Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.41\) and \(P(E)=0.23\). i. What is the value of \(P(E \cap F) ?\) ii. What is the value of \(P(E \cup F) ?\) b. Suppose that for events \(A\) and \(B, P(A)=0.26, P(B)=0.34\), and \(P(A \cup B)=0.47 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

Short Answer

Expert verified
i. \(P(E \cap F) = 0\), ii. \(P(E \cup F) = 0.64\); Events \(A\) and \(B\) are not mutually exclusive

Step by step solution

01

Find \(P(E \cap F)\) for mutually exclusive events

From the definition of mutually exclusive events, we know that two events cannot occur simultaneously. Thus, \(P(E \cap F) = 0\)
02

Find \(P(E \cup F)\) for mutually exclusive events

The union of two mutually exclusive events is found by adding their individual probabilities. Thus, \(P(E \cup F) = P(E) + P(F) = 0.41 + 0.23 = 0.64\)
03

Determine if events \(A\) and \(B\) are mutually exclusive

We know that if events are mutually exclusive then \(P(A \cup B) = P(A) + P(B)\). Let's compare \(0.26 + 0.34\) with \(0.47\). Since \(0.26 + 0.34 = 0.6\), which is not equal to \(0.47\), therefore events \(A\) and \(B\) are not 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 Vaia!

Key Concepts

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

Probability of Union of Events
Understanding the probability of the union of events involves evaluating the likelihood of either one event, another event, or both events occurring at the same time. Mathematically, for any two events A and B, this is represented as \(P(A \cup B)\), which reads as the probability of A or B occurring.

Now, to calculate \(P(A \cup B)\), we generally add the probabilities of A and B and subtract the probability of their intersection, which is \(P(A \cap B)\). This subtraction is necessary because if we simply added the probabilities of A and B without considering the intersection, we would be counting the intersection twice. However, there's an exception to this rule when dealing with mutually exclusive events, which we'll explore in other sections.
Probability of Intersection of Events
The probability of the intersection of events refers to the likelihood of two or more events occurring simultaneously. Represented as \(P(A \cap B)\), it illustrates the intersection of event A with event B.

However, to determine this intersection, we need to consider if the events can actually occur at the same time. For independent events, where one event's occurrence doesn't affect the other's, we can find the intersection by multiplying the probabilities of the individual events (i.e., \(P(A) \times P(B)\)). But the scenario is different for mutually exclusive events; their intersection probability is always zero because they cannot occur simultaneously.
Definition of Mutually Exclusive Events
Mutually exclusive events are cases in which the occurrence of one event completely prevents the occurrence of another. In other words, these events cannot happen at the same time.

For example, when you flip a coin, the event of landing tails is mutually exclusive from the event of landing heads. As a result, if two events E and F are mutually exclusive, then by definition, \(P(E \cap F) = 0\), meaning the probability of their intersection is zero. This also simplifies the calculation of their union, since \(P(E \cup F) = P(E) + P(F)\), with no need to subtract anything for their intersection because it's non-existent.
Probability Principles
There are core principles in probability that provide the foundation for understanding the behavior of random phenomena. One fundamental principle is the addition rule, which helps in determining the probability of the union of two events, considering if they are mutually exclusive or not.

The multiplication rule is another key principle, used to calculate the probability of the intersection of independent events. Moreover, it's important to distinguish between independent and mutually exclusive events, as this affects how we apply these principles: for mutually exclusive events, there is no overlap, while independent events can happen at the same time without influencing each other. Understanding and correctly applying these principles is essential for accurate probability calculations.

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

In a small city, approximately \(15 \%\) of those eligible are called for jury duty in any one calendar year. People are selected for jury duty at random from those eligible, and the same individual cannot be called more than once in the same year. What is the probability that an eligible person in this city is selected in both of the next 2 years? All of the next 3 years?

A small college has 2,700 students enrolled. Consider the chance experiment of selecting a student at random. For each of the following pairs of events, indicate whether or not you think they are mutually exclusive and explain your reasoning. a. the event that the selected student is a senior and the event that the selected student is majoring in computer science. b. the event that the selected student is female and the event that the selected student is majoring in computer science. c. the event that the selected student's college residence is more than 10 miles from campus and the event that the selected student lives in a college dormitory. d. the event that the selected student is female and the event that the selected student is on the college football team.

A mutual fund company offers its customers several different funds: a money market fund, three different bond funds, two stock funds, and a balanced fund. Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows: $$ \begin{array}{lr} \text { Money market } & 20 \% \\ \text { Short-term bond } & 15 \% \\ \text { Intermediate-term bond } & 10 \% \\ \text { Long-term bond } & 5 \% \\ \text { High-risk stock } & 18 \% \\ \text { Moderate-risk stock } & 25 \% \\ \text { Balanced fund } & 7 \% \end{array} $$ A customer who owns shares in just one fund is to be selected at random. a. What is the probability that the selected individual owns shares in the balanced fund? b. What is the probability that the individual owns shares in a bond fund? c. What is the probability that the selected individual does not own shares in a stock fund?

A friend who works in a big city owns two cars, one small and one large. Three-quarters of the time he drives the small car to work, and one-quarter of the time he takes the large car. If he takes the small car, he usually has little trouble parking and so is at work on time with probability \(0.9 .\) If he takes the large car, he is on time to work with probability 0.6 . Given that he was at work on time on a particular morning, what is the probability that he drove the small car?

The paper "Predictors of Complementary Therapy Use Among Asthma Patients: Results of a Primary Care Survey" (Health and Social Care in the Community [2008]: \(155-164)\) described a study in which each person in a large sample of asthma patients responded to two questions: Question 1: Do conventional asthma medications usually help your symptoms? Question 2: Do you use complementary therapies (such as herbs, acupuncture, aroma therapy) in the treatment of your asthma? Suppose that this sample is representative of asthma patients. Consider the following events: \(E=\) event that the patient uses complementary therapies \(F=\) event that the patient reports conventional medications usually help The data from the sample were used to estimate the following probabilities: $$P(E)=0.146 \quad P(F)=0.879 \quad P(E \cap F)=0.122$$ a. Use the given probability information to set up a "hypothetical 1000 " table with columns corresponding to \(E\) and \(n o t E\) and rows corresponding to \(F\) and not \(F\). b. Use the table from Part (a) to find the following probabilities: i. The probability that an asthma patient responds that conventional medications do not help and that patient uses complementary therapies. ii. The probability that an asthma patient responds that conventional medications do not help and that patient does not use complementary therapies. iii. The probability that an asthma patient responds that conventional medications usually help or the patient uses complementary therapies. c. Are the events \(E\) and \(F\) independent? Explain.

See all solutions

Recommended explanations on Math Textbooks

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.

Sign-up for free