/*! 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 34 Four women, \(\mathrm{A}, \mathr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Four women, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D},\) check their hats, and the hats are returned in a random manner. Let \(\Omega\) be the set of all possible permutations of \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\). Let \(X_{j}=1\) if the \(j\) th woman gets her own hat back and 0 otherwise. What is the distribution of \(X_{j} ?\) Are the \(X_{i}\) 's mutually independent?

Short Answer

Expert verified
X_j ~ Bernoulli(1/4) and X_i's are not mutually independent.

Step by step solution

01

Define the Probability

Each woman receives a randomly assigned hat. For any specific woman, say woman \( j \), there are four possible hats, but only one of these is her own hat. Thus the probability that woman \( j \) receives her own hat is \( \frac{1}{4} \).
02

Determine the Distribution of X_j

Because each woman can either receive her hat back (event happens with probability \( \frac{1}{4} \)) or not (event happens with probability \( \frac{3}{4} \)), \( X_j \) follows a Bernoulli distribution. Specifically, \( X_j \sim \text{Bernoulli}\left(\frac{1}{4}\right) \).
03

Evaluate Independence of X_i

Consider two women, \( A \) and \( B \). If \( A \) gets her hat, the number of available hats decreases by one, affecting the probability that \( B \) gets her own hat. Thus, the events that women \( A \) and \( B \) get their hats are not independent. Similarly, this correlation persists among all the women, indicating that \( X_i \)'s are not mutually independent.

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.

Permutations
Permutations are all about arranging objects in a specific order. When the order of arrangement matters, we talk about permutations. For example, if you have four distinct hats and four women, like in the exercise, each unique way the hats can be handed back is a permutation of these objects.
In this exercise, the total number of permutations for the four hats is the same as the number of ways to arrange them. To compute this, we use the formula for permutations of "n" distinct objects, which is given as \( n! \), where "!" signifies a factorial. So for the hats, it's \( 4! \), which equals 24 possible permutations.
This concept is important because it helps us understand the range of possible outcomes in this problem. By considering all permutations, we can examine each potential arrangement where each woman may or may not get her hat back. Understanding permutations lays the groundwork for evaluating other concepts like probability distribution in this exercise.
Bernoulli Distribution
A Bernoulli distribution describes outcomes where there are only two possible results, usually termed as "success" and "failure." Each trial or event in a Bernoulli distribution is independent of the other. To illustrate, when a woman checks to see if she gets her hat back, it is like making a binary decision: either she gets it back (success) or she doesn't (failure).

For a Bernoulli distribution, we denote it as \( X \sim \text{Bernoulli}(p) \), where "p" represents the probability of success. Here, the success is the event of a woman receiving her own hat back. In the given exercise, this probability is \( \frac{1}{4} \), as each woman has only one chance in four to get her original hat back given the random fashion of returning these items.

The expectation, or mean, of a Bernoulli random variable is equal to p (i.e., \( E[X] = p \)), while the variance is \( p(1-p) \). In this exercise, \( X_j \sim \text{Bernoulli}(\frac{1}{4}) \) which indicates a trial has a 25% chance of success.
Mutual Independence
Mutual independence is a crucial concept in probability. It extends beyond just simple independence; for a group of random variables to be mutually independent, each variable must be independent of every possible combination of others.

In the context of this hat problem, if the women receiving their respective hats were mutually independent events, then the probability of each woman getting her hat wouldn't affect others. However, this condition doesn't hold here.
When one woman successfully retrieves her hat, fewer hats remain for the others. For instance, if woman A gets her hat, it directly impacts the probability spaces—or possibilities available—for women B, C, and D. As a result, these events are interconnected.

Thus, in this problem, no mutual independence exists among the random variables \(X_i\), where each variable represents whether a particular woman gets her hat back. Recognizing the lack of mutual independence is vital because it informs how we evaluate events and calculate probabilities in a dependent system.

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

Given that \(P(X=a)=r, P(\max (X, Y)=a)=s,\) and \(P(\min (X, Y)=a)=\) \(t,\) show that you can determine \(u=P(Y=a)\) in terms of \(r, s,\) and \(t .\)

You are given two urns and fifty balls. Half of the balls are white and half are black. You are asked to distribute the balls in the urns with no restriction placed on the number of either type in an urn. How should you distribute the balls in the urns to maximize the probability of obtaining a white ball if an urn is chosen at random and a ball drawn out at random? Justify your answer.

A radioactive material emits \(\alpha\) -particles at a rate described by the density function \(f(t)=.1 e^{-.1 t}\) Find the probability that a particle is emitted in the first 10 seconds, given that (a) no particle is emitted in the first second. (b) no particle is emitted in the first 5 seconds. (c) a particle is emitted in the first 3 seconds. (d) a particle is emitted in the first 20 seconds.

One of the first conditional probability paradoxes was provided by Bertrand. \({ }^{23}\) It is called the Box Paradox. A cabinet has three drawers. In the first drawer there are two gold balls, in the second drawer there are two silver balls, and in the third drawer there is one silver and one gold ball. A drawer is picked at random and a ball chosen at random from the two balls in the drawer. Given that a gold ball was drawn, what is the probability that the drawer with the two gold balls was chosen?

Two cards are drawn from a bridge deck. What is the probability that the second card drawn is red?

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.