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Overbooking flights is a common practice of most airlines. A particular airline, believing that 3% of passengers fail to show for flights, overbooks (sells more tickets than there are seats). Suppose that for a particular flight involving a jumbo-jet with 267 seats, the airline sells 278 tickets. Question 1. What is the expected number of ticket holders that will fail to show for the flight

User EricZhao
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Answer:

The expected number of ticket holders that will fail to show for the flight is 8.34.

Explanation:

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

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:


E(X) = np

3% of passengers fail to show for flights

This means that
p = 0.03

The airline sells 278 tickets

This means that
n = 278

What is the expected number of ticket holders that will fail to show for the flight


E(X) = np = 278*0.03 = 8.34

The expected number of ticket holders that will fail to show for the flight is 8.34.

User Gowsikan
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