/*! 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 8 Consider two independent tosses ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 8

Consider two independent tosses of a fair coin. Let \(A\) be the event that the first toss lands heads, let \(B\) be the event that the second toss lands heads, and let \(C\) be the event that both land on the same side. Show that the events \(A\), \(B, C\) are pairwise independent-that is, \(A\) and \(B\) are independent, \(A\) and \(C\) are independent, and \(B\) and \(C\) are independent-but not independent.

Short Answer

Expert verified
The events A, B, and C are pairwise independent, as shown by calculating the probabilities of their intersections and comparing them with the products of their probabilities: - \(P(A \cap B) = P(A) \cdot P(B) = 0.25\) - \(P(A \cap C) = P(A) \cdot P(C) = 0.25\) - \(P(B \cap C) = P(B) \cdot P(C) = 0.25\) However, they are not mutually independent, since \(P(A \cap B \cap C) \neq P(A) \cdot P(B) \cdot P(C)\) (0 鈮� 0.125).

Step by step solution

01

Probability of A and B

Since the coin is fair, we can find the probability of events A and B as follows: - Probability of A (First toss landing heads): \(P(A) = 0.5\) - Probability of B (Second toss landing heads): \(P(B) = 0.5\)
02

Probability of C

To calculate the probability of both coins landing on the same side (either both heads or both tails), we will use the following formula: \[P(C) = P(\text{both heads}) + P(\text{both tails})\] \[P(C) = (P(A) \cdot P(B)) + ((1-P(A))\cdot(1-P(B)))\] Plug in the values of \(P(A)\) and \(P(B)\): \[P(C) = (0.5 \cdot 0.5) + ((1-0.5)\cdot(1-0.5))\] \[P(C)= 0.25 + 0.25 = 0.5\]
03

Pairwise Independence

For events A, B, and C to be pairwise independent, the following conditions should hold: - A and B are independent: \(P(A \cap B) = P(A) \cdot P(B)\) - A and C are independent: \(P(A \cap C) = P(A) \cdot P(C)\) - B and C are independent: \(P(B \cap C) = P(B) \cdot P(C)\) Let's check these conditions: 1) A and B are independent: \[P(A \cap B) = P(\text{both tosses land heads}) = 0.25\] Compare this with \(P(A) \cdot P(B) = 0.5 \cdot 0.5 = 0.25\). So, A and B are independent. 2) A and C are independent: \[P(A \cap C) = P(\text{first toss lands heads and both on the same side})\] Since "both on the same side" while the first coin lands heads means "both lands heads," this becomes: \[P(A \cap C) = 0.25\] Compare this with \(P(A) \cdot P(C) = 0.5 \cdot 0.5 = 0.25\). So, A and C are independent. 3) B and C are independent: \[P(B \cap C) = P(\text{second toss lands heads and both on the same side})\] This is similar to the scenario where A and C are independent: \[P(B \cap C) = 0.25\] Compare this with \(P(B) \cdot P(C) = 0.5 \cdot 0.5 = 0.25\). So, B and C are independent.
04

Checking for Independence

For events A, B, and C to be mutually independent, the following condition should also hold: - \(P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)\) \[P(A \cap B \cap C) = P(\text{all coins land heads}) = 0\] Compare this with \(P(A) \cdot P(B) \cdot P(C) = 0.5 \cdot 0.5 \cdot 0.5 = 0.125\). Since \(P(A \cap B \cap C) \neq P(A) \cdot P(B) \cdot P(C)\), the events A, B, and C are not mutually independent. So, the events A, B, and C are pairwise independent but 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 91影视!

