/*! 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 2 A sample space \(S\) consists of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 2

A sample space \(S\) consists of five simple events with these probabilities: $$\begin{array}{c}P\left(E_{1}\right)=P\left(E_{2}\right)=.15 \quad P\left(E_{3}\right)=.4 \\\P\left(E_{4}\right)=2 P\left(E_{5}\right)\end{array}$$ a. Find the probabilities for simple events \(E_{4}\) and \(E_{5}\). b. Find the probabilities for these two events: $$\begin{array}{l}A=\left\\{E_{1}, E_{3}, E_{4}\right\\} \\\B=\left\\{E_{2}, E_{3}\right\\}\end{array}$$ c. List the simple events that are either in event \(A\) or event \(B\) or both. d. List the simple events that are in both event \(A\) and event \(B\).

Short Answer

Expert verified
Answer: The probability of event \(E_5\) is 0.1, and the simple events in the union of events \(A\) and \(B\) are \(\\{E_1, E_2, E_3, E_4\\}\).

Step by step solution

01

a. Finding the probabilities of events \(E_{4}\) and \(E_{5}.\)

The sum of the probabilities of all simple events in the sample space must be 1. We can use this property to find the probabilities of events \(E_{4}\) and \(E_{5}\). We are given the probabilities of \(E_{1}, E_{2}\), and \(E_{3},\) and the relationship between probabilities of \(E_{4}\) and \(E_{5}\): $$P(E_1) + P(E_2) + P(E_3) + P(E_4) + P(E_5) = 1$$ Substitute the given values into this equation: $$0.15 + 0.15 + 0.4 + P(E_4) + 2P(E_5) = 1$$ Now, we have one equation with two unknowns, \(P(E_4)\) and \(P(E_5)\), so we need to find an additional equation in order to come to a solution.
02

a.1 - Finding an additional equation to solve for \(P(E_{4})\) and \(P(E_{5})\)

One simple event has a non-zero probability, so we can observe that: $$P(E_4) \neq 0 \space and \space P(E_5) \neq 0$$ This observation allows us to write an inequality together with the first equation: \( 0 < P(E_4) = 1 - (0.15 + 0.15 + 0.4 + 2P(E_5)) = 0.3 - 2P(E_5)\) \( 0 < 0.3 - 2P(E_5) \Rightarrow P(E_5) < 0.15\) Now we can find the possible range of values for \(P(E_5)\) and \(P(E_4)\) by solving the inequality for \(P(E_5).\)
03

a.2 - Solving for \(P(E_{4})\) and \(P(E_{5})\)

Given that \(0 < P(E_5) < 0.15\), we can find the value for \(P(E_4)\) and \(P(E_5)\). Setting \(P(E_5) = 0.1\), we get \(P(E_4) = 0.1\). We can then use this value to find the probability of event \(E_{5}\): $$ P(E_5) = 0.1$$ and the probability of event \(E_{4}\): $$P(E_4) = 1 - (0.15 + 0.15 + 0.4 + 2(0.1)) = 0.1$$
04

b. Finding the probabilities for events \(A\) and \(B\)

Event \(A = \\{E_1, E_3, E_4\\}\) and event \(B = \\{E_2, E_3\\}\). The probabilities for these compound events can be found by adding the probabilities of their simple events: $$P(A) = P(E_1) + P(E_3) + P(E_4) = 0.15 + 0.4 + 0.1 = 0.65$$ $$P(B) = P(E_2) + P(E_3) = 0.15 + 0.4 = 0.55$$
05

c. Listing the simple events in event \(A\) or event \(B\) or both

To list the simple events that are in either event \(A\) or event \(B\) or both, we can take the union of events \(A\) and \(B\): $$A \cup B = \\{E_1, E_2, E_3, E_4\\}$$
06

d. Listing the simple events in both event \(A\) and event \(B\)

To list the simple events that are in both event \(A\) and event \(B\), we can take the intersection of events \(A\) and \(B\): $$A \cap B = \\{E_3\\}$$

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.

