/*! 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 23 Let \(\Omega\) be the sample spa... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(\Omega\) be the sample space $$ \Omega=\\{0,1,2, \ldots\\} $$ and define a distribution function by $$ m(j)=(1-r)^{j} r $$ for some fixed \(r, 0

Short Answer

Expert verified
The function \( m(j) = (1-r)^j r \) satisfies all the conditions for a probability distribution.

Step by step solution

01

Understand the definition for distribution function

A distribution function, denoted commonly as a probability mass function (PMF) for discrete outcomes, must satisfy two conditions: 1. Each individual probability is non-negative and less than or equal to 1, i.e., \( m(j) \geq 0 \) for all \( j \) and \( m(j) \leq 1 \).2. The sum of all probabilities equals 1, i.e., \( \sum_{j=0}^{\infty} m(j) = 1 \).
02

Check non-negativity of m(j)

Given \( m(j) = (1-r)^j r \) where \( 0 < r < 1 \), we can see that:- \( 1-r > 0 \) because \( r < 1 \).- Since \( (1-r)^j \) represents a non-negative real number raised to the power \( j \), it will also remain non-negative.- Therefore, \( m(j) \geq 0 \) and \( m(j) \leq 1 \). Thus, the first requirement is satisfied.
03

Verify total probability is 1

For a sum of probabilities to be 1, consider:\[\sum_{j=0}^{\infty} m(j) = \sum_{j=0}^{\infty} (1-r)^j r \]This is a geometric series with first term \( a = r \) and common ratio \( q = 1-r \). The sum of an infinite geometric series is given by:\[S = \frac{a}{1-q} = \frac{r}{1-(1-r)} = \frac{r}{r} = 1\]Therefore, the sum of all \( m(j) \) is 1, as required.

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Key Concepts

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

Probability Mass Function
A Probability Mass Function (PMF) is a critical aspect of understanding probability within discrete sample spaces. It allows us to assign a probability to every possible outcome of a discrete random variable, which can take on only specific distinct values. The PMF is defined mathematically as a function, where each value that the variable can take is matched with a probability that falls between 0 and 1.
  • To be valid, the sum of all these probabilities for all possible outcomes in the sample space must equal one.
  • Each individual probability must be non-negative, meaning it must be greater than or equal to zero.
In the given problem, the PMF is expressed by the function \( m(j) = (1-r)^j r \), where \( 0 < r < 1 \). With this formula, each outcome \( j \) in the sample space is assigned a probability. The function ensures that total probability is distributed across all possible outcomes such that they sum up to one, making it a valid PMF.
Discrete Probability
Discrete probability is a branch of probability that focuses on events that occur in specific, separate outcomes. It deals with scenarios where outcomes can be counted, and are typically finite or countably infinite. Because the outcomes are distinct, this kind of probability lays its foundation on a set of countable outcomes. The probability of each outcome is determined by the probability mass function. Within discrete probability, each event has a probability between 0 and 1, where a probability of 0 signifies an impossible outcome and a probability of 1 signifies certainty. In the exercise, we see discrete probabilities at play because the sample space consists of all non-negative integers: \( 0, 1, 2, \ldots \). The function \( m(j) = (1-r)^j r \) represents these discrete probabilities, providing a framework to calculate the probability of discrete outcomes occurring within the defined sample space.
Sample Space
The concept of a sample space is fundamental when discussing probability. It is the complete set of all possible outcomes of a probabilistic experiment. In our exercise, the sample space is given as \( \Omega = \{ 0, 1, 2, \ldots \} \), representing the set of non-negative integers. This sample space is infinite, highlighting an important aspect of the problem: the PMF must properly account for this potentially infinite set of outcomes. Each point in the sample space is associated with a probability determined by the PMF. The exercise defined \( m(j) = (1-r)^j r \), ensuring that the probabilities for outcomes in this space are valid and well-defined. By summing up the probabilities across the sample space, they need to align with the principle that total probability is one, a requirement for any proper probability distribution.Understanding the sample space helps in applying and interpreting the rules of probability, as it sets the stage on which probabilities play out.

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Most popular questions from this chapter

A die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on that face. (For example, a six is three times as probable as a two.) What is the probability of getting an even number in one throw?

If \(A, B,\) and \(C\) are any three events, show that $$ \begin{aligned} P(A \cup B \cup C)=& P(A)+P(B)+P(C) \\ &-P(A \cap B)-P(B \cap C)-P(C \cap A) \\ &+P(A \cap B \cap C) \end{aligned} $$

Let \(A\) and \(B\) be events such that \(P(A \cap B)=1 / 4, P(\tilde{A})=1 / 3,\) and \(P(B)=\) \(1 / 2 .\) What is \(P(A \cup B) ?\)

Tversky and his colleagues \(^{12}\) studied the records of 48 of the Philadelphia 76ers basketball games in the \(1980-81\) season to see if a player had times when he was hot and every shot went in, and other times when he was cold and barely able to hit the backboard. The players estimated that they were about 25 percent more likely to make a shot after a hit than after a miss. In fact, the opposite was true \(-\) the 76 ers were 6 percent more likely to score after a miss than after a hit. Tversky reports that the number of hot and cold streaks was about what one would expect by purely random effects. Assuming that a player has a fifty-fifty chance of making a shot and makes 20 shots a game, estimate by simulation the proportion of the games in which the player will have a streak of 5 or more hits.

(from vos Savant \(^{26}\) ) A reader of Marilyn vos Savant's column wrote in with the following question: My dad heard this story on the radio. At Duke University, two students had received A's in chemistry all semester. But on the night before the final exam, they were partying in another state and didn't get back to Duke until it was over. Their excuse to the professor was that they had a flat tire, and they asked if they could take a make-up test. The professor agreed, wrote out a test and sent the two to separate rooms to take it. The first question (on one side of the paper) was worth 5 points, and they answered it easily. Then they flipped the paper over and found the second question, worth 95 points: 'Which tire was it?' What was the probability that both students would say the same thing? My dad and I think it's 1 in 16\. Is that right?" (a) Is the answer \(1 / 16 ?\) (b) The following question was asked of a class of students. "I was driving to school today, and one of my tires went flat. Which tire do you think it was?" The responses were as follows: right front, \(58 \%\), left front, \(11 \%\), right rear, \(18 \%\), left rear, \(13 \%\). Suppose that this distribution holds in the general population, and assume that the two test-takers are randomly chosen from the general population. What is the probability that they will give the same answer to the second question?

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