Key Concepts

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

Pairwise Independence
In probability theory, pairwise independence is a property that involves considering the independence between each pair of events within a given set. If you have events such as A, B, and C, these events are pairwise independent if every pair of these events is independent of each other. This means:- The joint probability of A and B is the product of their individual probabilities: \( P(A \cap B) = P(A) \times P(B) \).- The joint probability of A and C is the product of their individual probabilities: \( P(A \cap C) = P(A) \times P(C) \).- The joint probability of B and C is the product of their individual probabilities: \( P(B \cap C) = P(B) \times P(C) \).
It's essential to note that if a set of events is only pairwise independent, it doesn鈥檛 ensure that the entire group of events is mutually independent. In the example exercise with the fair coin tosses, the events A, B, and C were confirmed to meet pairwise independence but not mutual independence.
Mutual Independence
While pairwise independence considers only two events at a time, mutual independence involves all events in the set being considered together. Events A, B, and C are mutually independent if the probability of all occurring together is equal to the product of their individual probabilities: \( P(A \cap B \cap C) = P(A) \times P(B) \times P(C) \).
For mutual independence:- All possible subsets of the events, including pairs and single events, need to satisfy the independence condition.- This is a stricter condition than pairwise independence and less common in practice because the joint behavior of all events can complicate the pattern.
In the coin toss exercise, the events A, B, and C were shown not to be mutually independent because the joint probability \( P(A \cap B \cap C) \) did not equal the product \( P(A) \times P(B) \times P(C) \). This discrepancy means despite being pairwise independent, they don't meet the broader criteria for mutual independence.
Fair Coin Toss
A fair coin toss refers to a situation where a coin has an equal chance of landing on heads or tails. This means the probability of landing heads is \( P( ext{Heads}) = 0.5 \) and the probability of landing tails is \( P( ext{Tails}) = 0.5 \).
When analyzing fair coin tosses:- Each toss is independent of other tosses. This means the outcome of one toss does not affect another.- In a sequence of multiple tosses, the probabilities are easily calculated by raising the individual probabilities to the power of the number of tosses.
In scenarios such as the exercise, a fair coin is pivotal because it establishes a uniform probability model, allowing us to apply principles of probability with confidence and clarity. Modern examples include the use of fair coin tosses in decision-making processes where impartiality is crucial.
Probability of Events
The probability of events is a fundamental concept within probability theory. It quantifies the likelihood that a given event will occur. For any event, the probability is a number between 0 and 1:- A probability of 0 means the event cannot occur.- A probability of 1 means the event will certainly occur.
Calculating probabilities involves:- Determining individual probabilities of simple events, such as a single coin landing heads, calculated as \( P(A) = 0.5 \) in a fair coin example.- Calculating the probability of combined events like the probability that both tosses land on the same side, which is achieved through formulas like \( P(C) = P( ext{both heads}) + P( ext{both tails}) \).
Overall, understanding how to compute and interpret these probabilities enables predictions about outcomes, assessments of risk, and insights into the randomized processes seen in examples like coin tosses.

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

How can 20 balls, 10 white and 10 black, be put into two urns so as to maximize the probability of drawing a white ball if an urn is selected at random and a ball is drawn at random from it?

A certain organism possesses a pair of each of 5 different genes (which we will designate by the first 5 letters of the English alphabet). Each gene appears in 2 forms (which we designate by lowercase and capital letters). The capital letter will be assumed to be the dominant gene in the sense that if an organism possesses the gene pair \(x X\), then it will outwardly have the appearance of the \(X\) gene. For instance, if \(X\) stands for brown eyes and \(x\) for blue eyes, then an individual having either gene pair \(X X\) or \(x X\) will have brown eyes, whereas one having gene pair \(x x\) will have blue eyes. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus 2 organisms with respective genotypes \(a A, b B, c c\), \(d D, e e\) and \(A A, B B, c c, D D, e e\) would have different genotypes but the same phenotype.) In a mating between 2 organisms each one contributes, at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of its mate. In a mating between organisms having genotypes \(a A, b B, c C, d D, e E\) and \(a a, b B, c c, D d, e e\) what is the probability that the progeny will (i) phenotypically and (ii) genotypically resemble (a) the first parent; (b) the second parent; (c) either parent; (d) neither parent?

(a) A gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random; when he flips it, it shows heads. What is the probability that it is the fair coin? (b) Suppose that he flips the same coin a second time and again it shows heads. What is now the probability that it is the fair coin? (c) Suppose that he flips the same coin a third time and it shows tails. What is now the probability that it is the fair coin?

The following method was proposed to estimate the number of people over the age of 50 that reside in a town of known population 100,000 . "As you walk along the streets, keep a running count of the percentage of people that you encounter who are over \(50 .\) Do this for a few days; then multiply the obtained percentage by 100,000 to obtain the estimate." Comment on this method. HINT: Let \(p\) denote the proportion of people in this town who are over \(50 .\) Furthermore, let \(\alpha_{1}\) denote the proportion of time that a person under the age of 50 spends in the streets, and let \(\alpha_{2}\) be the corresponding value for those over \(50 .\) What quantity does the method suggested estimate? When is it approximately equal to \(p ?\)

An urn contains 5 white and 10 black balls. A fair die is rolled and that number of balls is randomly chosen from the um. What is the probability that all of the balls selected are white? What is the conditional probability that the die landed on 3 if all the balls selected are white?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation
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.