Airlines often overbook flights. Suppose that for a plane with 50 seats, 55 pass

Question

Airlines often overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable Y as the number of ticketed passengers who actually show up for the flight. The PMF of Y appears in the accompanying table: y 45 46 47 48 49 50 51 52 53 54 55 p(y0.05 0.10 0.12 0.14 0.25 0.17 0.06 0.05 0.03 0.02 0.01 What is the probability that the flight will accomodate all ticketed passengers who show up? What is the probability that not all ticketed passengers who show up can be accomo- dated? If you are the first person on the standby list (you will be the first one to get on the plane if there are any seats left after all ticketed passengers have been accomodated), what is the probability that you will be able to take the flight?

Explanation / Answer

a) Probability that the flight will accomodate all ticketed passengers who show up is computed as:

= 1 - P(X > = 51 )

= 1 - (0.06 + 0.05 + 0.03 + 0.02 + 0.01)

= 1 - 0.17

= 0.83

Therefore 0.83 is the required probability here.

b) Now the probability that not all ticketed passengers who show up can be accomodated is computed as:

= P(X > = 51 )

= 0.17

Therefore 0.17 is the required probability here

c) Given that we are on the first place on the standby list, we would clearly be able to get a seat in case, the number of ticketed passengers who arrive is less than or equal to 49. Therefore, the required probability here is computed as:

= 1 - P( X > = 50 )

= 1 - ( 0.17 + 0.17 )

= 1 - 0.34

= 0.66

Therefore 0.66 is the required probability here

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