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Chapter 3: Problem 37

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 it has a higher expected value (0.4083 versus 0.2857).

Step by step solution

01

Calculate the Expected Value of the Gamble

The random variable \(X\) represents the outcome when a fair die is rolled, so \(X\) can be any integer from 1 to 6. The probability of each outcome is \(\frac{1}{6}\). First, calculate the expected value of \(h(X) = \frac{1}{X}\). \[ E(h(X)) = \sum_{x=1}^{6} \frac{1}{x} \cdot P(X=x) = \frac{1}{6}\left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} \right) \] This simplifies to \[ E(h(X)) = \frac{1}{6} \cdot \frac{49}{20} = \frac{49}{120} \approx 0.4083 \] dollars.
02

Compare with the Guaranteed Amount

You are offered a guaranteed amount of \(\frac{1}{3.5}\) dollars, which equivalently is \(\frac{2}{7}\) dollars, approximately equal to 0.2857 dollars.
03

Decision

Now that we have both expected values, compare them: \(0.4083\) from gambling and \(0.2857\) as a guaranteed amount. Since \(0.4083 > 0.2857\), the expected value of the gamble is higher than the guaranteed offer.

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Key Concepts

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

Understanding Probability Distribution
Probability distribution is a fundamental concept when dealing with random variables. It describes how probabilities are assigned to different outcomes of an experiment. In the given exercise, we are rolling a fair die, which yields a distribution for the outcomes.
The die has six sides, numbered 1 to 6. When rolled, each side has an equal probability of landing face up. This probability is \(\frac{1}{6}\) for each side. Consequently, the probability distribution for the die roll is uniform. A uniform distribution means all outcomes are equally likely.
This setup allows us to predict and calculate expectations effectively. Using this distribution, we can explore the expected value of various functions of these outcomes, such as the function \(h(X) = \frac{1}{X}\) used in this exercise.
Exploring the Random Variable
A random variable is a variable whose values result from the outcomes of a random event. Here, the random variable \(X\) represents the result of one roll of a fair six-sided die. It can take any integer value from 1 to 6 inclusively.
To analyze scenarios like the one in this exercise, we often transform a random variable using a specific function. In this case, the function is \(h(X) = \frac{1}{X}\). This transformation helps us calculate the expected value by considering the possible outcomes of the die roll.
Each potential outcome of \(X\) is associated with a particular value of \(h(X)\), and since the probability of each outcome is the same (\(\frac{1}{6}\)), calculating the expected value of \(h(X)\) involves averaging these transformed outcome values based on their likelihood.
Decision Making with Expected Values
In decision making, expected value is a critical concept, particularly when evaluating different choices that involve uncertainty. In the exercise, you have two financial options: receiving a fixed sum of money or taking a gamble based on the outcome of the die roll.
To make a rational decision, compare the expected values of these options. We've computed the expected value of the gamble as approximately 0.4083 dollars and the guaranteed amount as approximately 0.2857 dollars.
Since 0.4083 is greater than 0.2857, the expected payoff from gambling exceeds the guaranteed amount. Consequently, if one's goal is to maximize the expected monetary return, the gamble offers a better decision despite the inherent risk involved.

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Most popular questions from this chapter

A student who is trying to write a paper for a course has a choice of two topics, \(A\) and \(B\). If topic \(A\) is chosen, the student will order two books through interlibrary loan, whereas if topic \(\mathrm{B}\) is chosen, the student will order four books. The student believes that a good paper necessitates receiving and using at least half the books ordered for either topic chosen. If the probability that a book ordered through interlibrary loan actually arrives in time is \(.9\) and books arrive independently of one another, which topic should the student choose to maximize the probability of writing a good paper? What if the arrival probability is only .5 instead of .9?

The number of people arriving for treatment at an emergency room can be modeled by a Poisson process with a rate parameter of \(5 / \mathrm{h}\). a. What is the probability that exactly four arrivals occur during a particular hour? b. What is the probability that at least four people arrive during a particular hour? c. How many people do you expect to arrive during a 45-min period?

Let \(X\) have a Poisson distribution with parameter 2. Show that \(E(X)=\lambda\) directly from the definition of expected value. [Hint: The first term in the sum equals 0 , and then \(x\) can be canceled. Now factor out \(\lambda\) and show that what is left sums to 1.]

The \(n\) candidates for a job have been ranked 1,2 , \(3, \ldots, n\). Let \(X=\) the rank of a randomly selected candidate, so that \(X\) has pmf $$ p(x)=\left\\{\begin{array}{cc} 1 / n & x=1,2,3, \ldots, n \\ 0 & \text { otherwise } \end{array}\right. $$ (this is called the discrete uniform distribution). Compute \(E(X)\) and \(V(X)\) using the shortcut formula. [Hint: The sum of the first \(n\) positive integers is \(n(n+1) / 2\), whereas the sum of their squares is \(n(n+1)(2 n+1) / 6\).]

Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of boards 1 and 2 for inspection. a. List the ten different possible outcomes. b. Suppose that boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define \(X\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(X\). c. Let \(F(x)\) denote the cdf of \(X\). First determine \(F(0)=P(X \leq 0), F(1)\), and \(F(2)\), and then obtain \(F(x)\) for all other \(x\).
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