/*! 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 53 Let \(E\) and \(F\) be independe... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 53

Let \(E\) and \(F\) be independent events; show that \(E\) and \(F^{c}\) are independent.

Short Answer

Expert verified
To show that events E and \(F^c\) are independent, we need to prove that \(P(E \cap F^c) = P(E) * P(F^c)\). First, we express \(P(E \cap F^c)\) as \(P(E) - P(E \cap F)\). Since E and F are independent, we know that \(P(E \cap F) = P(E) * P(F)\). Substituting this expression, we get \(P(E \cap F^c) = P(E) - P(E) * P(F)\). By rewriting the right-hand side of the equation, we obtain \(P(E \cap F^c) = P(E) * (1 - P(F))\), which is the same as \(P(E) * P(F^c)\). Therefore, E and \(F^c\) are independent events.

Step by step solution

01

Express P(E 鈭� F^c) in terms of E and F and their intersection

Let us start by writing down the probability of event E 鈭� F^c: \(P(E \cap F^c) = P(E - (E \cap F))\) Now let's recall the general formula for calculating the probability of the difference of two events: \(P(A - B) = P(A) - P(A \cap B)\) Using this formula, we rewrite the probability of event E 鈭� F^c in terms of E, F, and their intersection: \(P(E \cap F^c) = P(E) - P(E \cap F)\)
02

Use the independence of E and F to simplify the expression

Since E and F are given as independent events, we know that: \(P(E \cap F) = P(E) * P(F)\) Now we can substitute this expression into our original equation: \(P(E \cap F^c) = P(E) - P(E) * P(F)\)
03

Show that P(E 鈭� F^c) = P(E) * P(F^c)

Finally, we rewrite the right-hand side of the equation to express it as the product of P(E) and P(F^c): \(P(E \cap F^c) = P(E) * (1 - P(F))\) Since the complement of an event F is given by \(F^c = 1 - F\), we can replace (1 - P(F)) with P(F^c): \(P(E \cap F^c) = P(E) * P(F^c)\) This completes the proof. Since \(P(E \cap F^c) = P(E) * P(F^c)\), the events E and F^c are independent.

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 of Independent Events
Understanding the probability of independent events is essential when solving problems where two or more events do not influence each other. Independent events are those where the occurrence of one event does not affect the probability of the other event occurring. For example, flipping a coin and rolling a die are independent events because the outcome of the coin toss doesn't change the chances of rolling a particular number on the die.

To calculate the probability of both independent events occurring, you multiply the probability of one event by the probability of the other. The mathematical formula is simply: \[ P(E \cap F) = P(E) \times P(F) \]
Using this concept helps us understand complex probability scenarios by breaking them down into simpler, unrelated events.
Complement of an Event
The complement of an event, denoted as 'event complement,' is another essential concept which represents all outcomes that are not part of the original event. If you have an event 'A', the complement 'A^c' is everything that 'A' is not. For understanding probability, it's crucial to see the whole picture, which includes what could happen outside the desired event.

Mathematically, the probability of an event's complement is: \[ P(A^c) = 1 - P(A) \]
This equation expresses that the probability of not occurring (event complement) and the probability of occurring (event itself) always add up to 1, the certainty of all possible outcomes. This idea is instrumental when considering events and their non-occurrences together.
Intersection of Events
The intersection of events comes into play when you're looking for the combined occurrence of two events. When you see the symbol '\(\cap\)', it refers to the intersection, meaning 'and' in probability. It asks the question: What is the likelihood that both events happen simultaneously?

In practice, to find the probability of the intersection of two events 'E' and 'F', we usually multiply the probability of 'E' by the probability of 'F', but only if they are independent, as discussed earlier. For dependent events (those that affect each other), we must calculate this differently because the outcome of one event affects the outcome of the other.
Probability Formulas
Understanding a variety of probability formulas can arm students with the tools necessary to tackle various probability problems. Two essential formulas that have been discussed are:
  • The multiplication rule for independent events: \( P(E \cap F) = P(E) \times P(F) \)
  • The complement rule: \( P(A^c) = 1 - P(A) \)

There are additional formulas, such as the addition rule for mutually exclusive events, Bayes' theorem for conditional probability, and the formula for total probability, among others. These formulas are foundational to solving probability problems, and recognizing when and how to apply them is a vital skill in probability and statistics.

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

Suppose the probability that it will rain tomorrow is . 3 . a. What are the odds that it will rain tomorrow? b. What are the odds that it will not rain tomorrow?

Five black balls and four white balls are placed in an urn. Two balls are then drawn in succession. What is the probability that the second ball drawn is a white ball if a. The second ball is drawn without replacing the first? b. The first ball is replaced before the second is drawn?

Fifty people are selected at random. What is the probability that none of the people in this group have the same birthday?

Two dice are rolled. Let the random variable \(X\) denote the number that falls uppermost on the first die, and let \(Y\) denote the number that falls uppermost on the second die. a. Find the probability distributions of \(X\) and \(Y\). b. Find the probability distribution of \(X+Y\).

During the first year at a university that uses a 4 -point grading system, a freshman took ten 3 -credit courses and received two As, three Bs, four Cs, and one \(D\). a. Compute this student's grade-point average. b. Let the random variable \(X\) denote the number of points corresponding to a given letter grade. Find the probability distribution of the random variable \(X\) and compute \(E(X)\), the expected value of \(X\).
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Pure 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.