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Let \(X=\) the outcome when a fair die is rolled once. If before the die is rolled you are offered either \((1 / 3.5)\) dollars or \(h(X)=1 / X\) dollars, would you accept the guaranteed amount or would you gamble? [Note: It is not generally true that \(1 / E(X)=E(1 / X)\).]

Short Answer

Expert verified
Gamble, as the expected value of \(h(X)\) is greater.

Step by step solution

01

Understand the Problem

The problem asks to compare a guaranteed amount of \(\frac{1}{3.5}\) dollars against a gambling alternative based on a function \(h(X) = \frac{1}{X}\), where \(X\) is the outcome of a fair die roll. The decision depends on calculating the expected value of \(h(X)\) and comparing it with the guaranteed amount.
02

Calculate the Expected Value of a Die Roll

For a fair six-sided die, each outcome from 1 to 6 occurs with equal probability. Thus, the expected value \(E(X)\) of the die roll is given by:\[E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5\]
03

Calculate the Guaranteed Amount

The guaranteed amount offered is \(\frac{1}{3.5}\) dollars, which can be computed as follows:\[\frac{1}{3.5} = \frac{2}{7}\text{ dollars}\]
04

Compute the Expected Value of the Gamble

The expected value of \(h(X)\) is computed using the formula:\[E\left(\frac{1}{X}\right) = \sum_{i=1}^{6} \frac{1}{6} \cdot \frac{1}{i} = \frac{1}{6}\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\right)\]Evaluate the expression inside the parentheses:\[1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \approx 2.45\]Then, the expectation \(E\left(\frac{1}{X}\right)\) is:\[\frac{1}{6} \times 2.45 \approx 0.4083\text{ dollars}\]
05

Compare the Values

Compare the guaranteed amount \(\frac{2}{7} \approx 0.2857\) dollars with the expected value of the gamble \(0.4083\) dollars:Since \(0.4083 > 0.2857\), the expected value of the gamble \(E\left(\frac{1}{X}\right)\) is greater than the guaranteed amount.

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Key Concepts

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

Understanding the Fair Die
A fair die, commonly used in various games and probability exercises, is a six-sided cube with numbers from 1 to 6. Each side is equally likely to land face up when the die is thrown. This means that every face of the die has the same probability of appearing on top; hence the term "fair."

In probability theory, a fair die ensures each outcome occurs with equal likelihood. For a six-sided die, this translates to each number (1 through 6) having a probability of \(\frac{1}{6}\). This characteristic forms the basis for many calculations, including expected value, and fair probability models.
  • Ensures unbiased results
  • Equal probability for each side
  • Used widely in probability and statistics
Understanding this concept is essential when calculating other attributes like expected values in probability problems.
Exploring Probability
Probability measures the likelihood of an event occurring, and it ranges from 0 to 1. In the context of a fair die, the probability of any particular side showing up is \(\frac{1}{6}\). This uniform distribution is what makes probability calculations straightforward for a die.

The rules of probability help us calculate the outcomes of combined events. When rolling a die, for instance, if we want the probability of getting an even number, we count the possible outcomes (2, 4, 6) and find the probability, \(\frac{3}{6} = \frac{1}{2}\). These calculations are foundational in understanding broader statistical problems.
  • Probability of an event = number of favorable outcomes / total outcomes
  • Ranges from 0 (impossible event) to 1 (certain event)
  • Helps in predicting event outcomes
This concept is vital when considering whether to accept or deny a gamble based on expected return.
Delving into Expected Value Calculation
The expected value is a fundamental concept in probability that represents the average or mean value of a random variable over numerous trials. For a fair die, the expected value can be calculated using its outcomes and probabilities.

In this problem, rolling a six-sided fair die, each outcome (1 to 6) contributes equally to the expected value. Using the formula for expectation \(E(X) = \sum{(x_i \cdot p_i)}\), where \(x_i\) is the outcome and \(p_i = \frac{1}{6}\) for each outcome, the expected sum-value is computed as follows:
\[E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5\]
This tells us the average number rolled on a die, highlighting how expected value helps make informed decisions around uncertainty.
  • Represents the mean outcome of a random variable
  • Essential for decision-making in uncertain scenarios
  • Used to evaluate gambles and predicts outcomes over time
It’s crucial to understand the expected value to assess whether the gamble in the exercise offers a better return than the guaranteed amount.

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Most popular questions from this chapter

A new battery's voltage may be acceptable \((A)\) or unacceptable \((U)\). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that \(90 \%\) of all batteries have acceptable voltages. Let \(Y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is, \(P(Y=2)\) ? b. What is \(p(3)\) ? [Hint: There are two different outcomes that result in \(Y=3\).] c. To have \(Y=5\), what must be true of the fifth battery selected? List the four outcomes for which \(Y=5\) and then determine \(p(5)\). d. Use the pattern in your answers for parts (a)-(c) to obtain a general formula for \(p(y)\).

Starting at a fixed time, each car entering an intersection is observed to see whether it turns left \((L)\), right \((R)\), or goes straight ahead \((A)\). The experiment terminates as soon as a car is observed to turn left. Let \(X=\) the number of cars observed. What are possible \(X\) values? List five outcomes and their associated \(X\) values.

A small market orders copies of a certain magazine for its magazine rack each week. Let \(X=\) demand for the magazine, with pmf \begin{tabular}{l|llllll} \(x\) & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(\frac{1}{15}\) & \(\frac{2}{15}\) & \(\frac{3}{15}\) & \(\frac{4}{15}\) & \(\frac{3}{15}\) & \(\frac{2}{15}\) \end{tabular} Suppose the store owner actually pays \(\$ 2.00\) for each copy of the magazine and the price to customers is \(\$ 4.00\). If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine? [Hint: For both three and four copies ordered, express net revenue as a function of demand \(X\), and then compute the expected revenue.]

Give three examples of Bernoulli rv's (other than those in the text).

The College Board reports that \(2 \%\) of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities (Los Angeles Times, July 16, 2002). Consider a random sample of 25 students who have recently taken the test. a. What is the probability that exactly 1 received a special accommodation? b. What is the probability that at least 1 received a special accommodation? c. What is the probability that at least 2 received a special accommodation? d. What is the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated? e. Suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed \(4.5\) hours. What would you expect the average time allowed the 25 selected students to be?

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