/*! 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 Of all airline flight requests r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Of all airline flight requests received by a certain ticket broker, \(70 \%\) are for domestic travel \((\mathrm{D})\) and \(30 \%\) are for international flights (I). Define \(x\) to be the number that are for domestic flights among the next three requests received. Assuming independence of successive requests, determine the probability distribution of \(x\). (Hint: One possible outcome is DID, with the probability \((0.7)(0.3)(0.7)=0.147 .)\)

Short Answer

Expert verified
The probability distribution for the number of domestic flight requests \(x\) in the next three requests is: - P(x = 0) = 0.027 - P(x = 1) = 0.189 - P(x = 2) = 0.441 - P(x = 3) = 0.343

Step by step solution

01

List all possible combinations of requests

First, let's list down all the possible combinations of requests: 1. DDD 2. DDI 3. DID 4. IDD 5. IIDD 6. IDI 7. IID 8. III
02

Calculate the probability of each combination

Now, we will calculate the probability of each combination using the given probabilities of Domestic (D) and International (I) requests: 1. P(DDD) = (0.7)(0.7)(0.7) = 0.343 2. P(DDI) = (0.7)(0.7)(0.3) = 0.147 3. P(DID) = (0.7)(0.3)(0.7) = 0.147 4. P(IDD) = (0.3)(0.7)(0.7) = 0.147 5. P(IDI) = (0.3)(0.7)(0.3) = 0.063 6. P(DII) = (0.7)(0.3)(0.3) = 0.063 7. P(IID) = (0.3)(0.3)(0.7) = 0.063 8. P(III) = (0.3)(0.3)(0.3) = 0.027
03

Group probabilities by the number of domestic flight requests (x)

Now, we will group the probabilities based on the number of domestic flight requests: - x = 0 (No domestic flight requests): P(III) = 0.027 - x = 1 (One domestic flight request): P(IDI) + P(DII) + P(IID) = 0.063 + 0.063 + 0.063 = 0.189 - x = 2 (Two domestic flight requests): P(DDI) + P(DID) + P(IDD) = 0.147 + 0.147 + 0.147 = 0.441 - x = 3 (Three domestic flight requests): P(DDD) = 0.343
04

Write the probability distribution

Finally, we can write the probability distribution for the number of domestic flight requests \(x\) in the next three requests: - P(x = 0) = 0.027 - P(x = 1) = 0.189 - P(x = 2) = 0.441 - P(x = 3) = 0.343 This shows that there is a 2.7% chance that there are no domestic flight requests, a 18.9% chance that there is one domestic flight request, a 44.1% chance that there are two domestic flight requests, and a 34.3% chance that all three requests are for domestic flights.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Probability
Understanding binomial probability is essential when dealing with situations where there are two possible outcomes for a series of independent events. The scenario described in the exercise is a perfect example of a binomial distribution in action, with the outcomes being either a domestic flight request (\textbf{D}) or an international flight request (\textbf{I}).

Binomial probability can be calculated using the formula for any number of successes (\textbf{k}) out of a fixed number of trials (\textbf{n}), with a constant probability of success (\textbf{p}) on each trial. The formula is given by:
\[ P(k; n, p) = \binom{n}{k} p^k (1-p)^{n-k} \]
In our case, the probability of success (request for a domestic flight) is 0.7 and the number of trials (flight requests) is 3. So, by plugging in the values, we obtain the probability for each possible number of successes. Extra understanding can be achieved by visualizing the distribution on a graph, where the x-axis represents the number of successes and the y-axis the probability of those successes occurring.
Independent Events
The exercise assumes that each flight request is an independent event. Independent events in probability mean that the outcome of one event does not affect the outcome of another. For example, flipping a coin multiple times; each flip does not influence the next. Likewise, in the exercise, whether one request is for a domestic or an international flight has no bearing on the subsequent requests.

