/*! 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 11 Critical Thinking Suppose two ev... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 11

Critical Thinking Suppose two events \(A\) and \(B\) are mutually exclusive, with \(P(A) \neq 0\) and \(P(B) \neq 0 .\) By working through the following steps, you'll see why two mutually exclusive events are not independent. (a) For mutually exclusive events, can event \(A\) occur if event \(B\) has occurred? What is the value of \(P(A | B) ?\) (b) Using the information from part (a), can you conclude that events \(A\) and \(B\) are not independent if they are mutually exclusive? Explain.

Short Answer

Expert verified
Mutually exclusive events can't be independent since they can't occur together, making conditional probability zero.

Step by step solution

01

Understanding Mutually Exclusive Events

Two events, \(A\) and \(B\), are mutually exclusive if they cannot occur at the same time. This means that if event \(B\) occurs, then event \(A\) cannot occur, and vice versa. Mathematically, this is expressed as \(A \cap B = \emptyset\), where \(A \cap B\) is the intersection of events \(A\) and \(B\), and \(\emptyset\) represents an empty set.
02

Calculating Conditional Probability \(P(A | B)\)

Since \(A\) and \(B\) are mutually exclusive, if \(B\) occurs, \(A\) cannot occur. Therefore, the probability of \(A\) occurring given that \(B\) has occurred is zero. Thus, \(P(A | B) = 0\).
03

Understanding Independence of Events

Two events \(A\) and \(B\) are independent if the fact that \(A\) occurs does not affect the probability of \(B\) occurring. Mathematically, this is expressed as \(P(A \cap B) = P(A) \times P(B)\).
04

Conclusion about Independence

Since \(P(A | B) = 0\) and \(P(B) eq 0\), \(P(A | B) eq P(A)\), which contradicts the definition of independence. If \(A\) and \(B\) were independent, then \(P(A | B)\) should equal \(P(A)\). Thus, events \(A\) and \(B\) cannot be independent if they are mutually exclusive.

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.

Conditional Probability
Conditional probability explores the probability of an event occurring given that another event has already occurred. Written as \( P(A | B) \), it reads as "the probability of \( A \) given \( B \)." It's crucial for understanding relationships between different events.
Consider mutually exclusive events, such as our \( A \) and \( B \). If \( B \) occurs, \( A \) cannot鈥攎aking \( P(A | B) = 0 \). This zero probability highlights that the knowledge of \( B \) alters our understanding of \( A \)'s potential occurrence.
Conditional probability is a pivotal concept in statistics and everyday decision-making. By refining how probability is calculated based on prior knowledge, it enhances predictions and likelihood assessments.
Independent Events
Independent events do not influence each other's occurrence. If \((A)\) and \((B)\) are independent, the happening of \((A)\) has no bearing on the probability that \((B)\) might occur, and vice versa.
This relationship is mathematically described as \( P(A \cap B) = P(A) \times P(B) \). Simply put, this means the joint probability of both events occurring is the product of each event's individual probability.
Independent events behave differently from mutually exclusive ones. While mutually exclusive events cannot occur simultaneously, independent events can.Various scenarios, from rolling dice to business decision making, involve independence where events operate without affecting each other's risks or outcomes.
Probability Theory
Probability theory provides the foundation for evaluating likelihoods of events. It's an essential tool in fields ranging from science to economics.
Key concepts include:
  • Sample Space: The set of all possible outcomes, often denoted as \( S \).
  • Events: Subsets of the sample space, like events \( A \) and \( B \).
  • Probability Measure: Assigning probabilities to events, ensuring they sum to 1.
The central tenet of probability theory is understanding and calculating the chance of various combinations of events. For example, mutually exclusive events like \( A \) and \( B \) cannot simultaneously occur, affecting calculations and interpretations significantly.
Probability theory is integral in risk assessment, simulations, and determining event outcomes. Whether predicting weather patterns or managing project risks, it offers a structured approach for quantifying uncertainty.

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

For each of the following situations, explain why the combinations rule or the permutations rule should be used. (a) Determine the number of different groups of 5 items that can be selected from 12 distinct items. (b) Determine the number of different arrangements of 5 items that can be selected from 12 distinct items.

Survey: Customer Loyalty Are customers more loyal in the east or in the west? The following table is based on information from Trends in the United States, published by the Food Marketing Institute, Washington, D.C. The columns represent length of customer loyalty (in years) at a primary supermarket. The rows represent regions of the United States. $$\begin{array}{lccccccc} \hline & \begin{array}{c} \text { Less Than } \\ \text { 1 Year } \end{array} & \begin{array}{c} 1-2 \\ \text { Years } \end{array} & \begin{array}{c} 3-4 \\ \text { Years } \end{array} & \begin{array}{c} 5-9 \\ \text { Years } \end{array} & \begin{array}{c} 10-14 \\ \text { Years } \end{array} & \begin{array}{c} 15 \text { or More } \\ \text { Years } \end{array} & \begin{array}{c} \text { Row } \\ \text { Total } \end{array} \\ \hline \text { East } & 32 & 54 & 59 & 112 & 77 & 118 & 452 \\ \text { Midwest } & 31 & 68 & 68 & 120 & 63 & 173 & 523 \\ \text { South } & 53 & 92 & 93 & 158 & 106 & 158 & 660 \\ \text { West } & 41 & 56 & 67 & 78 & 45 & 86 & 373 \\ \text { Column Total } & 157 & 270 & 287 & 468 & 291 & 535 & 2008 \\ \hline \end{array}$$ What is the probability that a customer chosen at random (a) has been loyal 10 to 14 years? (b) has been loyal 10 to 14 years, given that he or she is from the east? (c) has been loyal at least 10 years? (d) has been loyal \(a t\) least 10 years, given that he or she is from the west? (e) is from the west, given that he or she has been loyal less than I year? (f) is from the south, given that he or she has been loyal less than 1 year? (g) has been loyal I or more years, given that he or she is from the east? (h) has been loyal I or more years, given that he or she is from the west? (i) Are the events "from the east" and "loyal 15 or more years" independent? Explain.

There are 15 qualified applicants for 5 trainee positions in a fast-food management program. How many different groups of trainees can be selected?

Sometimes probability statements are expressed in terms of odds. The odds in favor of an event \(A\) are the ratio \(\frac{P(A)}{P(n o t A)}=\frac{P(A)}{P\left(A^{c}\right)}\) For instance, if \(P(A)=0.60,\) then \(P\left(A^{C}\right)=0.40\) and the odds in favor of \(A\) are $$ \frac{0.60}{0.40}=\frac{6}{4}=\frac{3}{2}, \text { written as } 3 \text { to } 2 \text { or } 3: 2 $$ (a) Show that if we are given the odds in favor of event \(A\) as \(n: m,\) the probability of event \(A\) is given by \(P(A)=\frac{n}{n+m} .\) Hint: Solve the equation $$ \frac{n}{m}=\frac{P(A)}{1-P(A)} \text { for } P(A) $$ (b) A telemarketing supervisor tells a new worker that the odds of making a sale on a single call are 2 to \(15 .\) What is the probability of a successful call? (c) A sports announcer says that the odds a basketball player will make a free throw shot are 3 to \(5 .\) What is the probability the player will make the shot?

On a single toss of a fair coin, the probability of heads is 0.5 and the probability of tails is \(0.5 .\) If you toss a coin twice and get heads on the first toss, are you guaranteed to get tails on the second toss? Explain.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

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.