/*! 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 109 If \(P(A)=0.3\) and \(P(B)=0.4\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 109

If \(P(A)=0.3\) and \(P(B)=0.4\) and \(A\) and \(B\) are independent events, what is the probability of each of the following? a. \(\quad P(A \text { and } B)\) b. \(\quad P(\mathbf{B} | \mathbf{A})\) c. \(\quad P(\mathrm{A} | \mathrm{B})\)

Short Answer

Expert verified
The probability of both events A and B occurring is 0.12, the probability of event B given that A has occurred is 0.4, and the probability of event A given that B has occured is 0.3.

Step by step solution

01

Determine P(A and B)

Since events A and B are independent, the probability of both events occurring is the product of their individual probabilities. Therefore, P(A and B) = P(A) * P(B) = 0.3 * 0.4 = 0.12.
02

Determine P(B | A)

The probability of event B given that A has occurred is simply the probability of event B, since the events are independent. So, P(B | A) = P(B) = 0.4.
03

Determine P(A | B)

Similarly, the probability of event A given that B has occurred is just the probability of event A, as events A and B are independent. Therefore, P(A | B) = P(A) = 0.3.

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 is a key concept in probability theory that describes the likelihood of an event occurring given that another event has already occurred. This is represented as \( P(B | A) \), which reads as the probability of event B given event A. However, when the two events in question are independent, as outlined in the original exercise, things are simpler. For independent events, the occurrence of one event does not affect the likelihood of the other event occurring.
  • In the given problem, since events A and B are independent, the conditional probabilities \( P(B | A) \) and \( P(A | B) \) simplify to just \( P(B) \) and \( P(A) \) respectively.
  • This is because the occurrence of event A gives us no additional information about event B, and vice versa.
Understanding this simplification helps in calculating probabilities more efficiently when dealing with independent events.
Multiplication Rule for Independent Events
The multiplication rule for independent events provides a straightforward way to calculate the probability of two independent events both happening, denoted as \( P(A \text{ and } B) \). This involves multiplying the probabilities of each event:
  • The rule states that \( P(A \text{ and } B) = P(A) \times P(B) \) if A and B are independent.
  • In the original exercise, you were given \( P(A) = 0.3 \) and \( P(B) = 0.4 \). Applying the multiplication rule gives us \( P(A \text{ and } B) = 0.3 \times 0.4 = 0.12 \).
  • This calculation confirms that the probability of both events A and B occurring simultaneously is 0.12.
This rule highlights the simplicity involved when dealing with independent events, making it easier to manage calculations.
Probability Theory
Probability theory is the mathematical framework that allows us to quantify uncertainty. It is a branch of mathematics concerned with analyzing random phenomena and consists of various rules and principles that help in determining the likelihood of different outcomes.
  • In the context of the exercise, probability theory encompasses rules like the multiplication rule for independent events and concepts such as conditional probability.
  • The exercise demonstrates how independent events simplify these calculations, as their occurrence does not influence each other, allowing direct applications of basic probability rules.
  • Understanding these foundational principles provides a basis for exploring more complex probabilistic models and techniques.
By learning these concepts, students can better interpret and calculate the probabilities of events in various real-world situations.

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

A box contains four red and three blue poker chips. Three poker chips are to be randomly selected, one at a time. a. What is the probability that all three chips will be red if the selection is done with replacement? b. What is the probability that all three chips will be red if the selection is done without replacement? c. Are the drawings independent in either part a or b? Justify your answer.

The NCAA men's basketball season starts with 327 college teams all dreaming of making it to "the big dance" and attaining the National Championship. Sixtyfive teams are selected for the tournament and only one wins it all. a. What are the odds against a team being selected for the tournament? b. What are the odds of a team that is in the tournament winning the National Championship? c. Now wait a minute! What assumption did you make in order to answer the above questions? Does this seem realistic?

The probability that thunderstorms are in the vicinity of a particular Midwestern airport on an August day is \(0.70 .\) When thunderstorms are in the vicinity, the probability that an airplane lands on time is \(0.80 .\) Find the probability that thunderstorms are in the vicinity and that the plane lands on time.

Explain why an empirical probability, an observed proportion, and a relative frequency are actually three different names for the same thing.

A woman and a man (unrelated) each has two children. At least one of the woman's children is a boy, and the man's older child is a boy. Is the probability that the woman has two boys greater than, equal to, or less than the probability that the man has two boys? a. Demonstrate the truth of your answer by using a simple sample to represent each family. b. Demonstrate the truth of your answer by taking two samples, one from men with two-children families and one from women with two-children families. c. Demonstrate the truth of your answer using computer simulation. Using the Bernoulli probability function with \(p=0.5 \text { (let } 0=\text { girl and } 1=\text { boy })\), generate 500 "families of two children" for the man and the woman. Determine which of the 500 satisfy the condition for each and determine the observed proportion with two boys. d. Demonstrate the truth of your answer by repeating the computer simulation several times. Repeat the simulation in part c several times. e. Do the preceding procedures scem to yicld the same results? Explain.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Pure 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.