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.

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!

Attempt #2:

If we use the Poisson approximation we have

Attempt #3 is a simulation (Python code below.

Note the number of seats simulated is 10

^{8}!).

The program prints the probability that no one gets bumped, and the st dev for the simulated values.

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