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An exam consists of ten true-or-false questions. Assuming that every question is answered, in how many different ways can a student complete the exam? In how many ways can the exam be completed if a student can leave some questions unanswered because, say, a penalty is assessed for each incorrect answer?

Short Answer

Expert verified
A student can complete the exam in \(1024\) different ways when all questions must be answered and can complete the exam in \(59049\) different ways when they can leave some questions unanswered.

Step by step solution

01

Count the possibilities for each question

Since there are only two possible answers for each question (true or false), each question has 2 possible outcomes.
02

Calculate the total number of outcomes for all questions

To find the total number of outcomes for the entire exam, we need to apply the multiplication rule since the choices are independent. There are 10 questions in total, so the total number of outcomes would be calculated as follows: \(Total\; outcomes\; = (2 \ outcomes) ^ {10 \ questions} \)
03

Find the total number of outcomes

By using the formula in Step 2, we can find the total number of unique ways a student can complete the exam when they must answer all questions: \( Total\; outcomes\; = 2^{10} = 1024 \) So, a student can complete the exam in 1024 different ways when all questions must be answered. Scenario 2: The student can leave some questions unanswered
04

Recount the possibilities for each question

In this case, we must consider that for each question there are now 3 possible outcomes: true, false, or unanswered.
05

Calculate the total number of outcomes for all questions

We will apply the same multiplication rule as in Scenario 1, but this time considering the updated number of possible outcomes for each question: \(Total\; outcomes = (3 \ outcomes) ^ {10 \ questions} \)
06

Find the total number of outcomes

Using the formula in Step 2, we can find the total number of distinct ways a student can complete the exam when they can leave some questions unanswered: \(Total\; outcomes = 3^{10} = 59049 \) So, a student can complete the exam in 59049 different ways when they can leave some questions unanswered.

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Key Concepts

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

Multiplication Rule
The multiplication rule is a cornerstone concept in combinatorics and is often used when you want to find how many different ways a sequence of events can happen. Each event is considered independent, which means the outcome of one event doesn't affect the outcome of the others. In simpler terms, when you have multiple choices or events that happen in sequence, and you want to find out how many possible combinations there are, you multiply the number of choices for each event.

Let's apply this rule to the original exercise with the exam questions. Each of the 10 questions on the exam has two possible outcomes: true or false. Using the multiplication rule, we find the total number of ways to answer all questions by calculating \(2^{10}\), which results in 1024 different combinations.

If there's an option to leave questions unanswered, the number of possible outcomes per question becomes three: true, false, or unanswered. Again, using the multiplication rule gives us \(3^{10}\), resulting in 59049 different combinations. This elegantly shows the power of the multiplication rule in determining the number of possible outcomes.
Permutations
Permutations are about arranging elements in a specific order. When the order in which you arrange the items is important, permutations come into play. For example, organizing a set of books on a shelf or arranging digits in a number. It's important to note that in permutations, different orders count as different outcomes.

However, in the context of our exam example, permutations aren't directly applicable. The reason is that we are not looking to arrange questions or answers in any specific order. Instead, each question is a separate, independent choice. In this scenario, the primary tool we use is the multiplication rule, as discussed previously.

While permutations can solve problems with ordered arrangements, they're not needed when evaluating independent choices like in our true-or-false exams. Understanding when to use permutations versus other combinatorial techniques is vital in solving various math and probability problems.
Probability
Probability is the math of uncertainty and how we measure it. It tells us the likelihood of an event occurring. Understanding probability starts with understanding that the probability of any event is a number between 0 and 1. A probability of 0 means an event will not happen, while a probability of 1 indicates certainty.

In our context of exam questions, let's consider the simple case where a student guesses every answer. Assuming each answer is either true or false, without any strategy, the likelihood of getting one question correct is \(\frac{1}{2}\). Hence, the probability of answering all ten questions correctly by randomly guessing is \((\frac{1}{2})^{10}\), which can be calculated as \(\frac{1}{1024}\). This low probability illustrates why blind guessing isn't effective.

Probability becomes a powerful tool when connected with combinatorial methods like the multiplication rule. It helps us not only count possibilities but also measure the fairness and risks associated with different outcomes in practical scenarios such as exams.

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Most popular questions from this chapter

Let \(S=\left\\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}\right\\}\) be the sample space associated with an experiment having the following probability distribution: $$ \begin{array}{lcccccc} \hline \text { Outcome } & s_{1} & s_{2} & s_{3} & s_{4} & s_{5} & s_{6} \\ \hline \text { Probability } & \frac{1}{12} & \frac{1}{4} & \frac{1}{12} & \frac{1}{6} & \frac{1}{3} & \frac{1}{12} \\ \hline \end{array} $$ Find the probability of the event: a. \(A=\left\\{s_{1}, s_{3}\right\\}\) b. \(B=\left\\{s_{2}, s_{4}, s_{5}, s_{6}\right\\}\) c. \(C=S\)

In the opinion poll of Exercise 38, the voters were also asked to indicate their political affiliations-Democrat, Republican, or Independent. As before, let the letters \(L, M\), and \(U\) represent the low-, middle-, and upper-income groups, respectively. Let the letters \(D, R\) and \(I\) represent Democrat, Republican, and Independent, respectively. a. Describe a sample space corresponding to this poll. b. Describe the event \(E_{1}\) that a respondent is a Democrat. c. Describe the event \(E_{2}\) that a respondent belongs to the upper-income group and is a Republican. d. Describe the event \(E_{3}\) that a respondent belongs to the middle-income group and is not a Democrat.

A poll was conducted among 250 residents of a certain city regarding tougher gun-control laws. The results of the poll are shown in the table: $$ \begin{array}{lccccc} \hline & \begin{array}{c} \text { Own } \\ \text { Only a } \\ \text { Handgun } \end{array} & \begin{array}{c} \text { Own } \\ \text { Only a } \\ \text { Rifle } \end{array} & \begin{array}{c} \text { Own a } \\ \text { Handgun } \\ \text { and a Rifle } \end{array} & \begin{array}{c} \text { Own } \\ \text { Neither } \end{array} & \text { Total } \\ \hline \text { Favor } & & & & & \\ \text { Tougher Laws } & 0 & 12 & 0 & 138 & 150 \\ \hline \begin{array}{l} \text { Oppose } \\ \text { Tougher Laws } \end{array} & 58 & 5 & 25 & 0 & 88 \\ \hline \text { No } & & & & & \\ \text { Opinion } & 0 & 0 & 0 & 12 & 12 \\ \hline \text { Total } & 58 & 17 & 25 & 150 & 250 \\ \hline \end{array} $$ If one of the participants in this poll is selected at random, what is the probability that he or she a. Favors tougher gun-control laws? b. Owns a handgun? c. Owns a handgun but not a rifle? d. Favors tougher gun-control laws and does not own a handgun?

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.

Explain why the statement is incorrect. A red die and a green die are tossed. The probability that a 6 will appear uppermost on the red die is \(\frac{1}{6}\), and the probability that a 1 will appear uppermost on the green die is \(\frac{1}{6}\). Hence, the probability that the red die will show a 6 or the green die will show a 1 is \(\frac{1}{6}+\frac{1}{6}\).

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