/*! 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] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 23

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: \(P\left(E_{1}\right)=.05\). \\[ \begin{aligned} 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, \text { and } P\left(E_{7}\right)=.05 . \text { Let } \\ 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{aligned} \\] 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
P(A)=0.40, P(B)=0.50, P(C)=0.60; A \cup B = \{E_1, E_2, E_4, E_6, E_7\}, P(A \cup B)=0.65; A \cap B = \{E_4\}, P(A \cap B)=0.25; A and C are mutually exclusive; B^c = \{E_1, E_3, E_5, E_6\}, P(B^c)=0.50.

Step by step solution

01

Calculate Probability of A

To find the probability of event A, add the probabilities of the individual elements that make up A. The event A is given as \( A = \{E_1, E_4, E_6\} \). Compute as follows:\[P(A) = P(E_1) + P(E_4) + P(E_6) = 0.05 + 0.25 + 0.10 = 0.40.\]
02

Calculate Probability of B

To find the probability of event B, add the probabilities of the individual elements that make up B. The event B is given as \( B = \{E_2, E_4, E_7\} \). Compute as follows:\[P(B) = P(E_2) + P(E_4) + P(E_7) = 0.20 + 0.25 + 0.05 = 0.50.\]
03

Calculate Probability of C

To find the probability of event C, add the probabilities of the individual elements that make up C. The event C is given as \( C = \{E_2, E_3, E_5, E_7\} \). Compute as follows:\[P(C) = P(E_2) + P(E_3) + P(E_5) + P(E_7) = 0.20 + 0.20 + 0.15 + 0.05 = 0.60.\]
04

Find Union of A and B

The union of events A and B, \( A \cup B \), is the set of all elements that are in either A or B or both. This results in:\[A \cup B = \{E_1, E_2, E_4, E_6, E_7\}.\]
05

Calculate Probability of A Union B

To find \(P(A \cup B)\), we add the probabilities of the individual elements in \(A \cup B\), and make sure not to double-count any element:\[P(A \cup B) = P(E_1) + P(E_2) + P(E_4) + P(E_6) + P(E_7) = 0.05 + 0.20 + 0.25 + 0.10 + 0.05 = 0.65.\]
06

Find Intersection of A and B

The intersection of events A and B, \( A \cap B \), is the set of elements that are in both A and B. This results in:\[A \cap B = \{E_4\}.\]
07

Calculate Probability of A Intersection B

Since \( A \cap B = \{E_4\} \), then:\[P(A \cap B) = P(E_4) = 0.25.\]
08

Determine if A and C are Mutually Exclusive

Events A and C are mutually exclusive if they have no elements in common. Check the intersection:\[A \cap C = \{E_1, E_4, E_6\} \cap \{E_2, E_3, E_5, E_7\} = \varnothing.\]Since \( A \cap C = \varnothing \), A and C are mutually exclusive.
09

Find Complement of B and its Probability

The complement of B, \( B^c \), is the set of elements not in B. This results in:\[B^c = \{E_1, E_3, E_5, E_6\}.\]Now calculate the probability:\[P(B^c) = P(E_1) + P(E_3) + P(E_5) + P(E_6) = 0.05 + 0.20 + 0.15 + 0.10 = 0.50.\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Sample Space
A sample space is fundamental in probability theory, as it contains all the possible outcomes of a probability experiment. Imagine throwing a six-sided die. The sample space consists of all numbers that can appear on the die face, which are 1 through 6.
In this exercise, our sample space \( S \) is the set of events \( \{ E_1, E_2, E_3, E_4, E_5, E_6, E_7 \} \). Each \( E_i \) represents a distinct sample point.
When defining a sample space, ensure that all possible events are included. Each event in a sample space occurs with a probability, and the probability of all events combined in the sample space is always 1:
  • The probability of a single event \( P(S) = 1 \).
  • The sum of the probabilities of all sample points equals 1.
This concept ensures that something in the sample space will happen. It's essential because it lays the groundwork for building other probability-related concepts such as events and their probabilities.
Complement of an Event
The complement of an event in probability refers to all the outcomes in the sample space that are not part of the original event. If event \( B \) represents a set of outcomes, then the complement of \( B \), denoted as \( B^c \), includes all outcomes not in \( B \).
For example, if \( B = \{ E_2, E_4, E_7 \} \), then the complement \( B^c = \{ E_1, E_3, E_5, E_6 \} \).
Understanding complements is useful because it can often be easier to calculate the probability of the complement of an event and utilize the principle:
  • \( P(B^c) = 1 - P(B) \)
This concept is particularly helpful when it is simpler to define what isn鈥檛 included in an event, rather than what is. Complements are integral to solving problems efficiently by using the total probability principle.
Mutually Exclusive Events
Two events are defined as mutually exclusive if they cannot occur at the same time. In terms of sets, their intersection is empty, indicating no common outcomes exist between them.
For instance, in this exercise, events \( A \) and \( C \) are given as \( A = \{ E_1, E_4, E_6 \} \) and \( C = \{ E_2, E_3, E_5, E_7 \} \).
The intersection is \( A \cap C = \varnothing \), confirming that these events are mutually exclusive because they share no common elements.
It's essential in probability to recognize situations where two events cannot happen simultaneously because:
  • If events are mutually exclusive, then \( P(A \cap C) = 0 \).
  • The probability of either event occurring can be simplified as \( P(A \cup C) = P(A) + P(C) \) (since there's no overlap).
Recognizing mutually exclusive events help solve problems faster by simplifying computations due to their non-overlapping nature.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The prior probabilities for events \(A_{1}, A_{2},\) and \(A_{3}\) are \(P\left(A_{1}\right)=.20, P\left(A_{2}\right)=.50,\) and \(P\left(A_{3}\right)=\) \(.30 .\) The conditional probabilities of event \(B\) given \(A_{1}, A_{2},\) and \(A_{3}\) are \(P\left(B | A_{1}\right)=.50\) \(P\left(B | A_{2}\right)=.40,\) and \(P\left(B | A_{3}\right)=.30\). a. Compute \(P\left(B \cap A_{1}\right), P\left(B \cap A_{2}\right),\) and \(P\left(B \cap A_{3}\right)\). b. Apply Bayes' theorem, equation \((4,19),\) to compute the posterior probability \(P\left(A_{2} | B\right)\). c. Use the tabular approach to applying Bayes' theorem to compute \(P\left(A_{1} | B\right), P\left(A_{2} | B\right)\), and \(P\left(A_{3} | B\right)\) .

