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91Ó°ÊÓ

A decision maker subjectively assigned the following probabilities to the four outcomes of an experiment: \(P\left(E_{1}\right)=.10, P\left(E_{2}\right)=.15, P\left(E_{3}\right)=.40,\) and \(P\left(E_{4}\right)=.20 .\) Are these probability assignments valid? Explain.

Short Answer

Expert verified
No, the probabilities are not valid because they do not sum to 1.

Step by step solution

01

Understand Probability Properties

Probabilities must be non-negative and must sum to one. Therefore, check if each given probability is non-negative and if the sum of all probabilities equals one.
02

Check Non-Negativity

Examine each probability assignment: - \( P\left(E_1\right) = 0.10 \) - \( P\left(E_2\right) = 0.15 \) - \( P\left(E_3\right) = 0.40 \) - \( P\left(E_4\right) = 0.20 \) All these probabilities are greater than zero, satisfying the non-negativity condition.
03

Sum the Probabilities

Sum all the given probabilities: \[ 0.10 + 0.15 + 0.40 + 0.20 = 0.85 \]
04

Evaluate the Sum

The sum of the probabilities is \(0.85\), which should equal 1 according to the probability sum rule. Here, the sum does not equal 1.
05

Conclusion of Validity

Since the sum of the probabilities \(0.85\) does not equal 1, the probability assignments are not valid according to the basic rules of probability.

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Key Concepts

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

Probability Assignments
In probability theory, assigning a probability involves determining the likelihood of various outcomes in an experiment. We denote these probabilities as \( P(E) \), where \( E \) represents an event. Each probability assignment is a value between 0 and 1, indicating how likely it is for the event to occur. For example, in the exercise, four different outcomes have been assigned respective probabilities: \( P(E_1) = 0.10 \), \( P(E_2) = 0.15 \), \( P(E_3) = 0.40 \), and \( P(E_4) = 0.20 \).

It's crucial to understand that these probabilities reflect subjective measures of likelihood in this context, meaning they are assigned based on the decision maker's beliefs or past experiences. When assigning probabilities, one must ensure they align with certain mathematical conditions that ensure they represent real-world likelihoods accurately.
Sum of Probabilities
The sum of all probability assignments for an exhaustive list of outcomes must equal 1. This rule is a core principle in probability theory. It signifies that the total likelihood of every possible event happening is absolute; thus, if summed, they encompass all possible states of an experiment.

In our example, the probabilities \( 0.10 + 0.15 + 0.40 + 0.20 = 0.85 \) are expected to sum to 1. However, they only add up to 0.85, suggesting the assigned probabilities do not represent a complete set of possible outcomes.
  • The remaining probability is \( 1 - 0.85 = 0.15 \), which must be accounted for by either adding a missing outcome or reallocating the probabilities.
  • This deficiency typically indicates a miscalculation or oversight needing correction for the probability model to be valid.
Non-Negativity Condition
The non-negativity condition is a fundamental concept in probability theory stating that probabilities must be equal to or greater than zero. Simply put, you cannot have a negative chance of something happening, as this wouldn't make sense logically or mathematically.

In our exercise, it's important to confirm that all assigned probabilities fulfill the non-negativity condition. This means each probability value is checked to be \( \geq 0 \). We can see that \( P(E_1) = 0.10 \), \( P(E_2) = 0.15 \), \( P(E_3) = 0.40 \), and \( P(E_4) = 0.20 \) are all non-negative, which is a good start. Here's why non-negativity is essential:
  • Negative probabilities would imply an event is less likely than impossible, which defies logic.
  • Maintaining non-negativity ensures that the model reflects realistic likelihoods.
  • It upholds the integrity of the mathematical structure of probability.
Although these assignments meet the non-negativity requirement, they still fail to form a legitimate set of probabilities due to the incorrect sum.

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Most popular questions from this chapter

Two Wharton professors analyzed 1,613,234 putts by golfers on the Professional Golfers Association (PGA) Tour and found that 983,764 of the putts were made and 629,470 of the putts were missed. Further analysis showed that for putts that were made, \(64.0 \%\) of the time the player was attempting to make a par putt and \(18.8 \%\) of the time the player was attempting to make a birdie putt. And, for putts that were missed, \(20.3 \%\) of the time the player was attempting to make a par putt and \(73.4 \%\) of the time the player was attempting to make a birdie putt (Is Tiger Woods Loss Averse? Persistent Bias in the Face of Experience, Competition, and High Stakes, D.G. Pope and M. E. Schweitzer, June 2009, The Wharton School, University of Pennsylvania). a. What is the probability that a PGA Tour player makes a putt? b. Suppose that a PGA Tour player has a putt for par. What is the probability that the player will make the putt? c. Suppose that a PGA Tour player has a putt for birdie. What is the probability that the player will make the putt? d. Comment on the differences in the probabilities computed in parts (b) and (c).

Simple random sampling uses a sample of size \(n\) from a population of size \(N\) to obtain data that can be used to make inferences about the characteristics of a population. Suppose that, from a population of 50 bank accounts, we want to take a random sample of 4 accounts in order to learn about the population. How many different random samples of 4 accounts are possible?

A large consumer goods company ran a television advertisement for one of its soap products. On the basis of a survey that was conducted, probabilities were assigned to the following events. \(B=\) individual purchased the product \(S=\) individual recalls seeing the advertisement \(B \cap S=\) individual purchased the product and recalls seeing the advertisement The probabilities assigned were \(P(B)=.20, P(S)=.40,\) and \(P(B \cap S)=.12\). a. What is the probability of an individual's purchasing the product given that the individual recalls seeing the advertisement? Does seeing the advertisement increase the probability that the individual will purchase the product? As a decision maker, would you recommend continuing the advertisement (assuming that the cost is reasonable)? b. Assume that individuals who do not purchase the company's soap product buy from its competitors. What would be your estimate of the company's market share? Would you expect that continuing the advertisement will increase the company's market share? Why or why not? c. The company also tested another advertisement and assigned it values of \(P(S)=.30\) and \(P(B \cap S)=.10 .\) What is \(P(B | S)\) for this other advertisement? Which advertisement seems to have had the bigger effect on customer purchases?

A financial manager made two new investments-one in the oil industry and one in municipal bonds. After a one-year period, each of the investments will be classified as either successful or unsuccessful. Consider the making of the two investments as an experiment. a. How many sample points exist for this experiment? b. Show a tree diagram and list the sample points. c. Let \(O=\) the event that the oil industry investment is successful and \(M=\) the event that the municipal bond investment is successful. List the sample points in \(O\) and in \(M\). d. List the sample points in the union of the events \((O \cup M)\). e. List the sample points in the intersection of the events \((O \cap M)\). f. Are events \(O\) and \(M\) mutually exclusive? Explain.

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