/*! 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 24 Show that if one event \(A\) is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that if one event \(A\) is contained in another event \(B\) (i.e., \(A\) is a subset of \(B\) ), then \(P(A) \leq P(B)\). [Hint: For such \(A\) and \(B, A\) and \(B \cap A^{\prime}\) are disjoint and \(B=A \cup\left(B \cap A^{\prime}\right)\), as can be seen from a Venn diagram.] For general \(A\) and \(B\), what does this imply about the relationship among \(P(A \cap B)\), \(P(A)\), and \(P(A \cup B)\) ?

Short Answer

Expert verified
If \(A \subset B\), then \(P(A) \leq P(B)\). For any \(A\) and \(B\), \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).

Step by step solution

01

Understanding the Problem

We are given two events, \(A\) and \(B\), where \(A\) is a subset of \(B\) (\(A \subset B\)). We need to prove that if \(A\) is contained in \(B\), then the probability of \(A\), \(P(A)\), is less than or equal to the probability of \(B\), \(P(B)\). Additionally, we are asked to understand the relationship between \(P(A \cap B)\), \(P(A)\), and \(P(A \cup B)\) for any two events \(A\) and \(B\).
02

Using the Subset Relationship

Since \(A\) is a subset of \(B\), every outcome in \(A\) is also in \(B\). Therefore, the event \(B\) can be written as \(B = A \cup (B \cap A^\prime)\), where \(A^\prime\) is the complement of \(A\). This indicates that \(A\) and \(B \cap A^\prime\) are disjoint sets.
03

Probability of a Union of Disjoint Events

Using the properties of probability, the probability of a union of disjoint sets is the sum of their probabilities. Therefore, \(P(B) = P(A) + P(B \cap A^\prime)\). Since probabilities are non-negative, \(P(B \cap A^\prime) \geq 0\).
04

Conclusion of Part 1

From the equation \(P(B) = P(A) + P(B \cap A^\prime)\) and knowing that \(P(B \cap A^\prime) \geq 0\), it follows that \(P(A) \leq P(B)\). This proves the initial part of the exercise.
05

General Relationship Between Probabilities

For any events \(A\) and \(B\), we know that \(A \cap B\) is the intersection of \(A\) and \(B\), and \(A \cup B\) is the union. The relationship among these probabilities is given by the inclusion-exclusion principle: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Thus, \(P(A \cap B) \leq P(A)\) and \(P(A \cap B) \leq P(B)\). This also implies \(P(A) \leq P(A \cup B)\) and \(P(B) \leq P(A \cup B)\).

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.

Subset Relation in Events
In probability theory, understanding the concept of subsets in relation to events is crucial. When we say that one event, let's call it event \(A\), is a subset of another event \(B\) (represented as \(A \subset B\)), it essentially means that every possible outcome that leads to event \(A\) also leads to event \(B\). This sets the foundation for comparing the probabilities of these events.

Now, if \(A\) is a subset of \(B\), we must consider the probabilities associated with these events. Since all outcomes of \(A\) are included in the outcomes of \(B\), the likelihood or probability of \(A\) occurring cannot exceed the probability of \(B\) occurring. Mathematically, this is expressed as \(P(A) \leq P(B)\). This inequality forms a fundamental rule that helps in solving more complex problems related to probability subsets.

To visualize, think of a Venn diagram where the circle representing \(A\) is entirely contained within the circle representing \(B\). This visual shows how \(A\) shares all its outcomes with \(B\), reinforcing the idea of subset relationships.
Inclusion-Exclusion Principle
The inclusion-exclusion principle is a core concept for understanding probabilities involving unions of events. It's used to find the probability of either of two events occurring. However, it also accounts for the probability that both events happen simultaneously, to adjust for any double-counting.

For any two events \(A\) and \(B\), the principle is expressed with the formula:\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]This formula ensures that we count the overlap (intersection \(A \cap B\) only once by subtracting its probability from the sum of the individual probabilities.

Here’s how you can apply it:
  • Calculate the probability of each event occurring independently: \(P(A)\) and \(P(B)\).
  • Add these probabilities together.
  • Subtract the probability that both events happen together: \(P(A \cap B)\).
The result, \(P(A \cup B)\), gives you the probability that at least one of the events \(A\) or \(B\) occurs. This principle is especially useful in solving problems where overlapping parts of two events must be considered separately.
Probability of Union and Intersection
Understanding the probability of union and intersection is essential when dealing with multiple events. The union of two events \(A\) and \(B\), denoted as \(A \cup B\), represents the event that either \(A\) or \(B\) occurs (or both). Meanwhile, the intersection \(A \cap B\) focuses on scenarios where both \(A\) and \(B\) happen simultaneously.

