Poisson and airline reservations

In the problems for Chapter 5.1, Grinstead and Snell have the following:

An airline finds that 4 percent of the passengers that make reservations on a particular flight will not show up. Consequently, their policy is to sell 100 reserved seats on a plane that has only 98 seats. Find the probability that every person who shows up for the flight will find a seat available.

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Attempt #1:
Each seat that is sold is a Bernoulli trial with Prob(no-show) = 0.04.

N = 100, so the expected number of no-shows for a flight = N p = 4.
The variance is N p (1-p) = 3.84, sd = 1.95

If the number of no-shows exceeds ≈ 1 sd, there won't be a seat for someone.
I expect this to happen about 0.5*e^-1 of the time = 0.18.
The desired probability is 1 minus this.

The problem with this is that no-shows are not normally distributed!

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Attempt #2:

If we use the Poisson approximation we have

P = λk / k! e-λ lambda = 4 P(0 no-show) = e-4 = 0.018 P(1 no-show) = 4 * 0.018 = 0.073 P(nobody gets bumped) = 1 - 0.073 - 0.018 = 1 - 0.091 = 0.919

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Attempt #3 is a simulation (Python code below.
Note the number of seats simulated is 10⁸ !).
The program prints the probability that no one gets bumped, and the st dev for the simulated values.

$ python problem28.py 0.087023 0.00877316767194

import numpy as np import random def overbooked(): rL = list() for i in range(100): if random.random() > 0.96: pass else: rL.append(1) return sum(rL) > 98 N = 1000 L = list() for i in range(1000): S = sum([overbooked() for i in range(N)]) L.append(S*1.0/N) A = np.array(L) print np.mean(A), np.std(A)