Suppose that we have a sample space with five equally likely experimental outcomes: \(E_{1}\). \\[ \begin{aligned} E_{2}, E_{3}, E_{4}, E_{5} . \text { Let } \\ \qquad \begin{aligned} A &=\left\\{E_{1}, E_{2}\right\\} \\ B &=\left\\{E_{3}, E_{4}\right\\} \\ C &=\left\\{E_{2}, E_{3}, E_{5}\right\\} \end{aligned} \end{aligned} \\] a. \(\quad\) Find \(P(A), P(B),\) and \(P(C)\). b. Find \(P(A \cup B)\). Are \(A\) and \(B\) mutually exclusive? c. \(\quad\) Find \(A^{c}, C^{c}, P\left(A^{c}\right),\) and \(P\left(C^{c}\right)\). d. Find \(A \cup B^{c}\) and \(P\left(A \cup B^{c}\right)\). e. Find \(P(B \cup C)\).

Jerry Stackhouse of the National Basketball Association's Dallas Mavericks is the best freethrow shooter on the team, making \(89 \%\) of his shots (ESPN website, July, 2008 ). Assume that late in a basketball game, Jerry Stackhouse is fouled and is awarded two shots. a. What is the probability that he will make both shots? b. What is the probability that he will make at least one shot? c. What is the probability that he will miss both shots? d. Late in a basketball game, a team often intentionally fouls an opposing player in order to stop the game clock. The usual strategy is to intentionally foul the other team's worst free-throw shooter. Assume that the Dallas Mavericks' center makes \(58 \%\) of his free-throw shots. Calculate the probabilities for the center as shown in parts (a), (b), and (c), and show that intentionally fouling the Dallas Mavericks' center is a better strategy than intentionally fouling Jerry Stackhouse.

A financial manager made two new investments-one in the oil industry and one in municipal bonds. After a one-year period, each of the investments will be classified as either successful or unsuccessful. Consider the making of the two investments as an experiment. a. How many sample points exist for this experiment? b. Show a tree diagram and list the sample points. c. Let \(O=\) the event that the oil industry investment is successful and \(M=\) the event that the municipal bond investment is successful. List the sample points in \(O\) and in \(M\). d. List the sample points in the union of the events \((O \cup M)\). e. List the sample points in the intersection of the events \((O \cap M)\). f. Are events \(O\) and \(M\) mutually exclusive? Explain.

Statistics from the 2009 Major League Baseball season show that there were 157 players who had at least 500 plate appearances. For this group, 42 players had a batting average of 300 or higher, 53 players hit 25 or more home runs, and 14 players had a batting average of .300 or higher and hit 25 or more home runs. Only four players had 200 or more hits (ESPN website, January 10,2010 ). Use the 157 players who had at least 500 plate appearances to answer the following questions. a. What is the probability that a randomly selected player had a batting average of .300 or higher? b. What is the probability that a randomly selected player hit 25 or more home runs? c. Are the events having a batting average of .300 or higher and hitting 25 or more home runs mutually exclusive? d. What is the probability that a randomly selected player had a batting average of .300 or higher or hit 25 or more home runs? e. What is the probability that a randomly selected player had 200 or more hits? Does obtaining 200 or more hits appear to be more difficult than hitting 25 or more home runs? Explain.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.