/*! 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 33 Probability models? In each of t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Probability models? In each of the following sítuations, state whether or not the given assignment of probabilities to individual outcomes is legitimate, that is, satisfies the rules of probability. Remember, a legitimate model need not be a practically reasonable model. If the assignment of probabilities is not legitimate, give specific reasons for your answer. (a) Roll a six-sided die, and record the count of spots on the up-face: $$ \begin{gathered} \mathbf{P}(1)=0 \mathrm{P}(2)=1 / 6 P(3)=1 / 3 P(4)=1 / 3 P(5)=1 / 6 P(6)=0 \\\ P(1)=0 \quad P(2)=1 / 6 \quad P(3)=1 / 3 \\ P(4)=1 / 3 \quad P(5)=1 / 6 \quad P(6)=0 \end{gathered} $$ (b) Deal a card from a shuffled deck: $$ \begin{gathered} \mathrm{P}(\text { clubs })=12 / 52 P(\text { diamonds })=12 / 52 P(\text { hearts })=12 / 52 P(\text { spades })=16 / 52 \\ P(\text { clabs })=12 / 52 \quad P(\text { diamonis })=12 / 52 \\ P(\text { bearts })=12 / 52 \quad P(\text { spades })=16 / 52 \end{gathered} $$ (c) Choose a college student at random and record sex and enrollment status: \(P(\) female full-time \()=0.56 \mathrm{P}(\) male full-time \()=0.44 \mathrm{P}(\) female part- time \()=0.24 \mathrm{P}(\) male part-lime \()=0.17\)

Short Answer

Expert verified
(a) Yes (sum is 1, despite zero probabilities). (b) Yes (sum is exactly 1). (c) No (sum exceeds 1).

Step by step solution

01

Assess the Sum of Probabilities (a)

For a probability distribution to be legitimate, the sum of all the probabilities must be equal to 1. Here, we add up the probabilities: \( P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 0 + \frac{1}{6} + \frac{1}{3} + \frac{1}{3} + \frac{1}{6} + 0 \). Simplifying, this results in \( 0 + 0.1667 + 0.3333 + 0.3333 + 0.1667 + 0 = 1 \). So, the sum is 1. However, the probabilities for \( P(1) \) and \( P(6) \) are 0, which is unusual for a fair die, but it technically still sums to 1.
02

Evaluate Non-Negativity of Probabilities (a)

Each probability in a legitimate model must be between 0 and 1 inclusive. The given probabilities are: \( 0, \frac{1}{6}, \frac{1}{3}, \frac{1}{3}, \frac{1}{6}, 0 \). All values are within the range from 0 to 1, so this condition is satisfied.
03

Assess the Sum of Probabilities (b)

Again, sum the probabilities: \( P(\text{clubs}) + P(\text{diamonds}) + P(\text{hearts}) + P(\text{spades}) = \frac{12}{52} + \frac{12}{52} + \frac{12}{52} + \frac{16}{52} \). This simplifies to \( \frac{12 + 12 + 12 + 16}{52} = \frac{52}{52} = 1 \). The sum equals 1.
04

Evaluate Non-Negativity of Probabilities (b)

Check if each individual probability is non-negative: \( \frac{12}{52}, \frac{12}{52}, \frac{12}{52}, \frac{16}{52} \) are all positive, satisfying this condition.
05

Assess the Sum of Probabilities (c)

Sum the probabilities for each group: \( P(\text{female full-time}) + P(\text{male full-time}) + P(\text{female part-time}) + P(\text{male part-time}) = 0.56 + 0.44 + 0.24 + 0.17 \). The total is \( 1.41 \), which is greater than 1.
06

Evaluate Non-Negativity of Probabilities (c)

Ensure each probability is non-negative: \( 0.56, 0.44, 0.24, 0.17 \) are all positive, so this condition is met. However, the total exceeding 1 invalidates the model.

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.

Probability Distribution
Probability distribution is a fundamental concept in the study of probability. It describes how probabilities are assigned to each possible outcome in a sample space. Imagine rolling a six-sided die. The probability distribution would assign each side of the die a probability value, representing the likelihood of each side landing face up.

