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Each of the possible five outcomes of a random experiment is equally likely. The sample space is \(\\{a, b, c, d, e\\} .\) Let \(A\) denote the event \(\\{a, b\\},\) and let \(B\) denote the event \(\\{c, d, e\\} .\) Determine the following (a) \(P(A)\) (b) \(P(B)\) (c) \(P\left(A^{\prime}\right)\) (d) \(P(A \cup B)\) (e) \(P(A \cap B)\)

Short Answer

Expert verified
(a) \(\frac{2}{5}\), (b) \(\frac{3}{5}\), (c) \(\frac{3}{5}\), (d) 1, (e) 0.

Step by step solution

01

Understand the Sample Space

The sample space is the set of all possible outcomes, given as \(\{a, b, c, d, e\}\). Since all outcomes are equally likely, each has a probability of \(\frac{1}{5}\).
02

Calculate Probability of Event A

Event \(A\) is \(\{a, b\}\), which includes 2 outcomes. Therefore, \(P(A)\) is the probability of any one of these outcomes occurring, which is \(\frac{2}{5}\).
03

Calculate Probability of Event B

Event \(B\) is \(\{c, d, e\}\), which includes 3 outcomes. Therefore, \(P(B)\) is the probability of any one of these outcomes occurring, which is \(\frac{3}{5}\).
04

Calculate Probability of Complement of A

The event \(A^{\prime}\) includes all outcomes not in \(A\), which is \(\{c, d, e\}\). Since \(A^{\prime}\) includes 3 outcomes, \(P(A^{\prime})\) is \(\frac{3}{5}\).
05

Calculate Probability of Union of A and B

Event \(A \cup B\) includes all outcomes in either \(A\) or \(B\), which is \(\{a, b, c, d, e\}\). Since it covers all outcomes, \(P(A \cup B) = 1\).
06

Calculate Probability of Intersection of A and B

Events \(A\) and \(B\) are disjoint, meaning they have no common outcomes. Therefore, \(P(A \cap B) = 0\).

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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 a random experiment. Consider it as the complete set of possible results when you conduct an experiment or event. In our scenario, the sample space is given as \(\{a, b, c, d, e\}\). Each element in this set is an outcome of the random experiment; here, each letter represents a different potential result.

Understanding the sample space is crucial because it forms the basis for calculating probabilities. Once we know what all the possible outcomes are, we can talk about specific events and their probabilities. Importantly, the sample space must account for all possibilities and should be exhaustive, meaning there are no other outcomes not represented in the set. This comprehensive perspective ensures that our probability calculations are both accurate and meaningful.
Equally Likely Outcomes
Equally likely outcomes are a key principle in probability theory. When outcomes are equally likely, each outcome has the same chance of occurring. This simplifies probability calculations significantly. In our example, each outcome \(a, b, c, d, e\) is equally likely, meaning they each have the same probability of occurring, specifically \(\frac{1}{5}\), since there are 5 outcomes in total.

With equally likely outcomes, the probability of any specific event simply becomes the number of favorable outcomes divided by the total number of possible outcomes. For instance, the event \(A\) consisting of the outcomes \(\{a, b\}\) includes 2 outcomes. Thus, \(P(A) = \frac{2}{5}\). Similarly, the event \(B\) consisting of \(\{c, d, e\}\) includes 3 outcomes, leading to \(P(B) = \frac{3}{5}\). This symmetry and simplification are huge benefits when dealing with equally likely events.
Probability of Events
The probability of an event is a measure of how likely that event is to occur. For any event \(A\), its probability, \(P(A)\), ranges from 0 to 1. A probability of 0 means the event cannot happen, while a probability of 1 means the event is certain to happen. Events can be simple, involving a single outcome, or compound, involving multiple outcomes.

For instance, calculating the probability of events \(A\) and \(B\) is direct when dealing with equally likely outcomes. We determine the number of successful (favorable) outcomes and divide by the total number of outcomes in the sample space. Likewise, the complement event \(A^{\prime}\), consisting of all outcomes not in \(A\), highlights that probabilities must sum up to 1. Here, \(P(A^{\prime}) = \frac{3}{5}\). Understanding this helps us see that probability not only tells us what may happen but also is a tool for predicting and understanding the behavior of random processes.
Set Operations in Probability
Set operations in probability, such as union and intersection, provide powerful tools for analyzing events. The union of two events, \(A \cup B\), includes all outcomes that are in \(A\), in \(B\), or in both. For disjoint events like \(A\) and \(B\), where they have no outcomes in common, their union is simply the entire sample space, giving \(P(A \cup B) = 1\).

Conversely, the intersection of two events, \(A \cap B\), concerns outcomes common to both events. If events are mutually exclusive or disjoint, like \(A\) and \(B\), then their intersection is empty, leading to \(P(A \cap B) = 0\). These set operations facilitate more complex probability questions and are foundational in developing advanced probability models. Understanding these can greatly enhance your ability to dissect problems and derive the appropriate probabilities for complex combinations of events.

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Most popular questions from this chapter

In control replication, cells are replicated over a period of two days. Not until mitosis is completed can freshly synthesized DNA be replicated again. Two control mechanisms have been identified - one positive and one negative. Suppose that a replication is observed in three cells. Let \(A\) denote the event that all cells are identified as positive and let \(B\) denote the event that all cells are negative. Describe the sample space graphically and display each of the following events: (a) \(A\) (b) \(B\) (c) \(A \cap B\) (d) \(A \cup B\)

In the layout of a printed circuit board for an electronic product, there are 12 different locations that can accommodate chips. (a) If five different types of chips are to be placed on the board, how many different layouts are possible? (b) If the five chips that are placed on the board are of the same type, how many different layouts are possible?

Suppose \(A\) and \(B\) are mutually exclusive events. Construct a Venn diagram that contains the three events \(A, B\), and \(C\) such that \(P(A \mid C)=1\) and \(P(B \mid C)=0\).

Software to detect fraud in consumer phone cards tracks the number of metropolitan areas where calls originate each day. It is found that \(1 \%\) of the legitimate users originate calls from two or more metropolitan areas in a single day. However, \(30 \%\) of fraudulent users originate calls from two or more metropolitan areas in a single day. The proportion of fraudulent users is \(0.01 \%\). If the same user originates calls from two or more metropolitan areas in a single day, what is the probability that the user is fraudulent?

Six tissues are extracted from an ivy plant infested by spider mites. The plant in infested in \(20 \%\) of its area. Each tissue is chosen from a randomly selected area on the ivy plant. (a) What is the probability that four successive samples show the signs of infestation? (b) What is the probability that three out of four successive samples show the signs of infestation?

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