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

Question

Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets variable Y as the number of ticketed passengers who actually show up for the flight. The probability mass function of appears in the accompanying table. Define the random 45 4647 8 490 51 5253 54 55 p(y) 0.04 0.10 0.12 0.14 0.25 0.18 0.06 005 0.03 0.02 0.01 (a) What is the probability that the flight will accommodate all ticketed passengers who show up? (b) What is the probability that not all ticketed passengers who show up can be accommodated? (c) If you are the first person on the standby list (which means you will be the first one to get on the plane if there are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? What is this probability if you are the third person on the standby list?

Explanation / Answer

a) Probability that all passengers would be accomdated is computed as:

= P( 45 ) + P( 46 ) + P( 47 ) + P( 48 ) + P( 49 ) + P( 50 )

= 0.04 + 0.1 + 0.12 + 0.14 + 0.25 + 0.18

= 0.83

Therefore 0.83 is the required probability here.

b) This is just the complement of the above part.
So required probability here would be computed as: 1 - 0.83 = 0.17

Therefore 0.17 is the required probability here.

c) Probability that we would be able to take the flight is computed as:

= P( 45 ) + P( 46 ) + P( 47 ) + P( 48 ) + P( 49 )  

= 0.04 + 0.1 + 0.12 + 0.14 + 0.25

= 0.65

Therefore 0.65 is the required probability here.

d)

Probability that we would be able to take the flight is computed as:

= P( 45 ) + P( 46 ) + P( 47 )  

= 0.04 + 0.1 + 0.12

= 0.26

Therefore 0.26 is the required probability here.

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