In any legitimate probability distribution, two key conditions must always be met:
  • The sum of probabilities of all possible outcomes must be equal to 1. For example, if you're rolling a standard die, the sum of probabilities for landing on one particular side: 1/6 for each of the six sides, should sum up to 1 (i.e., 6/6).
  • Each probability value has to lie between 0 (never happens) and 1 (definitely happens), including 0 and 1 themselves. This guarantees that probabilities remain realistic and bounded.
Understanding these requirements helps ensure that the assignment of probabilities is sensible and conforms to the rules of probability. This is vital for creating and analyzing any probability distribution correctly.
Probability Model
A probability model is essentially a mathematical representation of a random phenomenon. It consists of two parts: the sample space, which includes all possible outcomes, and a method for assigning probabilities to events within this sample space. Consider a deck of cards as an example. A probability model would document each card's likelihood of being drawn—such as the probability of drawing a club, a diamond, a heart, or a spade.

The construction of a probability model needs to address the same foundational rules applicable to probability distributions: the total probability of all possible outcomes must sum to 1, and every individual probability must be non-negative, ranging between 0 and 1.

Crafting a probability model allows statisticians and scientists to predict and analyze potential outcomes. It is key, for example, in assessing the fairness of a game or any random process-driven scenario. While a probability model needs to obey mathematical laws, a legitimately constructed model might not always reflect reality or practical situations precisely.
Legitimacy of Probability
For a probability assignment to be considered legitimate, it must strictly adhere to the two essential rules of probability. Firstly, the total probability of all events or outcomes must add up to 1. This ensures that some outcome in the sample space will occur when an event happens. Take, for instance, dealing a card from a standard deck; the probabilities of each suit adding up to 52/52 satisfies this condition.

Secondly, each probability assigned cannot be negative, and it must not exceed 1. Each probability must realistically mirror the chance of its corresponding outcome occurring. For instance, rolling a fair die with probabilities of 0 for some sides still might be mathematically allowed if they sum to 1, but it doesn't accurately reflect fair play.

The exercise illustrates that without meeting these conditions, mathematical representations of probability can't be legitimized. When probabilities do not sum to 1, such as in situation (c) where the total exceeds 1, the model is invalidated. This could highlight either an error in calculations or an inaccurate representation of potential outcomes. These principles guide the creation, assessment, and improvement of probability models.

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

You read in a book on poker that the probability of being dealt a straight flush in a five-card poker hand is \(1 / 64,974\). This means that (a) if you deal millions of poker hands, the fraction of them that contain a straight flush will be very close to \(1 / 64,974 .\) (b) if you deal 64,974 poker hands, exactly one of them will contain a straight flush. (c) if you deal \(6,497,400\) poker hands, exactly 100 of them will contain a straight flush.

Running a Mile. A study of 12,000 able-bodied male students at the University of Illinois found that their times for the mile run were approximately Normal with mean \(7.11\) minutes and standard deviation \(0.74\) minute. \({ }^{11}\) Choose a student at random from this group and call his time for the mile \(Y\). (a) Is \(Y\) a finite or continuous random variable? Explain your answer. (b) Say in words what the meaning of \(P(Y \geq 3.0)\) is. What is this probability? (c) Write the event "the student could run a mile in less than 6 minutes" in terms of values of the random variable \(Y\). What is the probability of this event?

The Medical College Admission Test. The Normal distribution with mean \(\mu=499.6\) and standard deviation \(\sigma=10.4\) is a good description of the total score on the Medical College Admission Test (MCAT). \({ }^{9}\) This is a continuous probability model for the score of a randomly chosen student. Call the score of a randomly chosen srudent \(X\) for short. (a) Write the event "the student chosen has a score of 510 or higher" in terms of \(\boldsymbol{K}\). (b) Find the probability of this event.

1\. Choose a common fruit fly Drosophila melanogaster at random. Call the length of the thorax (where the wings and legs attach) \(Y\). The random variable \(Y\) has the Normal distribution with mean \(\mu=0.800\) millimeter (mm) and standard deviation \(\sigma=0.078 \mathrm{~mm}\). The probability \(P(Y>1)\) that the fly you choose has a thorax more than \(1 \mathrm{~mm}\) long is about (a) \(0.995\) (b) \(0.5\) (c) \(0.005\).

Nickels falling over. You may feel that it is obvious that the probability of a head in tossing a coin is about \(1 / 2\) because the coin has two faces. Such opinions are not always correct. Stand a nickel on edge on a hard, flat surface. Pound the surface with your hand so that the nickel falls over. What is the probability that it falls with heads upward? Make at least 50 trials to estimate the probability of a head.

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.