/*! 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 114 Show that if \(A_{1}, A_{2}\), a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 114

Show that if \(A_{1}, A_{2}\), and \(A_{3}\) are independent events, then \(P\left(A_{1} \mid A_{2} \cap A_{3}\right)=P\left(A_{1}\right)\) .

Short Answer

Expert verified
If \(A_1\), \(A_2\), and \(A_3\) are independent, then \(P(A_1 | A_2 \\ cap A_3) = P(A_1)\).

Step by step solution

01

Understand the concept of independence

Independence of events implies that the occurrence of one event does not affect the probability of the others. Specifically, if events \(A_1\), \(A_2\), and \(A_3\) are independent, then \(P(A_1 \ cap A_2) = P(A_1)P(A_2)\), \(P(A_1 \ cap A_3) = P(A_1)P(A_3)\), and \(P(A_2 \ cap A_3) = P(A_2)P(A_3)\). Furthermore, \(P(A_1 \ cap A_2 \ cap A_3) = P(A_1)P(A_2)P(A_3)\).
02

Recall the definition of conditional probability

The conditional probability \(P(A_1 | A_2 \ cap A_3)\) is defined as \(\frac{P(A_1 \ cap A_2 \ cap A_3)}{P(A_2 \ cap A_3)}\). This formula calculates the probability of \(A_1\) given that both \(A_2\) and \(A_3\) occur.
03

Substitute into the conditional probability formula using independence

Substitute the expression for \(P(A_1 \ cap A_2 \ cap A_3)\) as \(P(A_1)P(A_2)P(A_3)\) and \(P(A_2 \ cap A_3)\) as \(P(A_2)P(A_3)\) into the conditional probability formula: \[P(A_1 | A_2 \ cap A_3) = \frac{P(A_1 \ cap A_2 \ cap A_3)}{P(A_2 \ cap A_3)} = \frac{P(A_1)P(A_2)P(A_3)}{P(A_2)P(A_3)}\]
04

Simplify the expression

Cancel out \(P(A_2)P(A_3)\) in the numerator and denominator, yielding: \[P(A_1 | A_2 \ cap A_3) = P(A_1)\]Thus, it is shown that \(P(A_1 | A_2 \ cap A_3)\) equals \(P(A_1)\), confirming the independence statement.

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.

Probability Theory
Probability theory is the backbone of statistics and plays a vital role in understanding how likely events are to occur. It provides a mathematical framework to measure uncertainty and evaluate outcomes of random phenomena. Let's delve into some basic principles:
  • *Events*: Things or outcomes we want to measure. For example, rolling a dice.
  • *Probability*: A number between 0 and 1, indicating the likelihood of an event. 0 means impossible, 1 means certain.
  • *Sample Space*: All possible outcomes of a random experiment.
Imagine you are flipping a coin. The probability of getting heads is 0.5, because there are two possible outcomes (heads or tails) and each is equally likely.
Probability theory is not just about simple events; it extends to understanding the likelihood of multiple events happening together or sequentially. This foundation is essential for more advanced topics like conditional probability and event independence.
Conditional Probability
Conditional probability is a way to figure out how likely an event is, given that another event has already happened. It refines our predictions based on additional information. Consider the formula:\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]Here, \(P(A | B)\) means "the probability of event A occurring, given that B has occurred."
  • \(P(A \cap B)\): The probability that both events A and B occur.
  • \(P(B)\): The probability that event B alone occurs.
For instance, if the probability of it raining on any day is 0.3, and the probability that you carry an umbrella whenever it rains is 0.8, the probability that it rains and you carry an umbrella is adjusted by this condition.
In conditional probability, you only consider scenarios where the given condition is satisfied. This alteration in the view of the sample space aligns probabilities more closely with the real-world situations we wish to analyze.
Event Independence
Event independence is a critical concept in probability, describing situations where the occurrence of one event does not affect the occurrence of another. Mathematically, two events, A and B, are independent if:\[ P(A \cap B) = P(A) \times P(B) \]Independence tells us that knowing event A has occurred doesn't change the likelihood of event B and vice versa.
In the given exercise, the independence of three events, \(A_1\), \(A_2\), and \(A_3\) implies:
  • The probability of \(A_1\) occurring does not change whether \(A_2\) and \(A_3\) happen or not.
  • The formula \(P(A_1 | A_2 \cap A_3) = P(A_1)\) uses independence to simplify conditional probability, showing \(A_1\) is unaffected by \(A_2\) and \(A_3\).
This concept simplifies calculations and analysis significantly in complex scenarios. Understanding event independence helps you recognize which events can be analyzed separately without influencing each other, making it a cornerstone of probability theory.

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

As of April 2006, roughly 50 million .com web domain names were registered (e.g., yahoo.com). a. How many domain names consisting of just two letters in sequence can be formed? How many domain names of length two are there if digits as well as letters are permitted as characters? [Note: A character length of three or more is now mandated.] b. How many domain names are there consisting of three letters in sequence? How many of this length are there if either letters or digits are permitted? [Note: All are currently taken.] c. Answer the questions posed in (b) for four-character sequences. d. As of April \(2006,97,786\) of the four-character sequences using either letters or digits had not yet been claimed. If a four-character name is randomly selected, what is the probability that it is already owned?

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet (he only drinks red wine), all from different wineries. a. If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this? b. If 6 bottles of wine are to be randomly selected from the 30 for serving, how many ways are there to do this? c. If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety? d. If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen? e. If 6 bottles are randomly selected, what is the probability that all of them are the same variety?

A quality control inspector is inspecting newly produced items for faults. The inspector searches an item for faults in a series of independent fixations, each of a fixed duration. Given that a flaw is actually present, let \(p\) denote the probability that the flaw is detected during any one fixation (this model is discussed in "Human Performance in Sampling Inspection," Human Factors, 1979: 99-105). a. Assuming that an item has a flaw, what is the probability that it is detected by the end of the second fixation (once a flaw has been detected, the sequence of fixations terminates)? b. Give an expression for the probability that a flaw will be detected by the end of the \(n\)th fixation. c. If when a flaw has not been detected in three fixations, the item is passed, what is the probability that a flawed item will pass inspection? d. Suppose \(10 \%\) of all items contain a flaw \([P\) (randomly chosen item is flawed) \(=.1]\). With the assumption of part (c), what is the probability that a randomly chosen item will pass inspection (it will automatically pass if it is not flawed, but could also pass if it is flawed)? e. Given that an item has passed inspection (no flaws in three fixations), what is the probability that it is actually flawed? Calculate for \(p=.5\).

A subject is allowed a sequence of glimpses to detect a target. Let \(G_{i}=\\{\) the target is detected on the \(i\) th glimpse \(\\}\), with \(p_{i}=P\left(G_{i}\right)\). Suppose the \(G_{i}^{\prime}\) s are independent events, and write an expression for the probability that the target has been detected by the end of the \(n\)th glimpse. [Note: This model is discussed in "Predicting Aircraft Detectability" Human Factors, 1979: 277-291.]

An experimenter is studying the effects of temperature, pressure, and type of catalyst on yield from a certain chemical reaction. Three different temperatures, four different pressures, and five different catalysts are under consideration. a. If any particular experimental run involves the use of a single temperature, pressure, and catalyst, how many experimental runs are possible? b. How many experimental runs are there that involve use of the lowest temperature and two lowest pressures? c. Suppose that five different experimental runs are to be made on the first day of experimentation. If the five are randomly selected from among all the possibilities, so that any group of five has the same probability of selection, what is the probability that a different catalyst is used on each run?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

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.