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. Define the random variable Y as the number of ticketed passengers who actually show up for the flight. The probability mass function of Y appears in the accompanying table. 45 46 47 48 49 50 51 52 S3 54 () 0.05 0.10 0.13 0.14 0.24 0.17 0.06 0.05 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

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a. This will be the probability that the plan is able to accomodate all passengers , that is 55 passengers.

So, p(y<=50) = 1-p(y>50)

= 1- .06-.05-.03-.02-.01

= .83

B.this is basically p(y>50)

= 1-p(y<=50)

= 1-.83

= .17

C. If you are the 1st on standbys , you will be the 51st person with ticket. That means 49 showed up and you will be accommodated as the 50th person onboard

So, calculate p(y<=49) which is

.05+.10+.13+.14+.24 = .68

So, there is a chance of .68 for 1st standby to get a seat

Similarly , we have to calculate p(y<=47) to get the probability for 3rd standby to get a seat, which is

.05+.10+.13

= .28

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