Solving Airline Problem: Probability & Reservations

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The discussion revolves around calculating probabilities related to airline reservations and passenger attendance. It establishes that the probability of a passenger not showing up is 10%, leading to a binomial distribution for the number of passengers who do show up. For part (a), the goal is to find the probability that 90 or fewer passengers show up when 95 reservations are sold, using the binomial distribution X~Bin(95,0.9). Part (b) seeks to determine the number of reservations needed to ensure that 99% of the time, the airline can accommodate all passengers, requiring the calculation of N such that P(X<=90)=0.99. The discussion emphasizes the independence of passenger arrivals in these calculations.
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Ok, I was looking at some solutions and they don't seem correct. Here is the problem:
An airline knows that the probability a person holding a reservation on a certain flight will not appear is 10%. The plane holds 90 people.
a) If 95 reservations have been solf, find the prob. that the airline will be able o accommodate everyone appearing on the plane.
b) How many reservatiosn should be sold so that the ariline can accommodate everyone who appears for the flight 99% of the time?
 
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Is it not a homework problem?
Solution: Assume that arrival of a person with reservation is independent of other passengers.
Let X be the no. of passengers with reservation who finally turns up and N be the no. of reservations issued.
a) X~Bin(95,0.9). Find P(X<=90).
b) Find N such that P(X<=90)=0.99, when X~Bin(N,0.9).
 
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