Two competing airlnes determine that a randomly chosen passenger who booked a se

Question

Two competing airlnes determine that a randomly chosen passenger who booked a seat on their airplanes fails to show with probability 20 To ensure their planes are always full, both airlines try to account for passengers not showing and overbook their flights to ensure every flight isl. Airline A has a 19 seat plane and always books 20 people, while Airline B has a 38 seat plane and always books for 40 people. Let A be the random number of passengers who show to board Airline A's plane and let B be the random number of people who show to board Airline B's plane (a) With reasoning, determine the distribution of the random variables A and B, giving the correct PMF for each random variable (b) For each airline, what is the probability that too many people show and hence, someone doesn't have a seat? (c) Using Airline A's current policy, what is the minimum probability someone fails to show for their seat currently, this is = 0.05) which would ensure the probability that someone doesn't have a seat on Airline A's plane is less than 0.2? (d) Airline A's policy is to always book one extra seat. For example, for a 15 seat plane, they would book 16 people, or for a 30 seat plane, they would book 31. What size plane (at least how many seats) should Airline A use to ensure the probability someone doesn't have a seat is less than 0.1?

Explanation / Answer

A = No of passengers who show to board plane A

B = No of passengers who show to board plane B

a) A can take values from 0 to 20.

B can take values from 0 to 40.

Plane A issues tickets for 20 people.

P(person not showing) = 1/20=0.05 and p(person showing) = 19/20=0.95

The above is the same for both planes and both are binomial outcomes.

Hence P(A=r) = 20Cr (0.95)r(0.05)20-r, r =0,1,2......19,20, for plane A

P(B=r) = 40Cr(0.95)r(0.05)40-r, r =0,1,2......19,20,...,40 for plane B.

b) For plane A, if all 20 people turn out then one will not have a seat.

Hence P(A=20) = 20C20(0.95)20(0.05)0 = 0.3585

For plane B, 40 people turn out then one will not have a seat.

Hence P(B=40) = 40C40(0.95)40(0.05)0 = 0.1285

c) P(A=20)<0.2

i.e. (p)20<0.2 where p is the probability of a person booking showing up

p<0.9227

If probability of someone fails to show up is greater than 1-0.9227 = 0.0773, then prob for a person not getting a seat would be <0.2

d) Let no of people = n and Prob (fail) = 0.05

Then a person not having a seat = 0.95n<0.1

Or n ln 0.95 <ln 0.1

n(-0.05129) <(-2.3026)

n>44.89

Or n should be atleast 45.

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