Mr. Smith uses his 10-seater cars for shuttle-service from the city center to th

Question

Mr. Smith uses his 10-seater cars for shuttle-service from the city center to the airport. He makes overbooking up to 12 passengers for his car. This is because it has been seen that even after reserving a seat, there is 8% chance that a passenger may not show up. It is also noticed that showing up of a passenger is independent of other passengers. Let the random variable X be the number of passengers reserved a seat, with the probability mass function as given below:

x 8 9 10 11 12

p(x) 0.23 0.09 0.46 0.21 0.01

What is the chance that all the passengers who show-up can be accommodated?

What is the chance that the car will travel full (i.e. no empty seat)?

Explanation / Answer

Since there are 10 seats so all the passengers who show-up can be accommodated if 8 , 9 or 10 passengers will shows up. So the chance that all the passengers who show-up can be accommodated is

P(X=8) + P(X=9) + P(X=10) = 0.23 + 0.09 +0.46 = 0.78

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No emtpy seat means that 10 or more customer will shows up. So the chance that the car will travel full (i.e. no empty seat) is

P(X=10) + P(X=11) + P(X=12) = 0.46+0.21 +0.01 = 0.68

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