/*! 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 25 The following questions are from... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 25

The following questions are from ARTIST (reproduced with permission) You are to participate in an exam for which you had no chance to study, and for that reason cannot do anything but guess for each question (all questions being of the multiple choice type, so the chance of guessing the correct answer for each question is \(1 / \mathrm{d}, \mathrm{d}\) being the number of options (distractors) per question; so in case of a 4 -choice question, your guess chance is 0.25 ). Your instructor offers you the opportunity to choose amongst the following exam formats: I. 6 questions of the 4 -choice type; you pass when 5 or more answers are correct; II. 5 questions of the 5 -choice type; you pass when 4 or more answers are correct; III. 4 questions of the 10 -choice type; you pass when 3 or more answers are correct. Rank the three exam formats according to their attractiveness. It should be clear that the format with the highest probability to pass is the most attractive format. Which would you choose and why?

Short Answer

Expert verified
Choose Format II; it offers the highest probability of passing.

Step by step solution

01

Define the Probability of Guessing

Let's define the probability of guessing correctly for each exam format: For Format I, there are 4 options per question, so the probability of guessing correctly is \( p_1 = \frac{1}{4} = 0.25 \). For Format II, there are 5 options per question, so \( p_2 = \frac{1}{5} = 0.20 \). For Format III, there are 10 options per question, so \( p_3 = \frac{1}{10} = 0.10 \).
02

Calculate Probability of Passing for Format I

Format I requires 5 out of 6 questions to be correct. Use the binomial probability formula: \[ P(X \geq 5) = P(X = 5) + P(X = 6) \] where \( X \) is the number of correct answers in 6 questions. The binomial formula is \( P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \) where \( n = 6 \), \( p = 0.25 \), \( k \) is 5 or 6.Calculate each component:- \( P(X = 5) = \binom{6}{5} (0.25)^5 (0.75)^1 = 6 \times 0.0009765625 \times 0.75 = 0.00439453125 \)- \( P(X = 6) = \binom{6}{6} (0.25)^6 = 0.000244140625 \)Thus, \( P(X \geq 5) = 0.00439453125 + 0.000244140625 = 0.004638671875 \).
03

Calculate Probability of Passing for Format II

Format II requires 4 out of 5 questions to be correct. Calculate \( P(X \geq 4) \):- \( P(X = 4) = \binom{5}{4} (0.20)^4 (0.80)^1 = 5 \cdot 0.0016 \cdot 0.80 = 0.0064 \)- \( P(X = 5) = \binom{5}{5} (0.20)^5 = 0.00032 \)So, \( P(X \geq 4) = 0.0064 + 0.00032 = 0.00672 \).
04

Calculate Probability of Passing for Format III

Format III requires 3 out of 4 questions to be correct. Calculate \( P(X \geq 3) \):- \( P(X = 3) = \binom{4}{3} (0.10)^3 (0.90)^1 = 4 \cdot 0.001 \cdot 0.90 = 0.0036 \)- \( P(X = 4) = \binom{4}{4} (0.10)^4 = 0.0001 \)Thus, \( P(X \geq 3) = 0.0036 + 0.0001 = 0.0037 \).
05

Rank Formats by Probability of Passing

Compare the probabilities: - Format I: 0.00464 - Format II: 0.00672 - Format III: 0.0037 The highest probability of passing is with Format II, followed by Format I, and then Format III.

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 91影视!

Key Concepts

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

Binomial Probability
Binomial probability is a key concept in probability theory. It deals with the likelihood of a specific number of successes in a sequence of independent experiments. Each experiment has two possible outcomes, often labeled as 'success' or 'failure'.

This concept is especially useful for scenarios like multiple choice exams, where each question can be considered an experiment. You either get the question right (success) or wrong (failure). The probability of success remains constant for each question.

The binomial probability formula helps to calculate this. For any number of trials \( n \) and probability of success \( p \), the formula is:
  • \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
where \( \binom{n}{k} \) is the binomial coefficient, \( k \) is the number of successes, and \( (1-p) \) is the probability of failure. This formula calculates the probability of getting exactly \( k \) successes in \( n \) trials.

