/*! 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 26 One of the numbers 1 through 10 ... [FREE SOLUTION] | 91Ó°ÊÓ

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One of the numbers 1 through 10 is randomly chosen. You are to try to guess the number chosen by asking questions with "yes-no" answers. Compute the expected number of questions you will need to ask in each of the following two cases: (a) Your ith question is to be "Is it i?" \(i=\) 1,2,3,4,5,6,7,8,9,10 (b) With each question you try to eliminate onehalf of the remaining numbers, as nearly as possible.

Short Answer

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", "Is it 2?", etc.), the expected number of questions you'll need to ask is \(5.5\). For strategy (b), where you eliminate half of the remaining numbers with each question, the expected number of questions you'll need to ask is \(3\). Strategy (b) is more efficient in finding the correct number.

Step by step solution

01

Determine the Probability of Each Number of Questions Required

In this case, each number has an equal probability of being chosen, so the probability that the number you are guessing is 1 is 1/10, the probability that the number you are guessing is 2 is 1/10, and so on, up to 10.
02

Calculate the Expected Value for Strategy (a)

For each question, multiple the probability of that question (1/10) by the number of the question (i.e., \(i=1,2,3,...,10\)). Then, sum up these values to find the expected value: \[E_{a} = \sum_{i=1}^{10} \frac{i}{10}\] **Case (b) - Eliminating Half Remaining Numbers with Each Question**
03

Determine the Possible Questions and Outcomes

In this case, we ask questions to eliminate half of the remaining numbers with each question, narrowing down the possibilities. Start with "Is it less than or equal to 5?" to differentiate between numbers 1-5 and 6-10. Keep splitting the remaining numbers in half as much as possible.
04

Determine the Probability of Each Number of Questions Required

Now we'll calculate the probability for the number of questions: 1. 1 question: Probability = 0 (We cannot guess the number in just one question) 2. 2 questions: Probability = 2/10 (We have two equal subsets of numbers to guess) 3. 3 questions: Probability = 6/10 (We need 2 questions to choose between subsets, and 1 more to choose between the remaining 3 numbers) 4. 4 questions: Probability = 2/10 (We need 3 questions to choose between subsets and the remaining numbers, and 1 more to choose between the last 2 numbers)
05

Calculate the Expected Value for Strategy (b)

For each question, multiply the probability of that question by the number of questions required: \[E_{b} = \sum_{i=2}^{4} i \times \frac{1}{5}\] Now let's calculate the expected values for both strategies. **Strategy (a):** \[E_{a} = \sum_{i=1}^{10} \frac{i}{10} = \frac{1}{10} (1 + 2 + 3 + ... + 10) = \frac{1}{10} \frac{10(10+1)}{2} = \frac{55}{10} = 5.5\] **Strategy (b):** \[E_{b} = \sum_{i=2}^{4} i \times \frac{1}{5} = 2\times\frac{1}{5} + 3\times\frac{3}{5} + 4\times\frac{1}{5} = \frac{2+9+4}{5} = \frac{15}{5} = 3\] The expected number of questions for strategy (a) is 5.5, while for strategy (b), the expected number is 3.

