/*! 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 58 Show that for any three events \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 58

Show that for any three events \(A, B\), and \(C\) with \(P(C)>0\), \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C) .\)

Short Answer

Expert verified
The equation is derived using basic probability rules and conditional probability formulas, confirming the identity.

Step by step solution

01

Understanding Conditional Probability

For any event, the conditional probability is defined as the probability of one event occurring given that another event has occurred. The formula is given by \( P(A \mid C) = \frac{P(A \cap C)}{P(C)} \), provided that \( P(C) > 0 \).
02

Applying Conditional Probability Definition

We will express each term on the right-hand side of the equation using the definition of conditional probability: - \( P(A \mid C) = \frac{P(A \cap C)}{P(C)} \) - \( P(B \mid C) = \frac{P(B \cap C)}{P(C)} \) - \( P(A \cap B \mid C) = \frac{P(A \cap B \cap C)}{P(C)} \)
03

Rewriting the Union Probability Using Basic Probability

The probability of the union of two events \(A\) and \(B\) can be written in general form as:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] We want to find \(P(A \cup B \mid C)\). By the definition of conditional probability, it is:\[ P(A \cup B \mid C) = \frac{P((A \cup B) \cap C)}{P(C)} \]
04

Relating to Base Events

The intersection \((A \cup B) \cap C\) can be rewritten using distribution over union:\[ (A \cup B) \cap C = (A \cap C) \cup (B \cap C) \]So the numerator becomes:\[ P((A \cup B) \cap C) = P((A \cap C) \cup (B \cap C)) \]
05

Substituting Intersection into Conditional Probability

Using the union formula for base events:\[ P((A \cap C) \cup (B \cap C)) = P(A \cap C) + P(B \cap C) - P((A \cap B) \cap C) \]So the conditional probability becomes:\[ P(A \cup B \mid C) = \frac{P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)}{P(C)} \]
06

Substituting Back to Conditional Terms

Substitute the expressions derived from the definition of conditional probability:\[ \frac{P(A \cap C)}{P(C)} + \frac{P(B \cap C)}{P(C)} - \frac{P(A \cap B \cap C)}{P(C)} \]Simplified, it becomes:\[ P(A \mid C) + P(B \mid C) - P(A \cap B \mid C) \]
07

Concluding Proof

We have shown that each side of the original equation is equal, thus establishing the identity:\[ P(A \cup B \mid C) = P(A \mid C) + P(B \mid C) - P(A \cap B \mid C) \]

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.

Intersection and Union of Events
In probability, understanding the intersection and union of events is crucial. These concepts help us determine how events combine and interact with each other. The union of two events, like \(A\) and \(B\), denoted \(A \cup B\), represents the event that at least one of \(A\) or \(B\) occurs. On the other hand, the intersection, denoted \(A \cap B\), represents the event where both \(A\) and \(B\) occur simultaneously.

Using these definitions, the probability of the union of the events \(A\) and \(B\) is expressed by the formula:
  • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
This formula helps us calculate the likelihood of either event occurring without double-counting their intersection. Similarly, when considering multiple events, the intersections and unions can be expanded using set operations to simplify complex probability calculations.
Probability Formulas
Probability formulas form the foundation of solving problems in probability theory. One of the most fundamental formulas is the general rule for addition of probabilities, which is used to find the probability of the union of two events. As mentioned earlier:
  • \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\)
This is used when calculating the probability of either or both events occurring.

Another common aspect is using these formulas with conditional probabilities, where we must consider the effect of one event on another in the sample space. Remember, these formulas must be used alongside the conditions given in a problem to find the exact solution, as conditions may alter the sample space or event outcomes.
Conditional Probability Formula
Conditional probability allows us to calculate the probability of an event occurring given that another event has already happened. The conditional probability formula is expressed as:
  • \(P(A \mid C) = \frac{P(A \cap C)}{P(C)}\)
This formula is valid given \(P(C) > 0\), meaning \(C\) has indeed occurred. In solving probability problems, this formula helps adjust the likelihood based on known information.

Using conditional probability with intersections and unions can solve complex scenarios like the given exercise. For instance, the equation \(P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-P(A \cap B \mid C)\) shows how the union probability adjusts when given event \(C\). This is calculated using the definitions of conditional probability and requires understanding how intersections and unions change under those conditions.

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 customers purchasing a refrigerator at a certain appliance store, let \(A\) be the event that the refrigerator was manufactured in the U.S., \(B\) be the event that the refrigerator had an icemaker, and \(C\) be the event that the customer purchased an extended warranty. Relevant probabilities are $$ \begin{aligned} &P(A)=.75 \quad P(B \mid A)=.9 \quad P\left(B \mid A^{\prime}\right)=.8 \\ &P(C \mid A \cap B)=.8 \quad P\left(C \mid A \cap B^{\prime}\right)=.6 \\ &P\left(C \mid A^{\prime} \cap B\right)=.7 \quad P\left(C \mid A^{\prime} \cap B^{\prime}\right)=.3 \end{aligned} $$ a. Construct a tree diagram consisting of first-, second-, and third- generation branches, and place an event label and appropriate probability next to each branch. b. Compute \(P(A \cap B \cap C)\). c. Compute \(P(B \cap C)\). d. Compute \(P(C)\). e. Compute \(P(A \mid B \cap C)\), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased.

One percent of all individuals in a certain population are carriers of a particular disease. A diagnostic test for this disease has a \(90 \%\) detection rate for carriers and a \(5 \%\) detection rate for noncarriers. Suppose the test is applied independently to two different blood samples from the same randomly selected individual. a. What is the probability that both tests yield the same result? b. If both tests are positive, what is the probability that the selected individual is a carrier?

In five-card poker, a straight consists of five cards with adjacent denominations (e.g., 9 of clubs, 10 of hearts, jack of hearts, queen of spades, and king of clubs). Assuming that aces can be high or low, if you are dealt a five-card hand, what is the probability that it will be a straight with high card 10? What is the probability that it will be a straight? What is the probability that it will be a straight flush (all cards in the same suit)?

1,30 \%\( of the time on airline \)\\#鈥� # A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.; \(50 \%\) of the time she travels on airline \(\\# 1,30 \%\) of the time on airline \(\\# 2\), and the remaining \(20 \%\) of the time on airline \(\\# 3\). For airline \(\\# 1\), flights are late into D.C. \(30 \%\) of the time and late into L.A. \(10 \%\) of the time. For airline \(\\# 2\), these percentages are \(25 \%\) and \(20 \%\), whereas for airline \(\\# 3\) the percentages are \(40 \%\) and \(25 \%\). If we learn that on a particular trip she arrived late at exactly one of the two destinations, what are the posterior probabilities of having flown on airlines \(\\# 1, \\# 2\), and #3?Assume that the chance of a late arrival in L.A. is unaffected by what happens on the flight to D.C. [Hint: From the tip of each first-generation branch on a tree diagram, draw three second-generation branches labeled, respectively, 0 late, 1 late, and 2 late.]

Seventy percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, \(60 \%\) have an emergency locator, whereas \(90 \%\) of the aircraft not discovered do not have such a locator. Suppose a light aircraft has disappeared. a. If it has an emergency locator, what is the probability that it will not be discovered? b. If it does not have an emergency locator, what is the probability that it will be discovered?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

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.