Question

An airport limousine service (something like Super Shuttle) can accommodate up to 4 passengers on any one trip. A passenger must make an advanced reservation to make a trip (no walk-ups) and accepts a maximum of six reservations for a trip, knowing that 20% of all persons making reservations do not appear for the trip. The probability distribution for the number of reservations made is below:

Number of Reservations 3 4 5 6

Probability 0.1 0.2 0.3 0.4

Each passenger pays $25 for a reservation with no refund if they do not show. However, if a trip is overbooked (read: more passengers signed up for the trip than available space), the passenger who is “bumped” receives $75 (a refund of the original $25 plus “double punitive damages” of $50). What is the expected income per trip received by the limo service, rounded to the nearest dollar.

Answer 1

below is probability distribution of profit

P(X=75) =P(3 passenger reserved)+P(6 passenger reserved and 5 arrive)=0.1+0.4*(₆C₅)*(0.8)⁵(0.2)

=0.257286

P(X=100)=P(4 passenger reserved)=0.2

P(X=125)=P(5 passenger reserved and at most 4 arrive)=P(5 reserved)-P(5 reserved and 5 arrive)

=0.3-0.3*(₅C₅)*(0.8)⁵(0.2)⁰=0.3-0.3*0.32768=0.201696

P(X=150)=P(6 passenger reserved and at most 4 arrive)=P(6 reserved)-P(6 reserved and 5 or 6 arrive)

=0.4-0.4*((₆C₅)*(0.8)⁵(0.2)¹+(₆C₆)*(0.8)⁶(0.2)¹)=0.137856
P(X=50)=P(5 reserved and 5 arrive)=0.3*(₅C₅)*(0.8)⁵(0.2)⁰ =0.098304

P(X=0)=P(6 reserved and 6 arrive) =0.4*0.8⁶ =0.104858

therefore

expected income per trip received by the limo service E(X) =xP(x) =

=0*0.104858+50*0.098304+75*0.257286+100*0.2+125*0.201696+150*0.137856 =90.10