A small airline has a policy of booking as many as 60 persons on an airplane that can seat only 53. (Past studies have revealed that only 77% of the booked passengers actually arrive for the flight.) Find the probability that if Air-USA books 60 persons, not enough seats will be available.

Approximately , assuming that the presence of the passengers are all independently of one another.

Step-by-step explanation:

The question suggests that the event that each passenger shows up is a random event with a chance of . Let , , , represent whether each of the sixty passenger showed up, with denoting that the passenger showed up for the flight, and otherwise.

Let denote the sum of these sixty random variables. would represent the number of passengers that show up for the flight.

Each of these would be a Bernoulli random variable with probability .

Assume that whether one passenger show up does not depend on that of another passenger. In other words, assume that those sixty , , , are all independent of one another.

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, the sum of these independent Bernoulli random variables of the same probability , would be a binomial random variable with sixty trials () and a probability of at each trial (.) That is:

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For , the probability that would be:

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There would not be enough seats for those passengers whenever . Calculate the probability of that event:

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yes

step-by-step explanation:

so of course you want to start by plugging the 5 in which goes to 7(5)+3(5)=50 because 7*5=35 and 3*5=15 and then add them together, then plug in the 5 to the other equation as well 9(5)+5=50 because 9*5= 45 and then add 5 and so yes they are equivalent.