/*! 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 8 Experiment II A sample space con... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 8

Experiment II A sample space contains seven simple events: \(E_{1}, E_{2}, \ldots, E_{7} .\) Suppose that \(E_{1}, E_{2}, \ldots, E_{6}\) all have the same probability, but \(E_{7}\) is twice as likely as the others. Find the probabilities of the events. $$ B=\left\\{E_{1}, E_{3}, E_{5}, E_{7}\right\\} $$

Short Answer

Expert verified
Answer: For \(E_1, E_2,..., E_6\), the probability for each event is \(\frac{1}{8}\), and for \(E_7\), the probability is \(\frac{1}{4}\).

Step by step solution

01

Assign variables

Let's assign the probability of \(E_1, E_2,..., E_6\) as \(p\). Then, the probability of \(E_{7}\), which has double the probability of the others, will be \(2p\).
02

Write the formula for the total probability

The total probability in the sample space should be equal to 1. So, the sum of the probabilities of \(E_1, E_2,..., E_7\) should be equal to 1. Let's write this equation: $$p + p + p + p + p + p + 2p = 1$$
03

Solve the equation

Combine the terms in the equation: $$6p + 2p = 1$$ $$8p = 1$$ Now, we can solve for \(p\): $$p= \frac{1}{8} $$
04

Calculate the probabilities of each event

Now that we know the value of \(p\), we can easily find the probabilities of all simple events by using the information provided. For \(E_1, E_2,..., E_6\), the probability for each event is the same as \(p\): $$P(E_1) = P(E_2) = \ldots= P(E_6) = \frac{1}{8} $$ For \(E_7\), the probability is double the value of \(p\): $$P(E_7) = 2p = 2 \times \frac{1}{8}$$ $$P(E_7) = \frac{1}{4} $$ Now we have the probabilities of all the events.

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 Theory
Probability theory is a branch of mathematics concerned with the analysis of random phenomena. The central objects of probability theory are random variables, stochastic processes, and events. In simple terms, it's the study of uncertainty and how to measure the likelihood of various outcomes.

For example, if we consider the flipping of a fair coin, there are two possible outcomes: heads or tails. Each outcome is equally likely, giving a probability of 0.5 or 50% for heads and the same for tails. Probability theory allows us to quantify these outcomes in a systematic way. It is foundational for many fields, including statistics, finance, gambling, science, and philosophy.
Simple Events Probability
In the realm of probability, simple events are the most basic possible outcomes that cannot be broken down into further outcomes. Each simple event is unique and represents an individual outcome. The probability of a simple event occurring is a numerical value between 0 and 1, where 0 indicates impossibility, and 1 indicates certainty.

Considering our earlier sample space of seven simple events, each one, such as obtaining a specific face on a rolled die, is regarded as a simple event. The sum of the probabilities of all simple events in a sample space must equal 1, as they represent all possible outcomes of an experiment. Comprehending probabilities of simple events is crucial because complex events can often be expressed as combinations of these simple ones.
Total Probability Formula
The total probability formula is a key concept in understanding probability distributions in a finite sample space. It states that the total probability of all possible outcomes must add up to 1, reflecting the certainty that one of the outcomes will occur.

Our exercise example reflects this principle. With the probabilities of the first six events being equal and the last event having twice the probability, we theoretically distribute the total probability of 1 across all seven events proportionally. By setting up an equation as seen in the steps provided, we can resolve the value of each simple event's probability, ensuring the sum equals 1. This is essential, as the total probability formula serves as a validation tool for correctly calculating probabilities of events in a finite space.

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 worker-operated machine produces a defective item with probability .01 if the worker follows the machine's operating instructions exactly, and with probability .03 if he does not. If the worker follows the instructions \(90 \%\) of the time, what proportion of all items produced by the machine will be defective?

A sample space contains seven simple events: \(E_{1}, E_{2}, \ldots, E_{7} .\) Use the following three eventsA, \(B\), and \(C\) - and list the simple events in Exercises \(7-12\). \(A=\left\\{E_{3}, E_{4}, E_{6}\right\\} \quad B=\left\\{E_{1}, E_{3}, E_{5}, E_{7}\right\\} \quad C=\left\\{E_{2}, E_{4}\right\\}\) $$\text { Both } A \text { and } B$$

In how many ways can you select two people from a group of 20 if the order of selection is not important?

Professional basketball is now a reality for women basketball players in the United States. There are two conferences in the WNBA, each with six teams, as shown in the following table. \(^{3}\) $$ \begin{array}{ll} \hline \text { Western Conference } & \text { Eastern Conference } \\ \hline \text { Minnesota Lynx } & \text { Atlanta Dream } \\ \text { Phoenix Mercury } & \text { Indiana Fever } \\ \text { Dallas Wings } & \text { New York Liberty } \\ \text { Los Angeles Sparks } & \text { Washington Mystics } \\ \text { Seattle Storm } & \text { Connecticut Sun } \\ \text { San Antonio Stars } & \text { Chicago Sky } \end{array} $$ Two teams, one from each conference, are randomly selected to play an exhibition game. a. How many pairs of teams can be chosen? b. What is the probability that the two teams are Los Angeles and New York? c. What is the probability that the Western Conference team is not from California?

Suppose \(P(A)=.1\) and \(P(B)=.5 .\) $$\text { If } P(A \cap B)=0, \text { are } A \text { and } B \text { independent? }$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

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.