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Chapter 5: Problem 16

Your teacher gives a true-false pop quiz with 10 questions. a. Show that the number of possible outcomes for the sample space of possible sequences of 10 answers is \(1024 .\) b. What is the complement of the event of getting at least one of the questions wrong? c. With random guessing, show that the probability of getting at least one question wrong is approximately \(0.999 .\)

Short Answer

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a. 1024 outcomes. b. Getting at least one wrong. c. Probability is 0.999.

Step by step solution

01

Understanding the Sample Space for 10 Questions

Each question on the quiz has two possible answers: True (T) or False (F). Therefore, for each question, there are 2 options. Since there are 10 questions, the total number of possible sequences of answers is calculated by multiplying 2 by itself 10 times: \[ 2^{10} = 1024 \]So, there are \( 1024 \) possible outcomes in the sample space.
02

Identify the Complement of Getting All Questions Correct

To find the complement of getting at least one question wrong, first consider the event where all questions are answered correctly. This occurs if each answer matches the correct answer, which is just 1 specific sequence of 10 correct answers, hence there are 1 way. The complement event is that not all answers are correct, so at least one question is wrong. This means any sequence except the one where all are correct.
03

Calculate Probability of Getting All Questions Correct

When guessing answers, each question has a probability of \(\frac{1}{2}\) to be correct. For 10 questions to all be correct:\[ \left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \].This value represents the probability of getting all questions correct purely by guessing.
04

Calculate Probability of Getting At Least One Question Wrong

Use the complement rule, which states that the probability of an event happening plus the probability of it not happening equals 1. Therefore:\[ P(\text{at least one wrong}) + P(\text{all correct}) = 1 \]Substitute the probability of all correct:\[ P(\text{at least one wrong}) + \frac{1}{1024} = 1 \]\[ P(\text{at least one wrong}) = 1 - \frac{1}{1024} = \frac{1023}{1024} \approx 0.999 \]Thus, the probability of getting at least one wrong is approximately \(0.999\).

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Key Concepts

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

Sample Space
In probability theory, a sample space is a set that lists all possible outcomes of an experiment. For quizzes like this True-False quiz with 10 questions, the sample space helps us understand all the potential answer combinations. Each of the 10 questions has two possible answers: True (T) or False (F).
Multiplied across 10 questions, we calculate the total number of potential sequences by raising 2 (the number of choices per question) to the power of 10. This is expressed as:
  • There are 2 choices for each question: True or False.
  • For 10 questions: \(2^{10} = 1,024\).
This formula gives us the total of 1,024 different sequences of answers. Each sequence represents a different possible outcome for the quiz.
Complement Rule
The complement rule in probability refers to finding the probability of the complement of an event. The complement of an event is everything in the sample space that is not part of that event.
For instance, consider the event where all 10 questions are answered correctly. This is just 1 specific outcome, a single combination among the 1,024 possible sequences. This means the probability of all questions being correctly guessed is the complement of getting at least one question wrong.
  • If all questions were correct by guess, then probabilities would look like \(\left( \frac{1}{2} \right)^{10} = \frac{1}{1024}\).
  • The complement event here is that not all answers are correct, meaning at least one question is wrong.
By using the complement rule, finding the probability of getting at least one question wrong becomes simple once the probability for all correct outcomes is known.
Random Guessing
Random guessing happens when you answer each question without knowing the correct answer. In a True-False quiz with each guess having a 50% chance of being correct or incorrect, probability plays a crucial role.
For each individual question, when guessing, the chance of an answer being correct is \( \frac{1}{2} \). For 10 questions, to get them all correct by random guessing, the probability becomes:
  • All guesses are independent of each other.
  • Product of individual probabilities: \(\left( \frac{1}{2} \right)^{10} = \frac{1}{1024}\).
This means you have a 1 in 1,024 chance of guessing all 10 questions correctly by random guessing alone.
True-False Quiz
A True-False quiz provides a unique setting to explore basic probability concepts. With 10 questions, participants face a straightforward situation鈥攅ach question can only be either True (T) or False (F).
In such quizzes, participants鈥� ability to answer correctly relies heavily on their knowledge. However, it also becomes a simple yet effective demonstration of probabilities when random guessing occurs.
  • Each question has only 2 outcomes, simplifying conditional probabilities.
  • Sample space for 10 questions is easily calculated, making it a great learning tool.
  • The true power of probability rules, like the complement rule, is evident in calculating the chances of getting at least one question wrong by guess.
True-False quizzes are excellent tools to practice and understand probability because they blend simple rules with interesting mathematical insights.

