/*! 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 11 A sample space consists of five ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sample space consists of five simple events, \(E_{1}, E_{2}, E_{3}, E_{4},\) and \(E_{5}\) a. If \(P\left(E_{1}\right)=P\left(E_{2}\right)=0.15, P\left(E_{3}\right)=0.4,\) and \(P\left(E_{4}\right)=2 P\left(E_{5}\right),\) find the probabilities of \(E_{4}\) and \(E_{5}\) b. If \(P\left(E_{1}\right)=3 P\left(E_{2}\right)=0.3,\) find the probabilities of the remaining simple events if you know that the remaining simple events are equally probable.

Short Answer

Expert verified
\( P(E_4) = 0.2, P(E_5) = 0.1 \) in part (a); \( P(E_3) = P(E_4) = P(E_5) = 0.2 \) in part (b).

Step by step solution

01

Understand the Problem

Identify the given probabilities in part (a) of the problem: \( P(E_{1}) = P(E_{2}) = 0.15 \), \( P(E_{3}) = 0.4 \). Also, understand the relationship for \( E_{4} \) and \( E_{5} \): \( P(E_{4}) = 2P(E_{5}) \).
02

Apply the Total Probability Concept

Since the sample space is complete, the total of all probabilities must sum to 1. Therefore, apply the equation: \[ P(E_{1}) + P(E_{2}) + P(E_{3}) + P(E_{4}) + P(E_{5}) = 1 \] Substitute the known values: \[ 0.15 + 0.15 + 0.4 + P(E_{4}) + P(E_{5}) = 1 \] Simplify the equation to find \( P(E_{4}) + P(E_{5}) \).
03

Solve for Unknown Probabilities in (a)

Simplifying from Step 2, get:\[ P(E_{4}) + P(E_{5}) = 0.3 \] Use the relationship provided for \( P(E_{4}) \) and \( P(E_{5}) \): \[ P(E_{4}) = 2P(E_{5}) \] Substitute \( P(E_{4}) \) with \( 2P(E_{5}) \) in the equation: \[ 2P(E_{5}) + P(E_{5}) = 0.3 \] Thus, \( 3P(E_{5}) = 0.3 \), simplifying gives \( P(E_{5}) = 0.1 \). Then, \( P(E_{4}) = 2 \times 0.1 = 0.2 \).
04

Understand the Condition in (b)

For part (b), \( P(E_{1}) = 0.3 \), and \( P(E_{2}) = \frac{1}{3} \times 0.3 = 0.1 \). Recognize that \( E_{3}, E_{4}, \) and \( E_{5} \) are equally probable.
05

Apply Equal Distribution for (b)

The sum of the probabilities for \( E_{3}, E_{4}, \) and \( E_{5} \) is:\[ P(E_{3}) + P(E_{4}) + P(E_{5}) = 1 - 0.3 - 0.1 = 0.6 \] Since these events are equally probable, divide 0.6 equally: \[ P(E_{3}) = P(E_{4}) = P(E_{5}) = \frac{0.6}{3} = 0.2 \].

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 91Ó°ÊÓ!

Key Concepts

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

Sample Space
In probability theory, a sample space is the complete set of all possible outcomes of a random experiment. It's like the foundation of probability, where every possible outcome is listed, ensuring that we account for every eventuality in our analysis. - For instance, if you are tossing a fair coin, the sample space would be \[ \{\text{Heads, Tails}\} \]. This means that when you toss the coin, these are the only outcomes you can expect.- If you're rolling a six-sided die, the sample space is \[ \{1, 2, 3, 4, 5, 6\} \], representing all the numbers that could possibly show up on top.In our exercise, the sample space consists of five simple events: \( E_{1}, E_{2}, E_{3}, E_{4}, \text{ and } E_{5} \). This sample space is crucial because every probability we calculate must be in relation to these five outcomes. Each event denotes a potential individual outcome, all of which collectively represent all that could occur in this scenario.
Simple Events
Simple events are the individual possible outcomes or events that cannot be broken down further. They are the building blocks within a sample space. Each simple event is mutually exclusive, meaning the occurrence of one excludes the others. - For example, in the toss of a coin, "Heads" and "Tails" are simple events. Neither can happen simultaneously, and each is distinct from the other.In the given exercise, the simple events are \( E_{1}, E_{2}, E_{3}, E_{4}, \text{ and } E_{5} \). Each of these is treated as a unique outcome with a probability attached:- The known probabilities given are \( P(E_{1}) = P(E_{2}) = 0.15 \) and \( P(E_{3}) = 0.4 \).- We have the information that \( P(E_{4}) = 2P(E_{5}) \), which shows a relation between these two simple events.Understanding simple events helps us grasp the structure of the sample space, enabling the calculation of probabilities for each potential scenario.
Total Probability
The principle of total probability ensures that the probability of all possible outcomes of an experiment adds up to 1. This is because the sample space includes every potential event, and together they represent certainty—something in that space will occur.- For instance, with a die roll, the probabilities of each outcome \( \{1, 2, 3, 4, 5, 6\} \) must sum to 1.In the given exercise, we're using this principle by summing the probabilities of the simple events in the sample space:- The relationship is illustrated through the equation \[ P(E_{1}) + P(E_{2}) + P(E_{3}) + P(E_{4}) + P(E_{5}) = 1 \].- By substituting and solving, we derived values for \( P(E_{4}) \text{ and } P(E_{5}) \text{ where } P(E_{4}) = 0.2 \text{ and } P(E_{5}) = 0.1 \).Applying total probability not only confirms our calculations but also ensures logical consistency within the sample space. This concept is foundational because it guarantees that every calculation of an event's probability is valid within the context of the entire set of outcomes.

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

The symmetric difference between two events \(A\) and \(B\) is the set of all sample points that are in exactly one of the sets and is often denoted \(A \Delta B\). Note that \(A \Delta B=(A \cap B) \cup(\bar{A} \cap B)\). Prove that \(P(A \Delta B)=P(A)+P(B)-2 P(A \cap B)\)

Five cards are dealt from a standard 52 -card deck. What is the probability that we draw a. 3 aces and 2 kings? b. a "full house" (3 cards of one kind, 2 cards of another kind)?

A group of five applicants for a pair of identical jobs consists of three men and two women. The employer is to select two of the five applicants for the jobs. Let \(S\) denote the set of all possible outcomes for the employer's selection. Let \(A\) denote the subset of outcomes corresponding to the selection of two men and \(B\) the subset corresponding to the selection of at least one woman. List the outcomes in \(A, \bar{B}, A \cup B, A \cap B,\) and \(A \cap \bar{B}\). (Denote the different men and women by \(M_{1}, M_{2}, M_{3}\) and \(W_{1}, W_{2},\) respectively.)

A population of voters contains 40\% Republicans and 60\% Democrats. It is reported that 30\% of the Republicans and \(70 \%\) of the Democrats favor an election issue. A person chosen at random from this population is found to favor the issue in question. Find the conditional probability that this person is a Democrat.

An upscale restaurant offers a special fixe prix menu in which, for a fixed dinner cost, a diner can select from four appetizers, three salads, four entrees, and five desserts. How many different dinners are available if a dinner consists of one appetizer, one salad, one entree, and one dessert?

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.