The probability of availability of a seat for every passenger who shows up

Probability Level 3

An airline knows from past experience that 5% of people making reservations on a certain flight will not show up. Consequently, their policy is to sell 63 tickets for a flight that can hold only 60 passengers. What is the probability that there will be a seat available for every passenger who shows up?

0.798 0.616 0.957 0.459 0.863 0.559 0.761 0.892

1 solution

Winod Dhamnekar
Apr 24, 2019

Answer to this question is $1-(0.95)^{63}-\binom{63}{1}\times (0.95)^{62} \times(0.05) -\binom{63}{2} \times (0.95)^{61} \times (0.05)^2=0.61584$

This problem is not presented in a probabilistic manner, but we are asked to provide a probability as the solution. We are told "5% of people making reservations on a certain flight will not show up." Thus if 63 tickets are sold, we should have at most 60 people show up.
It is a different thing to say "each passenger shows up with probability 0.95" as opposed to saying "5% of people making reservation will not show up." The latter formulation is deterministic. It says that, if 60 tickets are sold, 3 people will not show up. Period. So there will be enough seats.

Richard Desper - 2 years, 1 month ago

@Richard Desper , The problem says that if 60 tickets are sold, 3 seats remains vacant. So in order to fill in those 3 vacant seats airline sells 3 additional tickets. So the correct probability that there will be a seat available to every passenger who shows up is as given in the solution.

Winod DHAMNEKAR - 2 years, 1 month ago

The probability of availability of a seat for every passenger who shows up

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