/*! 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 26 Let \(E\) and \(F\) be two event... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(E\) and \(F\) be two events of an experiment with sample space \(S\). Suppose \(P(E)=.6, P(F)=.4\), and \(P(E \cap F)=\) .2. Compute: a. \(P(E \cup F)\) b. \(P\left(E^{c}\right)\) c. \(P\left(F^{c}\right)\) d. \(P\left(E^{c} \cap F\right)\)

Short Answer

Expert verified
a. \(P(E \cup F) = 0.8\) b. \(P\left(E^{c}\right) = 0.4\) c. \(P\left(F^{c}\right) = 0.6\) d. \(P\left(E^{c} \cap F\right) = 0.2\)

Step by step solution

01

Recall the formula for probability of union

The probability of the union of two events E and F is given by: \(P(E \cup F) = P(E) + P(F) - P(E \cap F)\)
02

Plug in the given values and calculate

We are given \(P(E) = 0.6, P(F) = 0.4,\) and \(P(E \cap F) = 0.2.\) Plug these values into the formula: \(P(E \cup F) = 0.6 + 0.4 - 0.2 = 0.8\) b. Compute \(P\left(E^{c}\right)\)
03

Recall the formula for the probability of a complement

The probability of the complement of an event E is given by: \(P\left(E^{c}\right) = 1 - P(E)\)
04

Plug in the given value and calculate

We are given \(P(E) = 0.6.\) Plug this value into the formula: \(P\left(E^{c}\right) = 1 - 0.6 = 0.4\) c. Compute \(P\left(F^{c}\right)\)
05

Recall the formula for the probability of a complement

The probability of the complement of an event F is given by: \(P\left(F^{c}\right) = 1 - P(F)\)
06

Plug in the given value and calculate

We are given \(P(F) = 0.4.\) Plug this value into the formula: \(P\left(F^{c}\right) = 1 - 0.4 = 0.6\) d. Compute \(P\left(E^{c} \cap F\right)\)
07

Recognize the relationship between the given events

We can express \(P\left(E^{c} \cap F\right)\) as \(P(F) - P\left(E \cap F\right),\) because the intersection of \(E^c\) and F corresponds to the probability of F minus the common probability between E and F.
08

Plug in the given values and calculate

We are given \(P(F) = 0.4\) and \(P(E \cap F) = 0.2.\) Plug these values into the relationship: \(P\left(E^{c} \cap F\right) = 0.4 - 0.2 = 0.2\) To summarize, the answer to each part is a. \(P(E \cup F) = 0.8\), b. \(P\left(E^{c}\right) = 0.4\), c. \(P\left(F^{c}\right) = 0.6\), and d. \(P\left(E^{c} \cap F\right) = 0.2\).

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.

Probability of an Event
Understanding the probability of an event is crucial for grasping the basics of probability theory. Probability quantifies the likelihood that a specific event will occur within a set of conditions. It’s calculated by considering the ratio of the favorable outcomes to the total number of possible outcomes.

Let's consider an event, say 'E', within a larger set of possible outcomes — the sample space 'S'. If we know that event 'E' can happen in 'a' different ways, and there are 'b' total possible outcomes in the sample space, we express the probability of the event 'E' as:
\[ P(E) = \frac{a}{b} \]
If the probability of 'E' is close to 1, it means the event is very likely to occur, while a probability close to 0 indicates it is very unlikely to happen. In any case, probabilities will always fall between 0 and 1, inclusive.
Complement of an Event
The complement of an event offers another perspective to understanding the likelihood of outcomes. The complement of an event 'E' is denoted as \(E^c\) and includes all possible outcomes in the sample space that are not part of 'E'.

The probability of the complement of 'E', denoted \(P(E^c)\), tells us the likelihood of everything that doesn't satisfy the conditions of 'E' occurring. It's essentially like flipping the scenario: instead of looking for how likely 'E' is to occur, we're exploring the odds of it not occurring at all.
Given the probability of event 'E', we calculate the complement using the formula:\[ P(E^c) = 1 - P(E) \]
This formula makes intuitive sense because if we add the probabilities of an event and its complement, we should account for all possible outcomes, which sums to 1.
Probability of Intersection
In probability, when we're dealing with two events, we often want to know how likely they are to occur together. This is where the probability of intersection comes into play. It’s denoted as \(P(E \cap F)\) and describes the probability of events 'E' and 'F' both happening.

To find this probability, one must consider only the outcomes that are common to both events. It's an important concept because it allows us to evaluate the connections and interactions between events. For example, if 'E' represents 'having a sandwich' and 'F' means 'having soup', then \(P(E \cap F)\) would correspond to the likelihood of having both a sandwich and soup together.
When events are independent, their intersection probability is the product of their separate probabilities. However, events often affect each other, meaning we need to account for the overlap in their occurrence, as seen in the formula provided earlier:\[ P(E \cup F) = P(E) + P(F) - P(E \cap F) \]
This formula for the probability of the union of two events includes subtracting the intersection to avoid double-counting the cases where both events happen together.

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 accompanying data were obtained from a survey of 1500 Americans who were asked: How safe are American-made consumer products? Determine the empirical probability distribution associated with these data. $$ \begin{array}{lccccc} \hline \text { Rating } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } \\ \hline \text { Respondents } & 285 & 915 & 225 & 30 & 45 \\ \hline \end{array} $$ A: Very safe B: Somewhat safe C: Not too safe D: Not safe at all E: Don't know

According to a study conducted in 2003 concerning the participation, by age, of \(401(\mathrm{k})\) investors, the following data were obtained: $$ \begin{array}{lccccc} \hline \text { Age } & 20 \mathrm{~s} & 30 \mathrm{~s} & 40 \mathrm{~s} & 50 \mathrm{~s} & 60 \mathrm{~s} \\ \hline \text { Percent } & 11 & 28 & 32 & 22 & 7 \\ \hline \end{array} $$ a. What is the probability that a \(401(\mathrm{k})\) investor selected at random is in his or her 20 s or 60 s? b. What is the probability that a \(401(\mathrm{k})\) investor selected at random is under the age of 50 ?

According to Mediamark Research, 84 million out of 179 million adults in the United States correct their vision by using prescription eyeglasses, bifocals, or contact lenses. (Some respondents use more than one type.) What is the probability that an adult selected at random from the adult population uses corrective lenses?

Electricity in the United States is generated from many sources. The following table gives the sources as well as their share in the production of electricity: $$ \begin{array}{lcccccc} \hline \text { Source } & \text { Coal } & \text { Nuclear } & \text { Natural gas } & \text { Hydropower } & \text { Oil } & \text { Other } \\ \hline \text { Share, } \% & 50.0 & 19.3 & 18.7 & 6.7 & 3.0 & 2.3 \\ \hline \end{array} $$ If a source for generating electricity is picked at random, what is the probability that it comes from a. Coal or natural gas? b. Nonnuclear sources?

In a survey conducted in 2007 of 1004 adults 18 yr and older, the following question was asked: How are American companies doing on protecting the environment compared with companies in other countries? The results are summarized below: $$ \begin{array}{lcccc} \hline \text { Answer } & \text { Behind } & \text { Equal } & \text { Ahead } & \text { Don't know } \\ \hline \text { Respondents } & 382 & 281 & 251 & 90 \\ \hline \end{array} $$ If an adult in the survey is selected at random, what is the probability that he or she said that American companies are equal or ahead on protecting the environment compared with companies in other countries?

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.