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Chapter 7: Problem 68

A state Motor Vehicle Department requires learners to pass a written test on the motor vehicle laws of the state. The exam consists of ten true-or-false questions, of which eight must be answered correctly to qualify for a permit. In how many different ways can a learner who answers all the questions on the exam qualify for a permit?

Short Answer

Expert verified
There are 56 different ways a learner who answers all the questions on the exam can qualify for a permit, considering the possibilities of having 8, 9, or 10 correct answers. This is calculated using combinations: C(10, 8) = 45, C(10, 9) = 10, and C(10, 10) = 1, and summing them up, 45 + 10 + 1 = 56.

Step by step solution

01

Calculate the combinations for 8 correct answers.

To find the number of ways to get 8 correct answers out of 10 questions, we can use the formula for combinations: C(n, k) = \(\frac{n!}{k!(n-k)!}\), where n is the total number of questions and k is the number of correct answers. In this case, n = 10 and k = 8, so the formula becomes: C(10, 8) = \(\frac{10!}{8!(10-8)!}\)
02

Calculate the combinations for 9 correct answers.

To find the number of ways to get 9 correct answers out of 10 questions, we can use the same formula and set k = 9: C(10, 9) = \(\frac{10!}{9!(10-9)!}\)
03

Calculate the combinations for 10 correct answers.

To find the number of ways to get all 10 correct answers, we can use the same formula and set k = 10: C(10, 10) = \(\frac{10!}{10!(10-10)!}\)
04

Compute and sum the combinations for 8, 9, and 10 correct answers.

Now, compute the combination numbers and sum them: C(10, 8) = \(\frac{10!}{8!2!}\) = \(\frac{10 脳 9 脳 8 脳 7 脳 6 脳 5 脳 4 脳 3}{8 脳 7 脳 6 脳 5 脳 4 脳 3 脳 2}\) = 45 C(10, 9) = \(\frac{10!}{9!1!}\) = \(\frac{10 脳 9 脳 8 脳 7 脳 6 脳 5 脳 4 脳 3 脳 2}{9 脳 8 脳 7 脳 6 脳 5 脳 4 脳 3 脳 2}\) = 10 C(10, 10) = \(\frac{10!}{10!0!}\) = \(\frac{10!}{10!}\) = 1 Total ways to qualify for a permit = C(10, 8) + C(10, 9) + C(10, 10) = 45 + 10 + 1 = 56 So, there are 56 different ways a learner who answers all the questions on the exam can qualify for a permit.

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Key Concepts

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

Probability Theory
Probability theory is the branch of mathematics that deals with the likelihood of events occurring. It helps us understand and quantify uncertainty. In the context of true-or-false questions, like the ones in the Motor Vehicle Department's exam, probability can help determine the chances of passing if answers are chosen randomly.
  • Probability is expressed as a value between 0 and 1, where 0 indicates impossibility and 1 indicates certainty.
  • The probability of a single event occurring is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
For example, if a learner guesses on every question, the probability of exactly getting some questions right (such as 8 out of 10) can be found using combinatorial methods, often involving combinations or permutations, to enumerate the favorable outcomes.
Mathematical Combinations
Combinations are all about selecting items from a group, where the order of selection does not matter. This concept is vital when dealing with questions like those in the Motor Vehicle Department's exam.
  • The formula for combinations is given by \[ C(n, k) = \frac{n!}{k!(n-k)!} \]where \(n\) is the total number of items, and \(k\) is the number of items to choose.
  • In our exam example, we're interested in finding how many ways a learner can select certain questions to answer correctly out of ten.
  • For obtaining 8, 9, or 10 correct answers, combinations clearly show us how many distinct sets of questions can be answered correctly, regardless of sequence.
Understanding combinations helps in determining scenarios in which specific outcomes occur, especially when accuracy, rather than order, is important.
Permutations
Permutations involve arranging items where the order does matter. While permutations are not directly used in solving the given exercise since the order of answers does not matter (only the number of correct ones does), understanding them helps grasp how they differ from combinations.
  • The permutation formula is \[ P(n, k) = \frac{n!}{(n-k)!} \]where \(n\) is the total number of items, and \(k\) is the number being arranged.
  • In contexts where the sequence of events or selections matters, permutations become crucial.
  • Permutations would be applicable if, for instance, the order of correctly answered questions made a difference in the exam outcome, which it does not in our case.
Grasping permutations alongside combinations offers a fuller picture of how items can be arranged or selected, guiding deeper understanding of problem-solving techniques.

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Most popular questions from this chapter

List the simple events associated with each experiment. In a survey conducted to determine whether movie attendance is increasing \((i)\), decreasing \((d)\), or holding steady \((s)\) among various sectors of the population, participants are classified as follows: Group 1: Those aged 10-19 Group 2: Those aged 20-29 Group 3: Those aged 30-39 Group 4: Those aged 40-49 Group 5: Those aged 50 and older The response and age group of each participant are recorded.

The customer service department of Universal Instruments, manufacturer of the Galaxy home computer, conducted a survey among customers who had returned their purchase registration cards. Purchasers of its deluxe model home computer were asked to report the length of time \((t)\) in days before service was required. a. Describe a sample space corresponding to this survey. b. Describe the event \(E\) that a home computer required service before a period of 90 days had elapsed. c. Describe the event \(F\) that a home computer did not require service before a period of 1 yr had elapsed.

Two hundred workers were asked: Would a better economy lead you to switch jobs? The results of the survey follow: $$ \begin{array}{lccccc} \hline & \text { Very } & \text { Somewhat } & \text { Somewhat } & \text { Very } & \text { Don't } \\ \text { Answer } & \text { likely } & \text { likely } & \text { unlikely } & \text { unlikely } & \text { know } \\ \hline \text { Respondents } & 40 & 28 & 26 & 104 & 2 \\ \hline \end{array} $$ If a worker is chosen at random, what is the probability that he or she a. Is very unlikely to switch jobs? b. Is somewhat likely or very likely to switch jobs?

A study conducted by the Corrections Department of a certain state revealed that 163,605 people out of a total adult population of \(1,778,314\) were under correctional supervision (on probation, on parole, or in jail). What is the probability that a person selected at random from the adult population in that state is under correctional supervision?

Let \(S\) be a sample space for an experiment, and let \(E\) and \(F\) be events of this experiment. Show that the events \(E \cup F\) and \(E^{c} \cap F^{c}\) are mutually exclusive. Hint: Use De Morgan's law.
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