An airline has sold 560 tickets for a certain flight (with capacity 555 seats) in the assumption that not all passengers that purchased a ticket will arrive for the flight. assume that the probability that a passenger will not show up for the flight is 1%, independently of all other passengers. how likely is it that there are more passengers showing up for the flight than seats are available? calculate this probability by using a binomial distribution for the number of passengers that showed up for the flight.

There is a 34.09% probability that there are more passengers showing up for the flight than seats are available.

Step-by-step explanation:

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 combinatios of x objects from a set of n elements, given by the following formula.

And is the probability of X happening.

For this problem,

There are 560 tickets, so

There is a 1% probability that a passenger will not show up for the flight, so

How likely is it that there are more passengers showing up for the flight than seats are available?

An airline has sold 560 tickets for a certain flight (with capacity 555 seats).

So, this question is: What is the probability that at most 4 passengers do not show up for the flight.

So

There is a 34.09% probability that there are more passengers showing up for the flight than seats are available.

step-by-step explanation:

d would equal 12

actually what is a question i don't understand