/*! 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 22 Suppose that we have a sample sp... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that we have a sample space with five equally likely experimental outcomes: \(E_{1}\). \\[ \begin{aligned} E_{2}, E_{3}, E_{4}, E_{5} . \text { Let } \\ \qquad \begin{aligned} A &=\left\\{E_{1}, E_{2}\right\\} \\ B &=\left\\{E_{3}, E_{4}\right\\} \\ C &=\left\\{E_{2}, E_{3}, E_{5}\right\\} \end{aligned} \end{aligned} \\] a. \(\quad\) Find \(P(A), P(B),\) and \(P(C)\). b. Find \(P(A \cup B)\). Are \(A\) and \(B\) mutually exclusive? c. \(\quad\) Find \(A^{c}, C^{c}, P\left(A^{c}\right),\) and \(P\left(C^{c}\right)\). d. Find \(A \cup B^{c}\) and \(P\left(A \cup B^{c}\right)\). e. Find \(P(B \cup C)\).

Short Answer

Expert verified
P(A) = 2/5, P(B) = 2/5, P(C) = 3/5; A and B are mutually exclusive, so P(A ∪ B) = 4/5; P(A^c) = 3/5, P(C^c) = 2/5; P(A ∪ B^c) = 3/5; P(B ∪ C) = 4/5.

Step by step solution

01

Find Probability of Event A

Since the sample space has five equally likely outcomes, each outcome has a probability of \( \frac{1}{5} \). Event \( A \) consists of \( \{E_1, E_2\} \), so the probability \( P(A) \) is calculated as:\[P(A) = P(E_1) + P(E_2) = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}\]
02

Find Probability of Event B

Event \( B \) consists of \( \{E_3, E_4\} \). Similarly, the probability \( P(B) \) is:\[P(B) = P(E_3) + P(E_4) = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}\]
03

Find Probability of Event C

Event \( C \) includes \( \{E_2, E_3, E_5\} \). Thus, \( P(C) \) is calculated as:\[P(C) = P(E_2) + P(E_3) + P(E_5) = \frac{1}{5} + \frac{1}{5} + \frac{1}{5} = \frac{3}{5}\]
04

Determine Mutual Exclusivity of A and B

Events \( A \) and \( B \) are mutually exclusive if they have no outcomes in common. Since \( A = \{E_1, E_2\} \) and \( B = \{E_3, E_4\} \) share no common outcomes, they are mutually exclusive.
05

Find Probability of A Union B

Since \( A \) and \( B \) are mutually exclusive, \( P(A \cup B) = P(A) + P(B) \):\[P(A \cup B) = \frac{2}{5} + \frac{2}{5} = \frac{4}{5}\]
06

Find Complement of A and Calculate P(A^c)

The complement of \( A \), denoted \( A^c \), contains all outcomes not in \( A \): \( A^c = \{E_3, E_4, E_5\} \). So, \( P(A^c) \) is:\[P(A^c) = 1 - P(A) = 1 - \frac{2}{5} = \frac{3}{5}\]
07

Find Complement of C and Calculate P(C^c)

The complement of \( C \), denoted \( C^c \), has outcomes not in \( C \): \( C^c = \{E_1, E_4\} \). Hence, \( P(C^c) \) is:\[P(C^c) = 1 - P(C) = 1 - \frac{3}{5} = \frac{2}{5}\]
08

Find A Union B Complement and Its Probability

First, find \( B^c \), the complement of \( B \): \( B^c = \{E_1, E_2, E_5\} \). Then, \( A \cup B^c \) includes \( \{E_1, E_2, E_5\} \). Thus,\[P(A \cup B^c) = \frac{3}{5}\]
09

Find Probability of B Union C

\( B \cup C = \{E_2, E_3, E_4, E_5\} \) represents outcomes in either \( B \) or \( C \), or both. Therefore,\[P(B \cup C) = \frac{4}{5}\]

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Key Concepts

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

Mutually Exclusive Events
In probability theory, mutually exclusive events refer to events that cannot happen simultaneously. This means if one event occurs, the other cannot. For example, if you toss a six-sided die, the result can be either a 3 or a 4, but not both at the same time. In the context of our exercise, events \( A \) and \( B \) are defined as \( A = \{E_1, E_2 \} \) and \( B = \{E_3, E_4 \} \). Since there are no common outcomes in \( A \) and \( B \), they are called mutually exclusive.

🌟 **Key Takeaways:**
* Mutually exclusive events have no overlap in their outcomes.
* The probability of both events occurring simultaneously is zero.
* For such events, the probability of either event occurring is the sum of their individual probabilities: \( P(A \cup B) = P(A) + P(B) \).
Sample Space
The sample space in probability theory is the set of all possible outcomes of a random experiment. Think of it as the complete "inventory" of potential results when an experiment is conducted. For the given example with five equally likely outcomes \( E_1, E_2, E_3, E_4, \) and \( E_5 \), the sample space \( S \) is \( \{E_1, E_2, E_3, E_4, E_5 \} \). Each outcome in the sample space is equally likely, with a probability calculated as \( \frac{1}{5} \), since they are equally likely.

🧠 **Understanding Sample Space:**
* A comprehensive list of all possible individual outcomes of an experiment.
* Fundamental for calculating probabilities of events.
* Helps in identifying complements and unions of events.
Complementary Events
Complementary events are pairs of events where one event's occurrence means the other cannot occur, and vice versa. For any event \( A \) within a given sample space, the complement of \( A \) (denoted as \( A^c \)) includes all outcomes that are not part of \( A \).
In our exercise, the event \( A \) consists of \( \{E_1, E_2\} \), so its complement \( A^c \) is \( \{E_3, E_4, E_5\} \), as it includes all outcomes not in \( A \). Furthermore, the probability of the complement, \( P(A^c) \), is found by subtracting the probability of \( A \) from 1, resulting in \( P(A^c) = 1 - P(A) \).

🎯 **Essential Points About Complementary Events:**
* The sum of the probabilities of an event and its complement is always 1.
* Complementary events cover all possible scenarios together in a sample space.
* Finding the probability of an event's complement can often be simpler than finding the event's probability directly.

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Most popular questions from this chapter

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