/*! 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 99 Three of the most common types o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 99

Three of the most common types of pets are cats, dogs, and fish. Many families have more than one type of pet. Suppose that a family is selected at random and consider the following events and probabilities: \(F=\) event that the selected family has at least one fish \(D=\) event that the selected family has at least one dog \(C=\) event that the selected family has at least one cat $$ \begin{array}{rrr} P(F)=0.20 & P(D)=0.32 & P(C)=0.35 \\ P(F \cap D)=0.18 & P(F \cap C)=0.07 & P(D \mid C)=0.30 \end{array} $$ Calculate the following probabilities. a. \(P(F \mid D)\) b. \(P(F \cup D)\) c. \(P(C \cap D)\)

Short Answer

Expert verified
a. \(P(F \mid D) = 0.5625\) b. \(P(F \cup D) = 0.34\) c. \(P(C \cap D) = 0.105\)

Step by step solution

01

Recall the formula for conditional probability

The formula for conditional probability is given by: \(P(A|B) = \frac{P(A \cap B)}{P(B)}\) We will need this formula to solve part a and part c.
02

Recall the formula for the union of events

The formula for the union of events is given by: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) We will need this formula to solve part b.
03

Calculate \(P(F \mid D)\) (for part a)

Using the formula for conditional probability, we get: \[ P(F|D) = \frac{P(F \cap D)}{P(D)} = \frac{0.18}{0.32} = 0.5625 \]
04

Calculate \(P(F \cup D)\) (for part b)

Using the formula for the union of events, we get: \[ P(F \cup D) = P(F) + P(D) - P(F \cap D) = 0.20 + 0.32 - 0.18 = 0.34 \]
05

Calculate \(P(C \cap D)\) (for part c)

We know that: \[P(D|C) = \frac{P(C \cap D)}{P(C)}\] Rearranging the terms, we get: \[ P(C \cap D) = P(D|C) \cdot P(C) = 0.30 \cdot 0.35 = 0.105 \] Thus, the answers are: a. \(P(F \mid D) = 0.5625\) b. \(P(F \cup D) = 0.34\) c. \(P(C \cap D) = 0.105\)

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.

Probability of Events
Understanding how likely an event is to occur can be a head-scratcher, but it's the essence of probability. In our pet-loving families' scenario, when we talk about the probability of events, we're actually asking: 'What's the chance of this happening?' With single events, it's pretty straightforward. For example, the probability that a randomly chosen family has a dog is simply the number of dog-owning families divided by the total number of families.

But real magic happens when we start looking at two or more events simultaneously. We may want to know how likely two events are to occur together, like owning a fish and a dog. Or even what the chances are of at least one of the two events happening, which could be having either a fish or a dog. The formulas we apply here are the cornerstone of making these predictions accurately.
Union of Events
Imagine we've got one big group that includes any family with a fish, another with any family with a dog, and we're trying to form a supergroup. That's where the concept of the union of events comes into play. It's a bit like throwing two different parties and then deciding to join forces for an epic mega-party. When events are joined this way, we're interested in the probability that at least one of them will occur.

The formula for the union is a neat little package: for two events A and B, the probability of A or B (written as \(P(A \cup B)\)) is the sum of the probabilities of each event minus the probability that they both happen at the same time (because we don't want to count those twice). It's essential however to be aware that this formula assumes some overlap exists between the events; that is, they aren't mutually exclusive.
Probability Formulas
Diving into the world of probability formulas is like getting the keys to the probability kingdom. These formulas are your tools to unlock the secrets of chance and make sense of what might at first glance seem like random happenings.

We've discussed the basics like the formula for the probability of single events鈥攋ust a simple division. For understanding events happening together (their intersection), we use the multiplication rule. And when it's time to talk about conditional probability, we say stuff like 'What's the chance of finding Nemo (a fish), given we're already visiting a dog-owning family?' This is where \(P(A|B)\) comes into play, telling us how the occurrence of event B (having a dog) affects the likelihood of event A (having a fish).

