An airline sells 338 tickets for a flight to Manila which has 335 seats. It is estimated that 98% of all ticketed passengers show up for the flight. Find the probability that the flight will depart with (at least one) empty seats?

90.7% probability that the flight will depart with (at least one) empty seats

Step-by-step explanation:

For each passenger, there are only two possible outcomes. Either they show up for the flight, or they do not. The probability of a passenger showing up for the flight is independent of other passengers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

In this problem we have that:

Find the probability that the flight will depart with (at least one) empty seats?

This is the probability that there are at most 334 passengers showing up for the flight.

Either they are at most 334 passengers showing up for the flight, or there are at least 335. The sum of the probabilities of these events is decimal 1. So

We want . So

In which

90.7% probability that the flight will depart with (at least one) empty seats