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Let \(F\) denote the event that a randomly selected registered voter in a certain city has signed a petition to recall the mayor. Also, let \(E\) denote the event that a randomly selected registered voter actually votes in the recall election. Describe the event \(E \cap F\) in words. If \(P(F)=.10\) and \(P(E \mid F)=.80\), determine \(P(E \cap F)\).

Short Answer

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The probability that a randomly selected voter signed the petition and actually votes in the recall election is 0.08.

Step by step solution

01

Understanding the events

The event \(E \cap F\) stands for the situation where a randomly selected registered voter in this city has both signed the petition to recall the mayor (Event \(F\)) and actually votes in the recall election (Event \(E\)). In terms of probability, this is represented by \(P(E \cap F)\).
02

Using the conditional probability formula

The definition of conditional probability is \(P(A \mid B) = P(A \cap B)/P(B)\). We are given \(P(F)\) as 0.10 and \(P(E \mid F)\) as 0.80. We can use these values to calculate \(P(E \cap F)\) by rearranging the definition of conditional probability to \(P(E \cap F) = P(E \mid F) \times P(F)\).
03

Calculating the probability of E and F

Substitute the known values into the rearranged formula. So, \(P(E \cap F) = P(E \mid F) \times P(F) = 0.80 \times 0.10\).

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Key Concepts

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

Probability Theory
Probability theory forms the backbone of statistical analysis and is fundamental to understanding and predicting outcomes in a random process. The basics involve concepts such as random events, outcomes, and their respective probabilities.

When we define events in probability, they describe specific outcomes or sets of outcomes. For example, in our exercise, the event labeled as 'F' signifies the outcome in which a person has signed a petition, and 'E' represents that the same person has voted in the recall election.

The probability of an event is a measure of the likelihood that the event will occur, expressed as a number between 0 and 1, where 0 indicates the event will not occur, and 1 signifies certainty. In the context of our exercise, the given probabilities, such as the 10% chance that a voter signed the petition (\(P(F)=0.10\)) and the 80% chance that a voter who signed the petition will also vote in the election (\(P(E \text{ given } F)=0.80\)), provide us with numerical insight into the voters' behavior.
Statistical Analysis
Statistical analysis involves collecting, exploring, and presenting large quantities of data to discover underlying patterns and trends. Part of this analysis is determining the relationship between two events, which is where concepts like joint probability and conditional probability come into play.

To assess the connection between two events, the analysis might center around whether the occurrence of one event affects the probability of another event happening, known as conditional probability. A clear understanding of these relationships is crucial in many fields, such as social science, where our original exercise comes from. It allows statisticians to make predictions, such as voter turnout, based on known behaviors, like petition signing.

Our exercise illustrates how these probabilities can be applied by using the given data (\(P(F)\) and \(P(E \text{ given } F)\) to elucidate the intersection of two events, which in turn may provide valuable insights for political analysts.
Event Intersection
To discuss the intersection of events in probability theory, it's necessary to understand that it relates to the occurrence of two or more events simultaneously. In our exercise, the intersection (\(E \text{ intersect } F\text{ or } E \text{ cap } F\)) would be the group of people who have both signed the petition and voted in the election.

The probability of the intersection of two events can be found if we know the conditional probability of one event happening given that the other has already occurred. This is where the formula \[P(A \text{ intersect } B) = P(A \text{ given } B) \times P(B)\] comes into play. In simpler terms, this means that to find how likely it is for both events to happen, you would multiply the likelihood of one event by the probability that the other event happens too.

An accurate calculation of event intersections can provide valuable predictions in many scenarios, such as understanding the demographics of voters actively engaging in an election process, as highlighted by our textbook example.

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Most popular questions from this chapter

6.3 Consider the chance experiment in which the type of transmission- automatic (A) or manual (M) - is recorded for each of the next two cars purchased from a certain dealer. a. What is the set of all possible outcomes (the sample space)? b. Display the possible outcomes in a tree diagram. c. List the outcomes in each of the following events. Which of these events are simple events? i. \(B\) the event that at least one car has an automatic transmission ii. \(C\) the event that exactly one car has an automatic transmission iii. \(D\) the event that neither car has an automatic transmission d. What outcomes are in the event \(B\) and \(C\) ? In the event \(B\) or \(C\) ?

A library has five copies of a certain textbook on reserve of which two copies ( 1 and 2) are first printings and the other three \((3,4\), and 5 ) are second printings. A student examines these books in random order, stopping only when a second printing has been selected. a. Display the possible outcomes in a tree diagram. b. What outcomes are contained in the event \(A\), that exactly one book is examined before the chance experiment terminates? c. What outcomes are contained in the event \(C\), that the chance experiment terminates with the examination of book \(5 ?\)

Suppose that we define the following events: \(C=\) event that a randomly selected driver is observed to be using a cell phone, \(A=\) event that a randomly selected driver is observed driving a passenger automobile, \(V=\) event that a randomly selected driver is observed driving a van or SUV, and \(T=\) event that a randomly selected driver is observed driving a pickup truck. Based on the article "Three Percent of Drivers on Hand-Held Cell Phones at Any Given Time" (San Luis Obispo Tribune, July 24, 2001), the following probability estimates are reasonable: \(P(C)=.03\), \(P(C \mid A)=.026, P(C \mid V)=.048\), and \(P(C \mid T)=.019 .\) Ex- plain why \(P(C)\) is not just the average of the three given conditional probabilities.

In a school machine shop, \(60 \%\) of all machine breakdowns occur on lathes and \(15 \%\) occur on drill presses. Let \(E\) denote the event that the next machine breakdown is on a lathe, and let \(F\) denote the event that a drill press is the next machine to break down. With \(P(E)=.60\) and \(P(F)=.15\), calculate: a. \(P\left(E^{C}\right)\) b. \(P(E \cup F)\) c. \(P\left(E^{C} \cap F^{C}\right)\)

After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books at random to the four stu- dents \((1,2,3\), and 4\()\) who claim to have left books. One possible outcome is that 1 receives 2 's book, 2 receives 4 's book, 3 receives his or her own book, and 4 receives 1 's book. This outcome can be abbreviated \((2,4,3,1)\). a. List the 23 other possible outcomes. b. Which outcomes are contained in the event that exactly two of the books are returned to their correct owners? As- suming equally likely outcomes, what is the probability of this event? c. What is the probability that exactly one of the four students receives his or her own book? d. What is the probability that exactly three receive their own books? e. What is the probability that at least two of the four students receive their own books?

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