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Four bits are transmitted over a digital communications channel. Each bit is either distorted or received without distortion. Let \(A_{i}\) denote the event that the \(i\) th bit is distorted, \(i=1, \ldots, 4\) a. Describe the sample space for this experiment b. Are the \(A_{i}\) 's mutually exclusive? Describe the outcomes in each of the following events: c. \(A_{1}\) d. \(A_{1}^{\prime}\) e. \(A_{1} \cap A_{2} \cap A_{3} \cap A_{4}\) f. \(\left(A_{1} \cap A_{2}\right) \cup\left(A_{3} \cap A_{4}\right)\)

Short Answer

Expert verified
a. Sample space: 16 combinations of D, N; b. A_i's are not mutually exclusive; c. A_1: 8 outcomes; d. A_1': 8 outcomes; e. A_1 ∩ A_2 ∩ A_3 ∩ A_4: 1 outcome; f. (A_1 ∩ A_2) ∪ (A_3 ∩ A_4): 6 outcomes.

Step by step solution

01

Describe the Sample Space

The sample space for this experiment consists of all possible combinations of distorted (D) or not distorted (N) bits over four positions. For four bits, each can independently be D or N, so the sample space is: \( \{DDDD, DDDN, DDND, DDNN, DNDD, DNDN, DNND, DNNN, NDDD, NDDN, NDND, NDNN, NNDD, NNDN, NNND, NNNN\} \). This sample space contains all possible combinations of the four bits, which is \(2^4 = 16\) events.
02

Determine if A_i's are Mutually Exclusive

The events \(A_{i}\) represent individual bits being distorted. In the context of this experiment, these events are not mutually exclusive because more than one bit can be distorted at the same time. In fact, any combination of them can occur, which means they are not mutually exclusive; they can overlap.
03

Describe Event A_1

Event \(A_{1}\) represents the scenario where the first bit is distorted. The outcomes in this event will include any combination where the first bit is D; namely, \( \{DDDD, DDDN, DDND, DDNN, DNDD, DNDN, DNND, DNNN\} \).
04

Describe Event A_1'

Event \(A_{1}^{\prime}\) is the complement of \(A_{1}\), meaning the first bit is not distorted. The outcomes here will have the first bit as N, which gives us \( \{NDDD, NDDN, NDND, NDNN, NNDD, NNDN, NNND, NNNN\} \).
05

Describe Event A_1 ∩ A_2 ∩ A_3 ∩ A_4

The event \(A_{1} \cap A_{2} \cap A_{3} \cap A_{4}\) occurs when all four bits are distorted. There is only one outcome for this event: \(\{DDDD\}\).
06

Describe Event (A_1 ∩ A_2) ∪ (A_3 ∩ A_4)

The event \((A_{1} \cap A_{2}) \cup (A_{3} \cap A_{4})\) involves either the first two bits being distorted, or the last two bits being distorted, or both. You get the following outcomes: \(\{DDDD, NDDD, DDDN, DDND, DDNN, NDND\}\).

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Key Concepts

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

Sample Space
In probability theory, the concept of a sample space is fundamental. The sample space represents all possible outcomes of an experiment. Each outcome is an element of this space.

For example, when four bits are transmitted, and each bit can either be distorted (D) or not distorted (N), the sample space consists of all possible sequences of these four bits. We denote this as:
  • \( \{DDDD, DDDN, DDND, DDNN, DNDD, DNDN, DNND, DNNN, NDDD, NDDN, NDND, NDNN, NNDD, NNDN, NNND, NNNN\} \)
This means the total number of different outcomes is \(2^4 = 16\), because each of the four bits has two possible states.

Understanding the sample space is essential because it lays the foundation for determining the likelihood of various events.
Mutually Exclusive Events
Events are described as mutually exclusive if they cannot occur simultaneously. In other words, the occurrence of one event excludes the possibility of another.

In our four-bit transmission example, the events \(A_i\), which denote the distortion of individual bits, are **not** mutually exclusive. Multiple bits can be distorted at the same time.

This indicates that the events can overlap. For example, if event \(A_1\) means the first bit is distorted, and \(A_2\) means the second bit is distorted, they can both happen concurrently in an outcome like \(DDNN\). Therefore, these events overlap and are not mutually exclusive.
Complement Events
In probability, the complement of an event represents all outcomes not included in the event. It covers everything in the sample space that the event does not.

Taking the example of event \(A_1\), where the first bit is distorted, its complement \(A_1'\) consists of all outcomes where the first bit is not distorted. The complement event \(A_1'\) for our four-bit situation includes:
  • \( \{NDDD, NDDN, NDND, NDNN, NNDD, NNDN, NNND, NNNN\} \)
Understanding complements is crucial, especially when calculating probabilities, as their probabilities add up to 1. This is particularly useful for determining the probability that at least one event occurs by using its complement.
Intersection and Union of Events
The intersection and union are operations that allow us to evaluate and understand events involving two or more sets.

The intersection, denoted as \(A \cap B\), consists of outcomes that are common to both events \(A\) and \(B\). In our scenario, the intersection \(A_1 \cap A_2 \cap A_3 \cap A_4\) represents all bits being distorted, giving the single outcome \( \{DDDD\} \).

On the other hand, the union, \(A \cup B\), includes outcomes that belong to either of the events or both. For example, \((A_1 \cap A_2) \cup (A_3 \cap A_4)\) includes cases where either the first two bits, the last two bits, or both are distorted. This results in outcomes \( \{DDDD, NDDD, DDDN, DDND, DDNN, NDND\} \).

These operations are foundational in setting up complex event probabilities and comprehending how events correlate.

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Most popular questions from this chapter

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.

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