/*! 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 23 Suppose that we have a sample sp... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that we have a sample space \(S=\left\\{E_{1}, E_{2}, E_{3}, E_{4}, E_{5}, E_{6}, E_{7}\right\\},\) where \(E_{1}, E_{2}, \ldots,\) \(E_{7}\) denote the sample points. The following probability assignments apply: \(P\left(E_{1}\right)=.05\) \(P\left(E_{2}\right)=.20, P\left(E_{3}\right)=.20, P\left(E_{4}\right)=.25, P\left(E_{5}\right)=.15, P\left(E_{6}\right)=.10,\) and \(P\left(E_{7}\right)=.05 .\) Let $$\begin{array}{l} A=\left\\{E_{1}, E_{4}, E_{6}\right\\} \\ B=\left\\{E_{2}, E_{4}, E_{7}\right\\} \\ C=\left\\{E_{2}, E_{3}, E_{5}, E_{7}\right\\} \end{array}$$ a. Find \(P(A), P(B),\) and \(P(C)\) b. Find \(A \cup B\) and \(P(A \cup B)\) c. Find \(A \cap B\) and \(P(A \cap B)\) d. Are events \(A\) and \(C\) mutually exclusive? e. Find \(B^{c}\) and \(P\left(B^{c}\right)\)

Short Answer

Expert verified
a. P(A) = 0.40, P(B) = 0.50, P(C) = 0.60 b. A ∪ B = {E1, E2, E4, E6, E7}, P(A ∪ B) = 0.65 c. A ∩ B = {E4}, P(A ∩ B) = 0.25 d. Yes, A and C are mutually exclusive. e. B^c = {E1, E3, E5, E6}, P(B^c) = 0.50

Step by step solution

01

Find P(A)

To find the probability of event A, sum the probabilities of the sample points in A. \(A = \{E_1, E_4, E_6\}\). Thus, \( P(A) = P(E_1) + P(E_4) + P(E_6) = 0.05 + 0.25 + 0.10 = 0.40 \).
02

Find P(B)

To find the probability of event B, sum the probabilities of the sample points in B. \(B = \{E_2, E_4, E_7\}\). Thus, \( P(B) = P(E_2) + P(E_4) + P(E_7) = 0.20 + 0.25 + 0.05 = 0.50 \).
03

Find P(C)

To find the probability of event C, sum the probabilities of the sample points in C. \(C = \{E_2, E_3, E_5, E_7\}\). Thus, \( P(C) = P(E_2) + P(E_3) + P(E_5) + P(E_7) = 0.20 + 0.20 + 0.15 + 0.05 = 0.60 \).
04

Find A ∪ B

The union \( A \cup B \) consists of all elements that are in A, B, or in both sets. \(A \cup B = \{E_1, E_4, E_6, E_2, E_7\} = \{E_1, E_2, E_4, E_6, E_7\}\).
05

Find P(A ∪ B)

To find \( P(A \cup B) \), sum the probabilities of the sample points in \( A \cup B \). \( P(A \cup B) = P(E_1) + P(E_2) + P(E_4) + P(E_6) + P(E_7) = 0.05 + 0.20 + 0.25 + 0.10 + 0.05 = 0.65 \).
06

Find A ∩ B

The intersection \( A \cap B \) consists of all elements that are in both sets A and B. \( A \cap B = \{E_4\}\).
07

Find P(A ∩ B)

To find \( P(A \cap B) \), sum the probabilities of the sample points in \( A \cap B \). \( P(A \cap B) = P(E_4) = 0.25 \).
08

Check if A and C are mutually exclusive

Two events are mutually exclusive if they have no sample points in common. \( A = \{E_1, E_4, E_6\} \) and \( C = \{E_2, E_3, E_5, E_7\} \) have no common elements, hence A and C are mutually exclusive.
09

Find B^c

The complement \( B^c \) consists of all elements not in B. \( B^c = \{E_1, E_3, E_5, E_6\} \).
10

Find P(B^c)

To find \( P(B^c) \), sum the probabilities of the sample points in \( B^c \). \( P(B^c) = P(E_1) + P(E_3) + P(E_5) + P(E_6) = 0.05 + 0.20 + 0.15 + 0.10 = 0.50 \).

