Binomial Probability of Airline Overbooking 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.


*

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

*What is the probability that the flight departs with empty seats?
 A: 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 $P(X=125)$
A: 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


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 $\mathsf{Norm}(112.5, 3.354).$


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