger showing up for the flight = 1 - 0.0928 = 0.9072

Let X be the number of show up for the flight. Then X ~ Binomial(n = 56, p = 0.9072)

Probability that 55 or 56 passengers show up for the flight resulting in an overbooked​ flight = P(X = 55) + P(X = 56)

= ⁵⁶C₅₅ * 0.9072⁵⁵ * 0.0928¹ + ⁵⁶C₅₆ * 0.9072⁵⁶ * 0.0928⁰

= 56 * 0.9072⁵⁵ * 0.0928 + 1 * 0.9072⁵⁶

= 0.02879266

(b)

If 60 tickets are sold, X ~ Binomial(n = 60, p = 0.9072)

Probability that a passenger will have to be​ "bumped" = P(X > 54)

= P(X = 55) + P(X = 56) + P(X = 57) + P(X = 58) + P(X = 59) + P(X = 60)

= ⁶⁰C₅₅ * 0.9072⁵⁵ * 0.0928⁵ + ⁶⁰C₅₆ * 0.9072⁵⁶ * 0.0928⁴ + ⁶⁰C₅₇ * 0.9072⁵⁷ * 0.0928³ + ⁶⁰C₅₈ * 0.9072⁵⁸ * 0.0928² + ⁶⁰C₅₉ * 0.9072⁵⁹ * 0.0928¹ + ⁶⁰C₆₀ * 0.9072⁶⁰ * 0.0928⁰

= 0.5126057

(c)

Let N be the number of tickets sold. Then X ~ Binomial(n = N, p = 0.9072)

Probability of a passenger being​ "bumped" < 0.05

=> P(X > 59) < 0.05

For N = 60,

P(X > 59) = P(X = 60) = ⁶⁰C₆₀ * 0.9072⁶⁰ * 0.0928⁰ = 0.002898561

For N = 61,

P(X > 59) = P(X = 60) + P(X = 61) = ⁶¹C₆₀ * 0.9072⁶⁰ * 0.0928¹ +  ⁶¹C₆₁ * 0.9072⁶¹ * 0.0928⁰ = 0.01903775

For N = 62,

P(X > 59) = P(X = 60) + P(X = 61) + P(X = 62)

= ⁶²C₆₀ * 0.9072⁶⁰ * 0.0928² +  ⁶²C₆₁ * 0.9072⁶¹ * 0.09281 +  ⁶²C₆₂ * 0.9072⁶² * 0.0928⁰

= 0.06471811 which is greater than 005.

The maximum value of N = 61 for which P(X > 59) < 0.05

Thus, 61 tickets may be sold to keep the probability of a passenger being​ "bumped" below 5​%.