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Airlines sell more tickets for a flight than the number of available seats (overbooking). They do this because
they know from past experience that only 90% of ticketed passengers actually show up for the flight.
(a) A plane has 9 seats. If the airline sells 11 tickets for a flight, what is the probability that the flight
will be overbooked (the number of passengers who show up is greater than the number of available seats)?
(b) A plane has 235 seats. If the airline sells 250 tickets for a flight, what is the probability that the flight
will be overbooked (the number of passengers who show up is greater than the number of available seats)?
Use the normal approximation for this please
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Here I will solve part (a), ONLY.
It is a binomial experiment.
There are 11 potential passengers. The probability that every given passenger will show up is 0.9.
The events for individual passengers are INDEPENDENT.
The question is: find the probability that of 11 passengers that bought tickets, at least 10 will show up
(it is the event "the flight is overbooked").
n = 11 (the number of trials);
p = 0.9 (the probability of "success" to each individual trial);
k >= 10 (the event of overbooking flight).
P(n=11; p=0.9; k>=10) = P(n=11; p=0.9; k=10) + P(n=11; p=0.9; k=11) = + =
= = 0.697 (rounded). ANSWER
Solved.
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