/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 An airline knows that 5 percent ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An airline knows that 5 percent of the people making reservations on a certain flight will not show up. Consequently, their policy is to sell 52 tickets for a flight that can hold only 50 passengers. What is the probability that there will be a seat available for every passenger who shows up?

Short Answer

Expert verified
The probability that there will be a seat available for every passenger who shows up can be calculated using the binomial distribution formula: \(P(X \le 50) = \sum_{k=0}^{50} {52 \choose k} (0.95)^k (0.05)^{52-k}\). By calculating each term of the sum from k=0 to k=50 and summing them, we find the cumulative probability \(P(X \le 50)\), which represents the likelihood of having a seat for everyone who shows up on the flight.

Step by step solution

01

Define the variables

Let X be the number of passengers that show up for the flight. The probability of a passenger showing up \(p=0.95\) and the probability of a passenger not showing up \(q=0.05\). There are a total of \(n=52\) tickets sold and a seating capacity of 50 passengers.
02

Calculate the cumulative probability using binomial distribution

We will use the binomial distribution formula to calculate the probability of up to 50 people showing up to the flight: \(P(X \le 50) = \sum_{k=0}^{50} {n \choose k} p^k q^{n-k}\) Where \(n=52, p=0.95\), and \(q=0.05\).
03

Calculate the binomial coefficients

For each term in the sum, calculate the binomial coefficient {n \choose k}: \({n \choose k} = \frac{n!}{k!(n-k)!}\)
04

Calculate each term of the sum

Calculate each term of the sum for k=0 to k=50 using the values of n, p, and q, and their binomial coefficients: \(P(X=k)={n \choose k} p^k q^{n-k}\)
05

Calculate the cumulative probability

Now we sum the individual probabilities from \(k=0\) to \(k=50\) to find the cumulative probability: \(P(X \le 50) = P(X=0)+P(X=1)+...+P(X=50)\)
06

Interpret the result

The sum calculated in the previous step represents the probability that there will be a seat available for every passenger who shows up. The result will indicate the likelihood (in percentage form) that the airline's policy of overselling the flight by 2 tickets will not result in an overfull flight.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(X\) is normally distributed with mean 1 and variance 4 , use the tables to find \(P\\{2

Suppose that we want to generate a random variable \(X\) that is equally likely to be either 0 or 1 , and that all we have at our disposal is a biased coin that, when flipped, lands on heads with some (unknown) probability \(p\). Consider the following procedure: 1\. Flip the coin, and let \(0_{1}\), either heads or tails, be the result. 2\. Flip the coin again, and let \(0_{2}\) be the result. 3\. If \(0_{1}\) and \(0_{2}\) are the same, return to step 1 . 4\. If \(0_{2}\) is heads, set \(X=0\), otherwise set \(X=1\). (a) Show that the random variable \(X\) generated by this procedure is equally likely to be either 0 or 1 . (b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different, and then sets \(X=0\) if the final flip is a head, and sets \(X=1\) if it is a tail?

A coin having probability \(p\) of coming up heads is successively flipped until the \(r\) th head appears. Argue that \(X\), the number of flips required, will be \(n, n \geq r\), with probability $$ P[X=n\\}=\left(\begin{array}{c} n-1 \\ r-1 \end{array}\right) p^{T}(1-p)^{n-r}, \quad n \geq r $$ This is known as the negative binomial distribution. Hint: How many successes must there be in the first \(n-1\) trials?

Show that $$ \lim _{n \rightarrow \infty} e^{-n} \sum_{k=0}^{n} \frac{n^{k}}{k !}=\frac{1}{2} $$ Hint: Let \(X_{n}\) be Poisson with mean \(n\). Use the central limit theorem to show that \(P\left\\{X_{n} \leq n\right\\} \rightarrow \frac{1}{2}\)

An urn contains \(n+m\) balls, of which \(n\) are red and \(m\) are black. They are withdrawn from the urn, one at a time and without replacement. Let \(X\) be the number of red balls removed before the first black ball is chosen. We are interested in determining \(E[X]\). To obtain this quantity, number the red balls from 1 to \(n\). Now define the random variables \(X_{i}, i=1, \ldots, n\), by \(X_{i}=\left\\{\begin{array}{ll}1, & \text { if red ball } i \text { is taken before any black ball is chosen } \\ 0, & \text { otherwise }\end{array}\right.\) (a) Express \(X\) in terms of the \(X_{i}\). (b) Find \(E[X]\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.