Understanding binomial probability is essential for evaluating situations where guessing correctly is based on fixed probabilities, like with multiple choice exams.
Multiple Choice Exam
Multiple choice exams are assessments consisting of several questions, each accompanied by a set of possible answers. The task is to select the correct answer from these choices.

Such exams are not just about knowledge but also involve a fair amount of probability if one were to guess. The probability of guessing correctly depends on the number of choices provided. For instance:
  • If each question has 4 options, the probability of guessing correctly is \( \frac{1}{4} \) or 0.25.
  • With 5 options per question, it's \( \frac{1}{5} \) or 0.20.
  • For 10 options, the probability is \( \frac{1}{10} \) or 0.10.
Multiple choice exams feature prominently in educational settings because they provide a straightforward way to evaluate a wide range of knowledge topics quickly. However, when guessing, the number of answer choices significantly impacts the probability of correctly answering by chance.
Probability Calculation
Calculating probability effectively involves understanding the formulation and execution of mathematical principles concerning likelihoods. In the context of multiple choice exams, where one might resort to guessing, calculating the probability of passing requires the use of the binomial probability formula.

For example, if you have a test with multiple choice questions and you need a certain number of correct answers to pass, the process involves:
  • Identifying the total number of questions \( n \).
  • Determining the probability \( p \) for guessing an answer correctly.
  • Calculating the probability of getting a minimum number of answers right.
The binomial formula will help find the probability of getting a particular number of successes out of a certain number of trials (e.g., answering correctly in an exam setting).

This calculation can help in deciding the best strategy for answering questions when guessing is involved, balancing the need to get enough questions right with the lower probability of guessing correctly as the number of options increases. Thus, probability calculations enable one to statistically analyze and compare different exam formats based on the likelihood of passing.

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

The following questions are from ARTIST (reproduced with permission) A refrigerator contains 6 apples, 5 oranges, 10 bananas, 3 pears, 7 peaches, 11 plums, and 2 mangos. a. Imagine you stick your hand in this refrigerator and pull out a piece of fruit at random. What is the probability that you will pull out a pear? b. Imagine now that you put your hand in the refrigerator and pull out a piece of fruit. You decide you do not want to eat that fruit so you put it back into the refrigerator and pull out another piece of fruit. What is the probability that the first piece of fruit you pull out is a banana and the second piece you pull out is an apple? c. What is the probability that you stick your hand in the refrigerator one time and pull out a mango or an orange?

The probability that you will win a game is \(0.45 .\) (a) If you play the game 80 times, what is the most likely number of wins? (b) What are the mean and variance of a binomial distribution with \(\mathrm{p}=0.45\) and \(\mathrm{N}=80 ?\)

(a) What is the probability of rolling a pair of dice and obtaining a total score of 9 or more? (b) What is the probability of rolling a pair of dice and obtaining a total score of \(7 ?\)

The following questions are from ARTIST (reproduced with permission) A bowl has 100 wrapped hard candies in it. 20 are yellow, 50 are red, and 30 are blue. They are well mixed up in the bowl. Jenny pulls out a handful of 10 candies, counts the number of reds, and tells her teacher. The teacher writes the number of red candies on a list. Then, Jenny puts the candies back into the bowl, and mixes them all up again. Four of Jenny's classmates, Jack, Julie, Jason, and Jerry do the same thing. They each pick ten candies, count the reds, and the teacher writes down the number of reds. Then they put the candies back and mix them up again each time. The teacher's list for the number of reds is most likely to be (please select one): a. 8,9,7,10,9 b. 3,7,5,8,5 c. 5,5,5,5,5 d. 2,4,3,4,3 e. 3,0,9,2,8

The following questions are from ARTIST (reproduced with permission) An insurance company writes policies for a large number of newly-licensed drivers each year. Suppose \(40 \%\) of these are low-risk drivers, \(40 \%\) are moderate risk, and \(20 \%\) are high risk. The company has no way to know which group any individual driver falls in when it writes the policies. None of the low-risk drivers will have an at-fault accident in the next year, but \(10 \%\) of the moderate-risk and \(20 \%\) of the high-risk drivers will have such an accident. If a driver has an at-fault accident in the next year, what is the probability that he or she is high-risk?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.