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Most popular questions from this chapter

Discussion with students \(\quad\) In a statistics class of 30 stu- \(-\) dents, 20 students are from the business program and 10 students are from the science program. The instructor randomly select three students, successively and without replacement, to discuss a question. a. True or false: The probability of selecting three students from the business program is \((2 / 3) \times(2 / 3) \times(2 / 3)\). If true, explain why. If false, calculate the correct answer. b. Let \(\mathrm{A}=\) first student is from the business program and \(\mathrm{B}=\) second student is from the business program. Are A and B independent? Explain why or why not. c. Answer parts a and \(b\) if each student is replaced in the class after being selected.

The powerrank.com website (http:// thepowerrank.com/2014/06/06/world- cup-2014-winprobabilities-from-the-power-rank/) listed the probability of each team to win the 2014 World Cup in soccer as follows: 1\. Brazil, \(35.9 \%\). 2\. Argentina, \(10.0 \%\). 3\. Spain, \(8.9 \%\). 4\. Germany, \(7.4 \%\). 5\. Netherlands, \(5.7 \%\). 6\. Portugal, \(3.9 \%\). 7\. France, \(3.4 \%\). 8\. England, \(2.8 \%\). 9\. Uruguay, \(2.5 \%\). 10\. Mexico, \(2.5 \%\). 11\. Italy, \(2.3 \%\). 12\. Ivory Coast, \(2.0 \%\), 13\. Colombia, \(1.5 \%\). 14\. Russia, \(1.5 \%\). 15\. United States, \(1.1 \%\). 16\. Chile, \(1.0 \%\). 17\. Croatia, \(0.9 \%\) 18\. Ecuador, \(0.8 \%\). 19\. Nigeria, \(0.8 \%\). 20\. Switzerland, \(0.7 \%\). 21\. Greece, \(0.6 \%\) 22\. \(\operatorname{Iran}, 0.6 \%\). 23\. Japan, \(0.6 \%\). 24\. Ghana, \(0.6 \%\). 25\. Belgium, \(0.4 \%\). 26\. Honduras, \(0.3 \%\). 27\. South Korea, \(0.3 \%\). 28\. Bosnia-Herzegovina, \(0.3 \%\). 29\. Costa Rica, \(0.3 \%\). 30\. Cameroon, \(0.2 \%\). 31\. Australia, \(0.2 \%\). 32\. Algeria, \(0.1 \%\). a. Interpret Brazil's probability of \(35.9 \%,\) which was based on computer simulations of the tournament. Is it a relative frequency or a subjective interpretation of probability? b. Germany would emerge as the actual winner of the 2014 World Cup. Does this indicate that the \(7.4 \%\) chance of Germany winning, which was calculated before the tournament, should have been \(100 \%\) instead?

Every year the insurance industry spends considerable resources assessing risk probabilities. To accumulate a risk of about one in a million of death, you can drive 100 miles, take a cross country plane flight, work as a police officer for 10 hours, work in a coal mine for 12 hours, smoke two cigarettes, be a nonsmoker but live with a smoker for two weeks, or drink 70 pints of beer in a year (Wilson and Crouch, \(2001,\) pp. \(208-209)\). Show that a risk of about one in a million of death is also approximately the probability of flipping 20 heads in a row with a balanced coin.

Checking independence In each of three independent visits to a restaurant, you choose randomly between two of today's specials, TS1 and TS2, on tle memu. Let A denote \\{today's special on first visit is TS1\\}, B denote \\{ today's special on second visit is TS1\\}, C denote \\{ today's special on the first two visits are TS1\\}, and D denote \\{today's special on the three visits are TS1\\}. a. Find the probabilities of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\). b. Which, if any, pairs of these events are independent? Explain.

In the opening scene of Tom Stoppard's play Rosencrantz and Guildenstern Are Dead, about two Elizabethan contemporaries of Hamlet, Guildenstern flips a coin 91 times and gets a head each time. Suppose the coin was balanced. a. Specify the sample space for 91 coin flips, such that each outcome in the sample space is equally likely. How many outcomes are in the sample space? b. Show Guildenstern's outcome for this sample space. Show the outcome in which only the second flip is a tail. c. What's the probability of the event of getting a head 91 times in a row? d. What's the probability of at least one tail in the 91 flips? e. State the probability model on which your solutions in parts \(\mathrm{c}\) and \(\mathrm{d}\) are based.
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