The probability of the union can be computed using the inclusion-exclusion principle, previously discussed:\[P(A \cup B) = P(A) + P(B) - P(A \cap B)\]This calculation helps assess the likelihood of at least one event happening. It's crucial to consider the intersection to avoid overestimating the probability.

On the other hand, the probability of the intersection, \(P(A \cap B)\), is the chance of both events occurring. It's particularly useful in scenarios requiring the understanding of simultaneous outcomes.
  • The intersection's probability can never exceed the probability of each individual event: \(P(A \cap B) \leq P(A)\) and \(P(A \cap B) \leq P(B)\).
  • Similarly, each individual probability cannot exceed the union probability: \(P(A) \leq P(A \cup B)\) and \(P(B) \leq P(A \cup B)\).
Grasping these relationships equips you with powerful tools for solving complex probability problems.

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

A stereo store is offering a special price on a complete set of components (receiver, compact disc player, speakers, cassette deck). A purchaser is offered a choice of manufacturer for each component: Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood Compact disc player: Onkyo, Pioneer, Sony, Technics Speakers: Boston, Infinity, Polk Cassette deck: Onkyo, Sony, Teac, Technics A switchboard display in the store allows a customer to hook together any selection of components (consisting of one of each type). Use the product rules to answer the following questions: a. In how many ways can one component of each type be selected? b. In how many ways can components be selected if both the receiver and the compact disc player are to be Sony? c. In how many ways can components be selected if none is to be Sony? d. In how many ways can a selection be made if at least one Sony component is to be included? e. If someone flips switches on the selection in a completely random fashion, what is the probability that the system selected contains at least one Sony component? Exactly one Sony component?

An engineering construction firm is currently working on power plants at three different sites. Let \(A_{f}\) denote the event that the plant at site \(i\) is completed by the contract date. Use the operations of union, intersection, and complementation to describe each of the following events in terms of \(A_{1}, A_{2}\), and \(A_{3}\), draw a Venn diagram, and shade the region corresponding to each one. a. At least one plant is completed by the contract date. b. All plants are completed by the contract date. c. Only the plant at site 1 is completed by the contract date. d. Exactly one plant is completed by the contract date. e. Either the plant at site 1 or both of the other two plants are completed by the contract date.

There has been a great deal of controversy over the last several years regarding what types of surveillance are appropriate to prevent terrorism. Suppose a particular surveillance system has a \(99 \mathrm{~F}\) chance of correctly identifying a future terrorist and a \(99.9 \%\) chance of correctly identifying someone who is not a future terrorist. If there are 1000 future terrorists in a population of 300 million, and one of these 300 million is randomly selected, scrutinized by the system, and identified as a future terrorist, what is the probability that he/she actually is a future terrorist? Does the value of this probability make you uneasy about using the surveillance system? Explain.

Four universities- \(1,2,3\), and 4 -are participating in a holiday basketball tournament. In the first round, 1 will play 2 and 3 will play 4 . Then the two winners will play for the championship, and the two losers will also play. One possible outcome can be denoted by 1324 ( 1 beats 2 and 3 beats 4 in first- round games, and then 1 beats 3 and 2 beats 4 ). a. List all outcomes in \(\mathscr{S}\). b. Let \(A\) denote the event that 1 wins the tournament. List outcomes in \(A\). c. Let \(B\) denote the event that 2 gets into the championship game. List outcomes in \(B\). d. What are the outcomes in \(A \cup B\) and in \(A \cap B\) ? What are the outcomes in \(A^{\prime}\) ?

An academic department with five faculty membersAnderson, Box, Cox, Cramer, and Fisher-must select two of its members to serve on a personnel review committee. Because the work will be time-consuming, no one is anxious to serve, so it is decided that the representative will be selected by putting five slips of paper in a box, mixing them, and selecting two. a. What is the probability that both Anderson and Box will be selected? [Hint: List the equally likely outcomes.] b. What is the probability that at least one of the two members whose name begins with \(C\) is selected? c. If the five faculty members have taught for \(3,6,7,10\), and 14 years, respectively, at the university, what is the probability that the two chosen representatives have at least 15 years' teaching experience at the university?

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.