/*! 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 234 If the events \(A\) and \(B\) ar... [FREE SOLUTION] | 91Ó°ÊÓ

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If the events \(A\) and \(B\) are independent, show that \(A^{\prime}\) and \(B^{\prime}\) are independent.

Short Answer

Expert verified
Events \(A'\) and \(B'\) are independent.

Step by step solution

01

Define Independence for Events

For two events \(A\) and \(B\) in a probability space, they are said to be independent if the probability of both events occurring equals the product of their individual probabilities. Mathematically, \(P(A \cap B) = P(A) \times P(B)\).
02

Express Complementary Events

The complementary events of \(A\) and \(B\) are denoted as \(A'\) and \(B'\) respectively. By definition, \(P(A) + P(A') = 1\) and \(P(B) + P(B') = 1\).
03

Relate Complements to Original Events

Using De Morgan's laws, we know: \((A \cap B)' = A' \cup B'\) and consequently \((A' \cap B') = (A \cup B)'\).
04

Calculate Probability of Complementary Intersection

Using the formula for the probability of a complement, \(P((A \cup B)') = 1 - P(A \cup B)\). Since \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) and given \(P(A \cap B) = P(A) \cdot P(B)\), calculate \(P((A \cup B)')\).
05

Use Probability Independence for Complements

To establish independence for \(A'\) and \(B'\), check if \(P(A' \cap B') = P(A') \times P(B')\). Using previous steps: \(P(A' \cap B') = 1 - P(A \cup B) = (1 - P(A)) \cdot (1 - P(B)) = P(A') \cdot P(B')\).
06

Conclude the Independence of Complements

Since \(P(A' \cap B') = P(A') \times P(B')\), we conclude that events \(A'\) and \(B'\) are independent.

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Key Concepts

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

Independent Events
In probability theory, understanding independent events is fundamental. Two events, say \( A \) and \( B \), are considered independent if the occurrence of one does not affect the occurrence of the other. This is mathematically expressed by the equation \( P(A \cap B) = P(A) \times P(B) \). It essentially means the probability of both events occurring together is the product of their individual probabilities.
This might seem counterintuitive at first. However, consider the flip of two coins. The result of the first coin does not affect the result of the second coin. Thus, flipping heads on both is simply \( 0.5 \times 0.5 = 0.25 \), demonstrating their independence.
Recognizing independent events is crucial when solving probability problems, as it simplifies the calculation significantly. It can be particularly useful in situations involving multiple variables or numerous events.
Complementary Events
Complementary events are two opposite outcomes. If you have an event \( A \), its complement, denoted as \( A' \), encompasses all outcomes not in \( A \). The sum of probabilities of an event and its complement should always equal 1: \( P(A) + P(A') = 1 \).
Consider a die roll as an example. The probability of rolling a 4 is 1/6. Therefore, the probability of not rolling a 4, the complement, is \( 1 - 1/6 = 5/6 \).
Complementary events are extensively employed in various probability calculations, as they help in deducing the likelihood of an outcome not occurring based on its complementary event. This can streamline complex probability problems by allowing a focus on simpler secondary calculations.
De Morgan's Laws
De Morgan's laws are powerful tools in probability and set theory. They provide a way to express the complement of a union or intersection of events. These laws particularly state:
  • \((A \cap B)' = A' \cup B'\)
  • \((A \cup B)' = A' \cap B'\)
Using De Morgan's laws lets us find the probability of combined complementary events. For instance, if you want the probability that neither \( A \) nor \( B \) occurs, De Morgan’s laws allow rephrasing it to be about the intersection of their complements \( A' \) and \( B' \).
These transformations and insights are instrumental in simplifying the analysis of combined events across various probability scenarios. Utilizing De Morgan's laws can significantly assist in understanding probabilities in complex event combinations.
Probability Space
In the realm of probability, the concept of a probability space is foundational. A probability space is essentially the mathematical framework where probability calculations occur. It consists of three main elements:
  • A sample space \( S \), which is the set of all possible outcomes.
  • A set of events, each being a subset of the sample space.
  • A probability function \( P \), which assigns each event a likelihood, a value between 0 and 1.
Imagine a coin toss. Here, the sample space \( S = \{ \text{heads, tails} \} \). Events could be flipping a head or flipping a tail. The probability function usually assigns \( P(\text{heads}) = 0.5 \) and \( P(\text{tails}) = 0.5 \).
Probability spaces enable the systematic study and calculation of probabilities in diverse scenarios. They form the groundwork for more advanced theories such as random variables and expected values, which further deepen probability analyses.

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Most popular questions from this chapter

A researcher receives 100 containers of oxygen. Of those containers, 20 have oxygen that is not ionized, and the rest are ionized. Two samples are randomly selected, without replacement, from the lot. (a) What is the probability that the first one selected is not ionized? (b) What is the probability that the second one selected is not ionized given that the first one was ionized? (c) What is the probability that both are ionized? (d) How does the answer in part (b) change if samples selected were replaced prior to the next selection?

An article in the British Medical Journal ["Comparison of treatment of renal calculi by operative surgery, percutaneous nephrolithotomy, and extracorporeal shock wave lithotripsy" (1986, Vol. 82, pp. \(879-892\) ) ] provided the following discussion of success rates in kidney stone removals. Open surgery had a success rate of \(78 \%(273 / 350)\) and a newer method, percutaneous nephrolithotomy (PN), had a success rate of \(83 \%(289 / 350)\). This newer method looked better, but the results changed when stone diameter was considered. For stones with diameters less than 2 centimeters, \(93 \%(81 / 87)\) of cases of open surgery were successful compared with only \(83 \%(234 / 270)\) of cases of PN. For stones greater than or equal to 2 centimeters, the success rates were \(73 \%(192 / 263)\) and \(69 \%(55 / 80)\) for open surgery and PN, respectively. Open surgery is better for both stone sizes, but less successful in total. In \(1951,\) E. H. Simpson pointed out this apparent contradiction (known as Simpson's paradox), and the hazard still persists today. Explain how open surgery can be better for both stone sizes but worse in total.

Incoming calls to a customer service center are classified as complaints \((75 \%\) of calls) or requests for information \((25 \%\) of calls \() .\) Of the complaints, \(40 \%\) deal with computer equipment that does not respond and \(57 \%\) deal with incomplete software installation; in the remaining \(3 \%\) of complaints, the user has improperly followed the installation instructions. The requests for information are evenly divided on technical questions \((50 \%)\) and requests to purchase more products \((50 \%)\) (a) What is the probability that an incoming call to the customer service center will be from a customer who has not followed installation instructions properly? (b) Find the probability that an incoming call is a request for purchasing more products.

A sample of two items is selected without replacement from a batch. Describe the (ordered) sample space for each of the following batches: (a) The batch contains the items \(\\{a, b, c, d\\}\). (b) The batch contains the items \(\\{a, b, c, d, e, f, g\\}\). (c) The batch contains 4 defective items and 20 good items. (d) The batch contains 1 defective item and 20 good items.

Provide a reasonable description of the sample space for each of the random experiments in Exercises \(2-1\) to \(2-17\). There can be more than one acceptable interpretation of each experiment. Describe any assumptions you make. The time until a service transaction is requested of a computer to the nearest millisecond.

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