In application, you might notice that these formulas let us play a little math detective, piecing together the whole picture of events from just fragments of information. By using these formulas correctly, students like you can confidently predict all sorts of probabilities and deconstruct the most tangled of events.

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

Lyme disease is transmitted by infected ticks. Several tests are available for people with symptoms of Lyme disease. One of these tests is the EIA/IFA test. The paper "Lyme Disease Testing by Large Commercial Laboratories in the United States" (Clinical Infectious Disease [2014]: \(676-681\) ) found that \(11.4 \%\) of those tested actually had Lyme disease. Consider the following events: \(+\) represents a positive result on the blood test \- represents a negative result on the blood test \(L\) represents the event that the patient actually has Lyme disease \(L^{C}\) represents the event that the patient actually does not have Lyme disease The following probabilities are based on percentages given in the paper: $$ \begin{array}{r} P(L)=0.114 \\ P\left(L^{C}\right)=0.886 \end{array} $$ $$ \begin{array}{c} P(+\mid L)=0.933 \\ P(-\mid L)=0.067 \\ P\left(+\mid L^{C}\right)=0.039 \\ P\left(-\mid L^{C}\right)=0.961 \end{array} $$ a. For each of the given probabilities, write a sentence giving an interpretation of the probability in the context of this problem. b. Use the given probabilities to construct a hypothetical 1000 table with columns corresponding to whether or not a person has Lyme disease and rows corresponding to whether the blood test is positive or negative. c. Notice the form of the known conditional probabilities; for example, \(P(+\mid L)\) is the probability of a positive test given that a person selected at random from the population actually has Lyme disease. Of more interest is the probability that a person has Lyme disease, given that the test result is positive. Use information from the table constructed in Part (b) to calculate this probability.

The article "A Crash Course in Probability" from The Economist included the following information: The chance of being involved in an airplane crash when flying on an Airbus 330 from London to New York City on Virgin Atlantic Airlines is 1 in \(5,371,369 .\) This was interpreted as meaning that you "would expect to go down if you took this flight every day for 14,716 years." The article also states that a person could "expect to fly on the route for 14,716 years before plummeting into the Atlantic." Comment on why these statements are misleading.

Suppose that \(20 \%\) of all teenage drivers in a certain county received a citation for a moving violation within the past year. Assume in addition that \(80 \%\) of those receiving such a citation attended traffic school so that the citation would not appear on their permanent driving record. Consider the chance experiment that consists of randomly selecting a teenage driver from this county. a. One of the percentages given in the problem specifies an unconditional probability, and the other percentage specifies a conditional probability. Which one is the conditional probability, and how can you tell? b. Suppose that two events \(E\) and \(F\) are defined as follows: \(E=\) selected driver attended traffic school \(F=\) selected driver received such a citation Use probability notation to translate the given information into two probability statements of the form \(P(\underline{C})=\) probability value.

A company that offers roadside assistance to drivers reports that the probability that a call for assistance will be to help someone who is locked out of his or her car is \(0.18 .\) Give a relative frequency interpretation of this probability.

The following table summarizes data on smoking status and age group, and is consistent with summary quantities obtained in a Gallup Poll published in the online article "In U.S., Young Adults' Cigarette Use Is Down Sharply" $$ \begin{array}{|lcc|} \hline & {\text { Smoking Status }} \\ \hline { 2 - 3 } \text { Age Group } & \text { Smoker } & \text { Nonsmoker } \\\ \hline 18 \text { to } 29 & 174 & 618 \\ 30 \text { to } 49 & 333 & 1,115 \\ 50 \text { to } 64 & 384 & 1,445 \\ 65 \text { and older } & 211 & 1,707 \\ \hline \end{array} $$ Assume that it is reasonable to consider these data as representative of the American adult population. Consider the chance experiment or randomly selecting an adult American. a. What is the probability that the selected adult is a smoker? b. What is the probability that the selected adult is under 50 years of age? c. What is the probability that the selected adult is a smoker that is 65 or older? d. What is the probability that the selected adult is a smoker or is age 65 or older?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Logic and Functions

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.