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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, a sample space is the set of all possible outcomes or sample points of a particular experiment. Each outcome in this set is distinct and represents an individual result of the experiment or process of observation. For example, in the given exercise, the sample space is denoted as \(S = \{E_1, E_2, E_3, E_4, E_5, E_6, E_7\}\). This notation indicates that there are seven distinct events, each of which might occur as a result of the experiment. It is essential to understand that a well-defined sample space is crucial, as it forms the foundation for calculating probabilities of various related events. Each sample point in \(S\) has an assigned probability, such as \(P(E_1) = 0.05\) and \(P(E_2) = 0.20\), indicating the likelihood of occurrence for each specific outcome. Being clear about the sample space ensures that probabilities are assigned correctly and add up to a total of 1.
Mutually exclusive events
Mutually exclusive events are events that cannot happen at the same time. They have no sample points in common. If you consider two mutually exclusive events, the occurrence of one event prevents the occurrence of the other. In our given example, events \(A = \{E_1, E_4, E_6\}\) and \(C = \{E_2, E_3, E_5, E_7\}\) are mutually exclusive because they share no elements or outcomes in common. This means that whenever event \(A\) occurs, event \(C\) cannot occur and vice versa.

Understanding this principle helps when calculating combined probabilities, as the probability of their union is simply the sum of their individual probabilities, since they do not overlap. It’s important to identify these types of events early on, as they simplify the analysis and calculation of probabilities in a complex space.
Probability of events
The probability of an event is a measure of the likelihood that the event will occur. It is calculated by summing the probabilities of the individual sample points that make up the event. For example, to find the probability of event \(A\) in the given exercise, we add the probabilities of the sample points in \(A\), which are \(E_1, E_4, \) and \(E_6\). Therefore, \( P(A) = P(E_1) + P(E_4) + P(E_6) = 0.05 + 0.25 + 0.10 = 0.40 \).

Probabilities are always between 0 and 1, where 0 means the event cannot occur and 1 means the event is certain. Calculating probabilities involves adding the probabilities of mutually exclusive events when determining the probability of their union. For instance, the probability of \(A \cup B\) is found by considering all the unique outcomes in both \(A\) and \(B\) and summing their probabilities. By understanding how probabilities of events are determined, one can predict and analyze the likelihood of different scenarios occurring, making probability a powerful tool for decision-making.

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Most popular questions from this chapter

A telephone survey to determine viewer response to a new television show obtained the following data. $$\begin{array}{lc} \text { Rating } & \text { Frequency } \\ \text { Poor } & 4 \\ \text { Below average } & 8 \\ \text { Average } & 11 \\ \text { Above average } & 14 \\ \text { Excellent } & 13 \end{array}$$ a. What is the probability that a randomly selected viewer will rate the new show as average or better? b. What is the probability that a randomly selected viewer will rate the new show below average or worse?

A decision maker subjectively assigned the following probabilities to the four outcomes of an experiment: \(P\left(E_{1}\right)=.10, P\left(E_{2}\right)=.15, P\left(E_{3}\right)=.40,\) and \(P\left(E_{4}\right)=.20 .\) Are these probability assignments valid? Explain.

Consider the experiment of tossing a coin three times. a. Develop a tree diagram for the experiment. b. List the experimental outcomes. c. What is the probability for each experimental outcome?

Small cars get better gas mileage, but they are not as safe as bigger cars. Small cars accounted for \(18 \%\) of the vehicles on the road, but accidents involving small cars led to 11,898 fatalities during a recent year (Reader's Digest, May 2000 ). Assume the probability a small car is involved in an accident is .18. The probability of an accident involving a small car leading to a fatality is .128 and the probability of an accident not involving a small car leading to a fatality is .05. Suppose you learn of an accident involving a fatality. What is the probability a small car was involved? Assume that the likelihood of getting into an accident is independent of car size.

The U.S. Census Bureau provides data on the number of young adults, ages \(18-24,\) who are living in their parents' home. ' Let \(M=\) the event a male young adult is living in his parents' home \(F=\) the event a female young adult is living in her parents' home If we randomly select a male young adult and a female young adult, the Census Bureau data enable us to conclude \(P(M)=.56\) and \(P(F)=.42\) (The World Almanac, 2006 ). The probability that both are living in their parents' home is .24 a. What is the probability at least one of the two young adults selected is living in his or her parents' home? b. What is the probability both young adults selected are living on their own (neither is living in their parents' home)?

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