/*! 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 38 Let \(X=\) the outcome when a fa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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
Choose to gamble; \( E(1/X) \approx 0.4083 > 0.2857 \).

Step by step solution

01

Identify the Probability Distribution

When a fair die is rolled, the outcomes are 1, 2, 3, 4, 5, and 6. Each outcome has a probability of occurring equal to \( \frac{1}{6} \).
02

Calculate the Expected Value of X

The expected value \( E(X) \) for one roll of a die is calculated as follows:\[E(X) = \sum_{i=1}^{6} x_i P(x_i) = \left( 1 \times \frac{1}{6} \right) + \left( 2 \times \frac{1}{6} \right) + \left( 3 \times \frac{1}{6} \right) + \left( 4 \times \frac{1}{6} \right) + \left( 5 \times \frac{1}{6} \right) + \left( 6 \times \frac{1}{6} \right)\]\[E(X) = 3.5\]
03

Calculate the Expected Value of h(X)

We calculate \( E\left( \frac{1}{X} \right) \) by finding the expectation of the function \( h(X) = \frac{1}{X} \):\[E\left( \frac{1}{X} \right) = \frac{1}{6} \left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \right)\]\[E\left( \frac{1}{X} \right) = \frac{1}{6} \left( 1 + 0.5 + 0.333 + 0.25 + 0.2 + 0.167 \right)\]\[E\left( \frac{1}{X} \right) \approx \frac{1}{6} \times 2.45 = 0.4083\]
04

Compare Expected Values

We compare the two expected values: the guaranteed amount \( \frac{1}{3.5} \approx 0.2857 \) and the expected value of \( h(X) = \frac{1}{X} \) which is approximately \( 0.4083 \).
05

Decide Whether to Accept or Gamble

Since the expected value of earning \( 1/X \) dollars (\( 0.4083 \)) is greater than the guaranteed amount (\( 0.2857 \)), it is statistically better to gamble and choose \( h(X) = \frac{1}{X} \) dollars.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Expected Value
The expected value, often represented as \( E(X) \), provides the average outcome of a probability distribution if you were to repeat an experiment infinite times. In simpler terms, it gives you a measure of the center of the distribution – a value you might "expect" to get in each trial as an average in the long run.

In probability theory, calculating the expected value involves multiplying each possible outcome by the probability of that outcome and then summing all of these values. In the context of a fair die, each possible outcome (1 through 6) has an equal probability of \( \frac{1}{6} \). Therefore, the expected value \( E(X) \) of rolling a die is:

\[ E(X) = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + 3 \times \frac{1}{6} + 4 \times \frac{1}{6} + 5 \times \frac{1}{6} + 6 \times \frac{1}{6} = 3.5 \]

This means, on average, you can expect to roll a 3.5 with a fair six-sided die if rolled indefinitely.
Probability Theory
Probability theory is a branch of mathematics that deals with the likelihood of different outcomes. It is foundational for understanding how events occur and is pivotal in calculating probabilities for any random event that might take place.

When dealing with probability distributions like rolling a die, every outcome has a certain likelihood expressed as a probability between 0 and 1. The sum of the probabilities for all possible outcomes must equal 1 to ensure comprehensiveness.

In our die example, each face of the die is equally likely to land face up, with a probability of \( \frac{1}{6} \). This is an example of a uniform distribution—a core concept in probability theory where all outcomes have the same probability.
  • Every possible result has an assigned probability.
  • The total probability must account for 100% of possible outcomes, i.e., add to 1.
  • The theory aids in analyzing random events by quantifying uncertainty.
This comprehensive framework helps in modeling real-world random processes and understanding the underlying randomness.
Random Variables
A random variable is a fundamental concept in probability theory. It represents a numerical outcome of a random phenomenon, essentially a variable that can take on different values due to some random chance.

In the die-roll scenario, the random variable \( X \) represents the outcome of rolling the die. It can take one of the values {1, 2, 3, 4, 5, 6} corresponding to the sides of the die. Each specific value \( X \) can assume has an associated probability based on the mechanics of the experiment.
  • Random variables can be discrete or continuous.
  • Discrete random variables, like our die example, take on countable values.
  • Continuous random variables may take on any value within a range.
By understanding and manipulating these random variables, we can make informed predictions and understand the likelihood of different outcomes in randomized scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A small market orders copies of a certain magazine for its magazine rack each week. Let \(X=\) demand for the magazine, with pmf $$ \begin{array}{c|cccccc} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline p(x) & \frac{1}{15} & \frac{2}{15} & \frac{3}{15} & \frac{4}{15} & \frac{3}{15} & \frac{2}{15} \end{array} $$ Suppose the store owner actually pays \(\$ 1.00\) for each copy of the magazine and the price to customers is \(\$ 2.00\). 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?

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter \(\lambda=20\) (suggested in the article "Dynamic Ride Sharing: Theory and Practice," J. of Transp. Engr., 1997: 308-312). What is the probability that the number of drivers will a. Be at most 10 ? b. Exceed 20 ? c. Be between 10 and 20 , inclusive? Be strictly between 10 and 20 ? d. Be within 2 standard deviations of the mean value?

Suppose that trees are distributed in a forest according to a two-dimensional Poisson process with parameter \(\alpha\), the expected number of trees per acre, equal to 80 . a. What is the probability that in a certain quarter-acre plot, there will be at most 16 trees? b. If the forest covers 85,000 acres, what is the expected number of trees in the forest? c. Suppose you select a point in the forest and construct a circle of radius \(.1\) mile. Let \(X=\) the number of trees within that circular region. What is the pmf of \(X\) ? [Hint: 1 sq mile \(=640\) acres. \(]\)

An automobile service facility specializing in engine tune-ups knows that \(45 \%\) of all tune-ups are done on four-cylinder automobiles, \(40 \%\) on six- cylinder automobiles, and \(15 \%\) on eight-cylinder automobiles. Let \(X=\) the number of cylinders on the next car to be tuned. a. What is the pmf of \(X\) ? b. Draw both a line graph and a probability histogram for the pmf of part (a). c. What is the probability that the next car tuned has at least six cylinders? More than six cylinders?

Individual A has a red die and B has a green die (both fair). If they each roll until they obtain five "doubles" \((1-1, \ldots\), 6-6), what is the pmf of \(X=\) the total number of times a die is rolled? What are \(E(X)\) and \(V(X)\) ?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.