To consider events independent, the probability of the occurrence of one event must remain constant regardless of the outcomes of previous events. This assumption allows us to use the multiplication rule for independent events, where the probability of multiple independent events occurring is the product of their individual probabilities:
\[ P(A \text{ and } B) = P(A) \times P(B) \]
This principle was used in the solution to calculate the probability of each sequence of flight requests, which formed the basis of the probability distribution.
Probability Calculations
Probability calculations can sometimes be complex, but the method used in the exercise helps simplify the process. By listing all possible combinations of domestic and international flight requests and assigning probabilities to them, we turn what could be a complex scenario into manageable parts. This step-by-step approach is crucial for clarity and accuracy.

Once all possible events and their probabilities have been determined, the next step is to group and sum these probabilities to find the distribution of probabilities across different numbers of successes (or in our case, domestic flight requests). Understanding how to execute these calculations and interpret the results is key for tasks involving probability distributions.

Exercise Improvement Advice

For students who require further assistance or to enhance conceptual understanding, it's beneficial to:
  • Demonstrate more clearly the concept of independent events and their impact on probability calculation.
  • Provide additional practice problems to solidify understanding of binomial probability calculations.
  • Visualize the probability distribution graphically to provide a visual representation of the probabilities of different outcomes.
  • Include real-world examples to show how these concepts apply outside textbook problems.

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

Determine the following standard normal (z) curve areas: a. The area under the \(z\) curve to the left of 1.75 b. The area under the \(z\) curve to the left of -0.68 c. The area under the \(z\) curve to the right of 1.20 d. The area under the \(z\) curve to the right of -2.82 e. The area under the \(z\) curve between -2.22 and 0.53 f. The area under the \(z\) curve between -1 and 1 g. The area under the \(z\) curve between -4 and 4

Studies have found that women diagnosed with cancer in one breast also sometimes have cancer in the other breast that was not initially detected by mammogram or physical examination ("MRI Evaluation of the Contralateral Breast in Women with Recently Diagnosed Breast Cancer," The New England Journal of Medicine [2007]: 1295-1303). To determine if magnetic resonance imaging (MRI) could detect missed tumors in the other breast, 969 women diagnosed with cancer in one breast had an MRI exam. The MRI detected tumors in the other breast in 30 of these women. a. Use \(p=\frac{30}{969}=0.031\) as an estimate of the probability that a woman diagnosed with cancer in one breast has an undetected tumor in the other breast. Consider a random sample of 500 women diagnosed with cancer in one breast. Explain why it is reasonable to think that the random variable \(x=\) number in the sample who have an undetected tumor in the other breast has a binomial distribution with \(n=500\) and \(p=0.031\). b. Is it reasonable to use the normal distribution to approximate probabilities for the random variable \(x\) defined in Part (a)? Explain why or why not. c. Approximate the following probabilities: i. \(\quad P(x<10)\) ii. \(\quad P(10 \leq x \leq 25)\) iii. \(P(x>20)\) d. For each of the probabilities calculated in Part (c), write a sentence interpreting the probability.

An experiment was conducted to investigate whether a graphologist (a handwriting analyst) could distinguish a normal person's handwriting from that of a psychotic. A well known expert was given 10 files, each containing handwriting samples from a normal person and from a person diagnosed as psychotic, and asked to identify the psychotic's handwriting. The graphologist made correct identifications in 6 of the 10 trials. Does this indicate that the graphologist has an ability to distinguish the handwriting of psychotics?

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean 45 minutes and standard deviation 5 minutes. a. If 50 minutes is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if you wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all students to complete the exam?

Twenty-five percent of the customers of a grocery store use an express checkout. Consider five randomly selected customers, and let \(x\) denote the number among the five who use the express checkout. a. Calculate \(p(2),\) that is, \(P(x=2)\). b. Calculate \(P(x \leq 1)\). c. Calculate \(P(x \geq 2)\). (Hint: Make use of your answer to Part (b).) d. Calculate \(P(x \neq 2)\).

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.