Binomial Distribution Problem on airline overbooking
1) How many tickets should an airline company sell for a 120 seat plane if they have averaged 15 percent of the people not showing up?
2) What if in a simplified world all tickets are 250dollars and a person with a ticket who cannot fly due to overbooking receives 400dollars. Does this change the answer to number 1?
3) Now on this same 120 seat plane, If Airline1 partners with Airline2 with 100 seats on the flight alloted for A1 and 20 seats alloted for A2
and the two companies sell tickets independently and still the same 15
percent do not show up, how many tickets should each company sell?
For question one, at first I believe to find n, which is the number of tickets i should sell would be n x .85 = 120. So i believe n = 142. Is this correct for question number one?
I am very confused for the second question. I am not sure how to find if the airline still makes a profit from selling 142 tickets.
if x = number of people who DON'T show up then
P (x<22) = binomcdf(142, .15, 21) = .6206
So the probability that fewer than 22 people (15) don't show up then we would have overbooking and have to pay back 400 dollars... Except I dont know how to figure out how many would be overbooked. I'm not sure if i'm heading in the right direction... Can somebody please help me out? (Thinking)
Re: Binomial Distribution Problem on airline overbooking
There is no answer to part 1 as there is no cost model associated with the question. Without further qualification they should not sell more than 120 tickets.
Originally Posted by yoyostatplease
The other two parts you solve by maximising the expected revenue.