Sample Space
When we talk about probability, one of the most foundational concepts is the sample space. The sample space, denoted usually by the symbol \( S \), is essentially the set of all possible outcomes of a probabilistic experiment. For example, when you toss a coin, the sample space consists of two simple events: \( \{ \text{Heads}, \text{Tails} \} \).
Each outcome is unique and well-defined, making it possible to assign a probability to each event.
  • The probabilities of all the events in the sample space should always sum up to 1.
  • This means that something from the sample space will definitely occur when the experiment takes place.
Simple Events
Simple events are the individual outcomes within the sample space. Each simple event can happen as a result of the experiment, and it is assigned a probability. In a lottery draw, for instance, a single ticket being drawn is a simple event.
  • Each simple event in a sample space has a unique probability assigned to it.
  • The probability of a simple event must be a number between 0 and 1, where 0 means the event cannot happen, and 1 means the event must happen.
Understanding simple events is crucial as they form the building blocks of more complex events like compound events.
Compound Events
Compound events are collections of two or more simple events. These are the events we often refer to when we describe more complex probabilistic phenomena. For example, rolling an even number with a die is a compound event, consisting of the simple events \( \{2, 4, 6\} \).
  • The probability of a compound event is found by combining the probabilities of the simple events that make it up.
  • Depending on the scenario, this combination could be an addition or another operation based on the rules of probability.
Compound events allow us to explore different outcomes of interest in probability experiments, making them quite versatile and useful in statistical analysis.
Intersection and Union of Events
In probability, understanding how events overlap or combine is essential. The intersection of events, denoted by \( A \cap B \), represents the set of outcomes that are common to both events \( A \) and \( B \). This is akin to finding elements common to both lists in a database.
  • Events that have no outcomes in common are termed "mutually exclusive," meaning their intersection is empty.
On the other hand, the union of events, represented by \( A \cup B \), includes all outcomes that are in either event or both. This helps in determining the probability of either or both events occurring.
  • For instance, the union of rolling a number less than 4 or a number greater than 3 on a die represents all possible outcomes, which is the whole sample space \( \{1, 2, 3, 4, 5, 6\} \).
  • Understanding these concepts aids significantly in calculating the resulting probabilities in various scenarios.

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

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 table below. \(^{3}\) 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?

A company has five applicants for two positions: two women and three men. Suppose that the five applicants are equally qualified and that no preference is given for choosing either gender. Let \(x\) equal the number of women chosen to fill the two positions. a. Find \(p(x)\). b. Construct a probability histogram for \(x\).

A key ring contains four office keys that are identical in appearance, but only one will open your office door. Suppose you randomly select one key and try it. If it does not fit, you randomly select one of the three remaining keys. If it does not fit, you randomly select one of the last two. Each different sequence that could occur in selecting the keys represents one of a set of equiprobable simple events. a. List the simple events in \(S\) and assign probabilities to the simple events. b. Let \(x\) equal the number of keys that you try before you find the one that opens the door \((x=1,2,3,4)\). Then assign the appropriate value of \(x\) to each simple event. c. Calculate the values of \(p(x)\) and display them in a table. d. Construct a probability histogram for \(p(x)\).

Suppose that at a particular supermarket the probability of waiting 5 minutes or longer for checkout at the cashier's counter is . \(2 .\) On a given day, a man and his wife decide to shop individually at the market, each checking out at different cashier counters. They both reach cashier counters at the same time. a. What is the probability that the man will wait less than 5 minutes for checkout? b. What is probability that both the man and his wife will be checked out in less than 5 minutes? (Assume that the checkout times for the two are independent events.) c. What is the probability that one or the other or both will wait 5 minutes or longer?

Let \(x\) equal the number observed on the throw of a single balanced die. a. Find and graph the probability distribution for \(x\). b. What is the average or expected value of \(x ?\) c. What is the standard deviation of \(x ?\) d. Locate the interval \(\mu \pm 2 \sigma\) on the \(x\) -axis of the graph in part a. What proportion of all the measurements would fall into this range?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Probability and Statistics

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.