/*! 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 42 An experiment can result in one ... [FREE SOLUTION] | 91Ó°ÊÓ

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An experiment can result in one of five equally likely simple events, \(E_{1}, E_{2}, \ldots, E_{5} .\) Events \(A, B,\) and \(C\) are defined as follows: $$A: E_{1}, E_{3} \quad P(A)=.4$$ $$B: E_{1}, E_{2}, E_{4}, E_{5} \quad P(B)=.8$$ $$C: E_{3}, E_{4} \quad P(C)=.4$$. Find the probabilities associated with these compound events by listing the simple events in each. a. \(A^{c}\) b. \(A \cap B\) c. \(B \cap C\) d. \(A \cup B\) e. \(B \mid C\) f. \(A \mid B\) g. \(A \cup B \cup C\) h. \((A \cap B)^{c}\)

Short Answer

Expert verified
Answer: The probability of the complement of event A, \(P(A^c)\), is \(\frac{3}{5}\). 2. What is the probability of the intersection of events A and B? Answer: The probability of the intersection of events A and B, \(P(A\cap B)\), is \(\frac{1}{5}\). 3. What is the probability of the intersection of events B and C? Answer: The probability of the intersection of events B and C, \(P(B\cap C)\), is \(\frac{1}{5}\). 4. What is the probability of event B given event C? Answer: The probability of event B given event C, \(P(B|C)\), is 0.5. 5. What is the probability of event A given event B? Answer: The probability of event A given event B, \(P(A|B)\), is 0.25.

Step by step solution

01

1. Identify the sample space and simple event probabilities

There are 5 equally likely simple events: \(E_1, E_2, E_3, E_4,\) and \(E_5\). As there are 5 equally likely events, the probability of any single event is \(\frac{1}{5}\). We need this information to calculate the probabilities of the specified compound events.
02

2. Find \(A^c\) and its probability

The complement of event A, denoted by \(A^c\), includes all simple events that are not part of A. In this case, \(A^c\) consists of the simple events \(E_2, E_4,\) and \(E_5\). Since all simple events are equally likely, \(P(A^c) = P(E_2) + P(E_4) + P(E_5) = \frac{3}{5}\).
03

3. Find \(A\cap B\) and its probability

The intersection of events A and B, denoted by \(A\cap B\), is the set of simple events that belong to both A and B. In this case, \(A\cap B = E_1\). So, \(P(A\cap B) = P(E_1) = \frac{1}{5}\).
04

4. Find \(B\cap C\) and its probability

The intersection of events B and C, denoted by \(B\cap C\), is the set of simple events that belong to both B and C. In this case, \(B\cap C = E_4\). Hence \(P(B\cap C) = P(E_4) = \frac{1}{5}\).
05

5. Find \(A\cup B\) and its probability

The union of events A and B, denoted by \(A\cup B\), includes all simple events that belong to either A or B or both. In this case, \(A\cup B = E_1, E_2, E_3, E_4,\) and \(E_5\). Since this is the entire sample space, \(P(A\cup B) = 1\).
06

6. Find \(B|C\) and its probability

The probability of event B given C, denoted by \(P(B|C)\), is the ratio of the probability of the intersection of B and C to the probability of event C: $$P(B|C) = \frac{P(B\cap C)}{P(C)} = \frac{\frac{1}{5}}{0.4} = \frac{1}{2}$$. So, \(P(B|C) = 0.5\).
07

7. Find \(A|B\) and its probability

The probability of event A given B, denoted by \(P(A|B)\), is the ratio of the probability of the intersection of A and B to the probability of event B: $$P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{\frac{1}{5}}{0.8} = \frac{1}{4}$$. Therefore, \(P(A|B) = 0.25\).
08

8. Find \(A\cup B\cup C\) and its probability

The union of events A, B, and C, denoted by \(A\cup B\cup C\), includes all simple events that belong to A or B or C or any combination of the three. In this case, \(A\cup B\cup C\) consists of all five simple events \(E_1, E_2, E_3, E_4,\) and \(E_5\). As this is the entire sample space, \(P(A\cup B\cup C) = 1\).
09

9. Find \((A\cap B)^c\) and its probability

The complement of the intersection of events A and B, denoted by \((A\cap B)^c\), includes all simple events that are not part of the intersection A and B. We found in part 3 that \(A\cap B = E_1\), and so \((A\cap B)^c = E_2, E_3, E_4,\) and \(E_5\). Thus, \(P((A\cap B)^c) = P(E_2) + P(E_3) + P(E_4) + P(E_5) = \frac{4}{5}\). The probabilities for the specified compound events are as follows: a. \(P(A^c) = \frac{3}{5}\) b. \(P(A\cap B) = \frac{1}{5}\) c. \(P(B\cap C) = \frac{1}{5}\) d. \(P(A\cup B) = 1\) e. \(P(B|C) = 0.5\) f. \(P(A|B) = 0.25\) g. \(P(A\cup B\cup C) = 1\) h. \(P((A\cap B)^c) = \frac{4}{5}\)

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Key Concepts

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

Compound Events
In probability theory, a compound event is an event made up of two or more simple events. In our case, the simple events are represented as \(E_1, E_2, E_3, E_4,\) and \(E_5\), each with an equal probability. To construct compound events, we use operations like union and intersection. For instance, to find the probability of \(A \cup B\), which includes all outcomes in either event A or B or both, we list all unique events from both sets. Here, both A and B include themselves, so \(A \cup B = E_1, E_2, E_3, E_4, E_5\), covering the entire sample space with a probability of 1. Other compound events include intersections like \(A \cap B\), where only outcomes common to both are considered, such as \(E_1\). Understanding these operations helps us calculate probabilities for more complex scenarios.
Sample Space
The sample space is a fundamental concept in probability, representing the set of all possible outcomes of an experiment. In our example, the sample space consists of \(E_1, E_2, E_3, E_4,\) and \(E_5\), indicating all potential simple events that can result from the experiment. Each event in the sample space is equally likely, with a probability of \(\frac{1}{5}\). It's important to list all possible outcomes in the sample space when determining the probabilities of compound events, as any event or combination of events must be a subset of it. This ensures that the probabilities of all possible events within the sample space add up to 1, reflecting the certainty that one of these outcomes will occur.
Conditional Probability
Conditional probability measures the likelihood of an event occurring, given that another event has already occurred. This is represented by \(P(A|B)\), which reads as the probability of A occurring given B is true. To calculate this, we use the formula \(P(A|B) = \frac{P(A \cap B)}{P(B)}\). In our exercise, for \(P(B|C)\), we find the intersection events, \(B \cap C\), and their probability, followed by dividing this by the probability of C: \(\frac{\frac{1}{5}}{0.4} = 0.5\). This indicates there's a 50% chance B will occur if C has happened. Conditional probability highlights the dependency between events and is crucial for evaluating scenarios where outcomes are not independent.
Complementary Events
A complementary event refers to the occurrence of all outcomes in the sample space that are not part of a certain event. We denote the complement of an event, say A, as \(A^c\). In practical terms, \(A^c\) comprises all simple events except those found in A. For example, if \(A = E_1, E_3\), then \(A^c = E_2, E_4, E_5\). The probabilities of complementary events can be easily determined once the probability of the original event is known, since the sum of probabilities of an event and its complement equals 1: \(P(A) + P(A^c) = 1\). In our case, \(P(A^c) = \frac{3}{5}\) confirms this relationship, filling in the gap left by events not part of A.

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