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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 a fundamental part of mathematics that deals with the likelihood of events happening. In the exercise context, we are dealing with a very straightforward probability model where each number from 1 to 10 has an equal chance of being selected.
This is a classic example of a uniform distribution. Each number is equally likely, so the probability of choosing any specific number is \( rac{1}{10}\). Probability is all about predicting the chances of an occurrence and, in this exercise, understanding the probabilities helps us calculate the expected number of questions we need to ask.
When applying probability to determine the expected value of a random event, multiply the outcome by the probability of that outcome and sum these values up. It's like calculating the 'average' result you would expect if the experiment was repeated many times.
  • This concept is used daily, whether you are betting on sports or choosing a route to work based on traffic levels.
  • By understanding it, you can make more informed decisions and predictions over time.
  • It's important to remember that just because something has a low probability doesn't mean it can't happen. It's about weighing the likelihood of different outcomes.
Binary Search
Binary Search is a technique used in computer science to efficiently locate a value within a sorted list. In probability and question-asking scenarios, like in this exercise, binary search is a strategy where each question aims to cut the possibilities in half.
Imagine looking for a number between 1 and 10. Instead of asking "Is it 1?", you could ask "Is it less than or equal to 5?". This question immediately halves the choices to 1-5 or 6-10.
This method significantly reduces the number of questions needed to find the answer because each question provides maximum information by eliminating half of the leftover possibilities.
  • Binary search is effective when you have an ordered dataset and want to minimize search time.
  • It's especially helpful when the search set is large, as it reduces the time complexity from linear (checking each one) to logarithmic.
  • Using this method, you tackle big problems by breaking them down into smaller, more manageable pieces.
Binary search's precise nature is what makes it an efficient tool in both computing and logical deduction tasks like our guessing game here.
Random Selection
Random Selection is the foundation of our guessing game exercise. Here, every number from 1 to 10 has an equal probability of being chosen, representing a random selection process.
Random selection ensures that there is no bias, and each choice is made independently of the others. This is crucial when aiming for fairness and objectivity in experiments and plays a critical role in forming samples for statistical studies.
In practice, understanding random selection allows one to design more accurate studies and make predictions with a known level of uncertainty. It applies to scientific experiments, polling, computer algorithms, etc.
  • Random selection is used when you need a true mix of potential outcomes, like drawing names from a hat.
  • Ensures each participant or outcome has a fair and equal chance of selection.
  • Fosters objectivity and removes bias from experiments and decision-making processes.
By recognizing how random selection operates, you set the groundwork for understanding more complex statistical methods and ensure your approach is as unbiased as possible.

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Most popular questions from this chapter

Each night different meteorologists give us the probability that it will rain the next day. To judge how well these people predict, we will score each of them as follows: If a meteorologist says that it will rain with probability \(p,\) then he or she will receive a score of \(1-(1-p)^{2} \quad\) if it does rain \(1-p^{2} \quad\) if it does not rain We will then keep track of scores over a certain time span and conclude that the meteorologist with the highest average score is the best predictor of weather. Suppose now that a given meteorologist is aware of our scoring mechanism and wants to maximize his or her expected score. If this person truly believes that it will rain tomorrow with probability \(p^{*},\) what value of \(p\) should he or she assert so as to maximize the expected score?

\(A\) and \(B\) play the following game: \(A\) writes down either number 1 or number \(2,\) and \(B\) must guess which one. If the number that \(A\) has written down is \(i\) and \(B\) has guessed correctly, \(B\) receives \(i\) units from \(A .\) If \(B\) makes a wrong guess, \(B\) pays \(\frac{3}{4}\) unit to A. If \(B\) randomizes his decision by guessing 1 with probability \(p\) and 2 with probability \(1-p,\) determine his expected gain if (a) \(A\) has written down number 1 and (b) \(A\) has written down number 2 What value of \(p\) maximizes the minimum possible value of \(B^{\prime}\) s expected gain, and what is this maximin value? (Note that \(B\) 's expected gain depends not only on \(p,\) but also on what \(A\) does. Consider now player \(A .\) Suppose that she also randomizes her decision, writing down number 1 with probability \(q\). What is \(A\) 's expected loss if (c) \(B\) chooses number 1 and \((\text { d) } B \text { chooses number } 2 ?\) What value of \(q\) minimizes \(A\) 's maximum expected loss? Show that the minimum of \(A\) 's maximum expected loss is equal to the maximum of \(B\) 's minimum expected gain. This result, known as the minimax theorem, was first established in generality by the mathematician John von Neumann and is the fundamental result in the mathematical discipline known as the theory of games. The common value is called the value of the game to player \(B\).

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