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Chapter 6: Problem 74

An airline finds that \(5 \%\) of the persons making reservations on a certain flight will not show up for the flight. If the airline sells 160 tickets for a flight that has only 155 seats, what is the probability that a seat will be available for every person holding a reservation and planning to fly?

Short Answer

Expert verified
Answer: The probability that a seat will be available for every person holding a reservation and planning to fly is approximately 98.40%.

Step by step solution

01

Define the variables

Let's define the variables: - \(n\): number of trials (reservations) = 160 - \(p\): probability of success (not showing up) = \(5\% = 0.05\) - \(q\): probability of failure (showing up) = \(1 - p = 0.95\) - \(k\): number of successes needed (people not showing up) to have enough seats Since there are 155 seats and 160 tickets sold, we need at least 5 people not showing up for this to happen. So we want to find the probability that \(k = 5, 6, \dots, 160\) people don't show up.
02

Calculate the binomial probabilities

We'll use the binomial probability formula to find the probability of having \(k\) people not showing up: $$P(X=k) = \binom{n}{k} p^k q^{n-k}$$ We'll sum the probabilities for having enough seats available, meaning that at least 5 people don't show up: $$P(X\geq5) = \sum_{k=5}^{160} \binom{160}{k} (0.05)^k (0.95)^{160-k}$$
03

Calculate the final probability

Use a calculator or computational software to evaluate the sum: $$P(X\geq5) = \sum_{k=5}^{160} \binom{160}{k} (0.05)^k (0.95)^{160-k} \approx 0.9840$$ So the probability that a seat will be available for every person holding a reservation and planning to fly is approximately \(98.40\%\).

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