Binomial Probability of Airline Overbooking

Question

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds 120 passengers. The probability that a passenger does not show up is 0.10, and the passengers behave independently.

1. What is the probability that every passenger who shows up can take the flight?

2. What is the probability that the flight departs with empty seats?

Answer 1

Here are general methods of approach and some answers from software. I will leave it to you to do whatever hand or calculator computations may be expected of you.

Let the number of people who show up be $X \sim \mathsf{Binom}(n=125,\, p=.9)$

(1) The probability that all who show will have a seat on the flight is $P(X \le 120) = 0.9961$ (to four places).

This exact answer is from R statistical software, but other software or a statistical calculator might also be used.

pbinom(120, 125, .9)
## 0.9961414

If you are doing this computation by hand, using the formula for the PDF (or PMF) of the binomial distribution, it is easier to find

$$P(X > 120) = P(X = 121) + P(X = 122) + P(X = 123) + P(X = 124) + P(X = 125)$$ and subtract that from $1.$

(2) The probability that there are one or more empty seats is $P(X < 120) = P(X \le 119) = 0.9886$ (to four places).

pbinom(119, 125 ,.9)
## 0.9885678

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I suppose there is a chance that you are intended to use a normal approximation. Then you would note that $\mu = E(X) = np = 125(.9) = 112.5$ and $SD(X) = \sqrt{125(.9)(.1)} = 3.354.$ Then the normal approximation would use $Y \sim \mathsf{Norm}(\mu, \sigma)$ and for (1) you would find $P(Y \le 120) = P(Y < 120.5) = 0.9915,$ instead of the exact 0.9961. To compute that you might use software or standardize and use printed normal CDF tables, but notice that this method does not quite give 2-place accuracy.

This problem meets some criteria for using the normal approximation, but normal approximations when binomial $p$ is far from $1/2$ should be avoided if it is important to have an exact answer.

pnorm(120.5, 112.5, 3.354)  # for (1)
## 0.9914654

pnorm(119.5, 112.5, 3.354)  # for (2)
## 0.9815587 

The figure below compares the PDFs of the discrete distribution $\mathsf{Binom}(125,.9)$ and the continuous distribution $\mathsf{Norm}(112.5, 3.354).$

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Note: You could also work this problem in terms of the number $N$ of no-shows where $N \sim \mathsf{Binom}(n = 126, p=.01).$

Answer 2

Let $X$ denote the random variable representing the number of passengers which did not show up. As their showing up is independent, $X$ is binomially distributed $X\sim(n=125,p=0.1)$.

Thus we have $P(X=x)={n\choose x}(p)^n(1-p)^{n-x}={125\choose x}(0.1)^{125}(0.9)^{125-x}$

For details about the binomial distribution see: Wikipedia

The probability that all passengers can take the flight is $P(X>4)=1-P(X<5)=1-P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)$

The probability that the flight departs with any empty seats is $P(X>5)$.

The probability that the flight departs with only empty seats is $P(X=125)$