/*! 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 69 If \(A\) and \(B\) are independe... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 69

If \(A\) and \(B\) are independent events, show that \(A^{\prime}\) and \(B\) are also independent. [Hint: First establish a relationship among \(P\left(A^{\prime} \cap B\right), P(B)\), and \(P(A \cap B)\).]

Short Answer

Expert verified
\(A^{\prime}\) and \(B\) are independent because \(P(A^{\prime} \cap B) = P(A^{\prime}) \cdot P(B)\).

Step by step solution

01

Define Independence

Two events, \(A\) and \(B\), are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other. Mathematically, this can be expressed as \(P(A \cap B) = P(A) \cdot P(B)\).
02

Understand the Hint

We need to establish a relationship between \(P(A^{\prime} \cap B)\), \(P(B)\), and \(P(A \cap B)\). The goal is to demonstrate independence between \(A^{\prime}\) and \(B\).
03

Express \(A^{\prime} \cap B\) Using Set Operations

Given that \(A^{\prime}\) is the complement of \(A\), it means \(A^{\prime}\) consists of all outcomes not in \(A\). Thus, \(A^{\prime} \cap B\) corresponds to the outcomes that are in \(B\) but not in \(A\). This can be expressed as \(A^{\prime} \cap B = B - (A \cap B)\).
04

Calculate \(P(A^{\prime} \cap B)\)

We use the formula for the probability of a set difference, which tells us \(P(A^{\prime} \cap B) = P(B) - P(A \cap B)\).
05

Use Independence of \(A\) and \(B\)

Since \(A\) and \(B\) are independent, \(P(A \cap B) = P(A) \cdot P(B)\). Substituting this into our expression from Step 4 gives \(P(A^{\prime} \cap B) = P(B) - P(A) \cdot P(B)\).
06

Factor Out \(P(B)\)

Factor \(P(B)\) out of the expression: \(P(A^{\prime} \cap B) = P(B)(1 - P(A))\). Notice that \(1 - P(A) = P(A^{\prime})\).
07

Show Independence of \(A^{\prime}\) and \(B\)

Since \(P(A^{\prime} \cap B) = P(A^{\prime}) \cdot P(B)\), by definition \(A^{\prime}\) and \(B\) are 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.

Complementary Events
In probability theory, complementary events are an essential concept to understand. When dealing with an event, say event \(A\), its complement, denoted as \(A'\), encompasses all the outcomes that are not part of \(A\). Essentially, if the probability of an event \(A\) happening is \(P(A)\), then the probability that \(A\) does not happen, or the probability of \(A'\), is \(1 - P(A)\).

Here鈥檚 a friendly reminder: the total probability of all possible outcomes in an experiment is always 1. This makes complementary events quite straightforward to work with as they together cover the whole sample space.

Why are complements important? They allow you to reason about the occurrence of an event by considering its non-occurrence, which is often easier to handle in calculations.
Set Operations
Understanding set operations is crucial when working with probabilities. The key operations include union, intersection, and complement. Here鈥檚 how they work in a probability context:

  • Union (\(A \cup B\)): This operation represents all outcomes that are in either event \(A\), event \(B\), or both.
  • Intersection (\(A \cap B\)): These are the outcomes that are common to both events \(A\) and \(B\).
  • Complement (\(A'\)): Covers all outcomes not in event \(A\).


In the exercise we discussed, we applied the difference operation derived from complement and intersection, such as \(A' \cap B = B - (A \cap B)\). This technique is particularly handy when calculating the probabilities involving complementary events and helps us understand how outcomes overlap within sets.
Probability of Intersection
The probability of intersection represents the likelihood that two events, \(A\) and \(B\), happen simultaneously and is denoted \(P(A \cap B)\). For independent events, the formula is straightforward: \(P(A \cap B) = P(A) \cdot P(B)\). This formula holds due to the independence assumption, meaning one event happening does not alter the probability of the other event occurring.

In our exercise, we discovered that the intersection probability for events involving complements can be expressed using set difference. Specifically, \(P(A' \cap B) = P(B) - P(A \cap B)\). This relation stems from the concept of subtracting the probability of overlapping outcomes in sets, and helps ascertain the probability of \(A\) not happening while \(B\) still occurs.
Probability Theory
Probability theory is a branch of mathematics concerned with analyzing random phenomena. It gives us the tools to quantify how likely events are to occur and is foundational for fields such as statistics, finance, and science. Some fundamental principles include:

  • Probability Ranges: The probability of any event falls between 0 and 1.
  • Sum of Probabilities: The total probability of all mutually exclusive outcomes in an experiment is 1.
  • Complementary Rule: \(P(A') = 1 - P(A)\) provides a way to calculate the probability of an event not occurring.


By applying these principles, probability theory enables us to make informed decisions based on the likelihood of different outcomes. In our example, we showed how independence in probability theory means two events do not influence each other, using this to extend independence from events \(A\) and \(B\) to their complements, \(A'\) and \(B\). Understanding these concepts allows for a solid foundation in reasoning about real-world uncertainties.

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

For any events \(A\) and \(B\) with \(P(B)>0\), show that \(P(A \mid B)+P\left(A^{\prime} \mid B\right)=1 .\)

Three molecules of type \(A\), three of type \(B\), three of type \(C\), and three of type \(D\) are to be linked together to form a chain molecule. One such chain molecule is \(A B C D A B C D A B C D\), and another is \(B C D D A A A B D B C C\). a. How many such chain molecules are there? [Hint: If the three A's were distinguishable from one another- \(A_{1}, A_{2}, A_{3}\)-and the \(B\) 's, \(C\) 's, and \(D\) 's were also, how many molecules would there be? How is this number reduced when the subscripts are removed from the \(A\) 's?] b. Suppose a chain molecule of the type described is randomly selected. What is the probability that all three molecules of each type end up next to each other (such as in \(B B B A A A D D D C C C\) )?

An individual is presented with three different glasses of cola, labeled \(C, D\), and \(P .\) He is asked to taste all three and then list them in order of preference. Suppose the same cola has actually been put into all three glasses. a. What are the simple events in this ranking experiment, and what probability would you assign to each one? b. What is the probability that \(C\) is ranked first? c. What is the probability that \(C\) is ranked first and \(D\) is ranked last?

Individual A has a circle of five close friends (B, C, D, E, and F). A has heard a certain rumor from outside the circle and has invited the five friends to a party to circulate the rumor. To begin, A selects one of the five at random and tells the rumor to the chosen individual. That individual then selects at random one of the four remaining individuals and repeats the rumor. Continuing, a new individual is selected from those not already having heard the rumor by the individual who has just heard it, until everyone has been told. a. What is the probability that the rumor is repeated in the order B, C, D, E, and F? b. What is the probability that \(F\) is the third person at the party to be told the rumor? c. What is the probability that \(\mathrm{F}\) is the last person to hear the rumor?

Consider a woman whose brother is afflicted with hemophilia, which implies that the woman's mother has the hemophilia gene on one of her two X chromosomes (almost surely not both, since that is generally fatal). Thus there is a \(50-50\) chance that the woman's mother has passed on the bad gene to her. The woman has two sons, each of whom will independently inherit the gene from one of her two chromosomes. If the woman herself has a bad gene, there is a \(50-50\) chance she will pass this on to a son. Suppose that neither of her two sons is afflicted with hemophilia. What then is the probability that the woman is indeed the carrier of the hemophilia gene? What is this probability if she has a third son who is also not afflicted?
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

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.