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Chapter 3: Problem 100

A college finds that 10% of students have taken a distance learning class and that 40% of students are part time students. Of the part time students, 20% have taken a distance learning class. Let D = event that a student takes a distance learning class and E = event that a student is a part time student a. Find P(D AND E). b. Find P(E|D). c. Find P(D OR E). d. Using an appropriate test, show whether D and E are independent. e. Using an appropriate test, show whether D and E are mutually exclusive.

Short Answer

Expert verified
a. 0.08 b. 0.80 c. 0.42 d. Not independent e. Not mutually exclusive.

Step by step solution

01

Understanding the Problem

Let's define the events more clearly:- Event \(D\): A student takes a distance learning class.- Event \(E\): A student is a part-time student.Given probabilities:- \(P(D) = 0.10\)- \(P(E) = 0.40\)- \(P(D|E) = 0.20\), which represents the probability of taking a distance learning class given that the student is part-time.
02

Calculating \(P(D \cap E)\)

To find \(P(D \cap E)\), use the conditional probability formula: \[ P(D \cap E) = P(D|E) \times P(E) \]Substituting the given probabilities:\[ P(D \cap E) = 0.20 \times 0.40 = 0.08 \]
03

Calculating \(P(E|D)\)

To find \(P(E|D)\), apply Bayes' theorem:\[ P(E|D) = \frac{P(D \cap E)}{P(D)} \]Substituting known values:\[ P(E|D) = \frac{0.08}{0.10} = 0.80 \]
04

Calculating \(P(D \cup E)\)

Use the formula for the union of two events:\[ P(D \cup E) = P(D) + P(E) - P(D \cap E) \]Substitute the known values:\[ P(D \cup E) = 0.10 + 0.40 - 0.08 = 0.42 \]
05

Testing for Independence

The events \(D\) and \(E\) are independent if \(P(D \cap E) = P(D) \times P(E)\). Calculate this:\[ P(D) \times P(E) = 0.10 \times 0.40 = 0.04 \]Since \(0.04 eq 0.08\), the events are not independent.
06

Testing for Mutual Exclusivity

The events \(D\) and \(E\) are mutually exclusive if \(P(D \cap E) = 0\). Since \(P(D \cap E) = 0.08\), the events are not mutually exclusive.

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Key Concepts

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

Independence of Events
In probability theory, two events are deemed independent when the occurrence of one event does not affect the probability of the other event occurring. This is a significant concept because it helps us understand the relationships between different events.
In the given exercise, we assess whether two events, namely, a student taking a distance learning class (Event D) and being a part-time student (Event E), are independent. To determine this, we can use the formula for independence:
  • The events D and E are independent if and only if \(P(D \cap E) = P(D) \times P(E)\).
We calculated \(P(D) \times P(E) = 0.10 \times 0.40 = 0.04\). However, \(P(D \cap E)\) was found to be 0.08, which is not equal to 0.04. Hence, D and E are not independent.
This implies that knowing whether a student is part-time does influence the likelihood of them taking a distance learning class, indicating some form of relationship between the two events.
Mutually Exclusive Events
Mutually exclusive events are a pair of events that cannot occur at the same time. For example, if Event A occurs, Event B cannot happen, and vice versa.
In our context, we need to determine if the events of a student taking a distance learning class (Event D) and being a part-time student (Event E) are mutually exclusive. The condition for mutual exclusivity is:
  • Events D and E are mutually exclusive if \(P(D \cap E) = 0\).
From the calculations provided, \(P(D \cap E) = 0.08\). This indicates that there is a non-zero probability of a student being both part-time and taking a distance learning class. Therefore, the events are not mutually exclusive.
Understanding that these events are not mutually exclusive helps us realize that a student can simultaneously be part-time and enrolled in a distance learning class, making the analysis of such events more complex.
Probability Calculations
Probability calculations enable us to quantify uncertainties and determine the likelihood of different events occurring.
For this exercise, several key probabilities are calculated, such as:
  • Joint Probability \(P(D \cap E)\): This represents the probability of both events occurring. Calculated as \(P(D \cap E) = P(D|E) \times P(E) = 0.08\).
  • Conditional Probability \(P(E|D)\): This is the probability of a student being part-time given that they are taking a distance learning class. Derived using Bayes' theorem: \(P(E|D) = \frac{P(D \cap E)}{P(D)} = 0.80\).
  • Union of Events \(P(D \cup E)\): This tells us the probability of at least one of the events occurring, calculated as \(P(D) + P(E) - P(D \cap E) = 0.42\).
These calculated probabilities help unravel the interconnectedness of the student characteristics in the scenario, offering a clearer understanding of how these events occur in the population.

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Most popular questions from this chapter

In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts a through j. (Note that you cannot find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.) 鈥� Let F be the event that a student is female. 鈥� Let M be the event that a student is male. 鈥� Let S be the event that a student has short hair. 鈥� Let L be the event that a student has long hair. a. The probability that a student does not have long hair. b. The probability that a student is male or has short hair. c. The probability that a student is a female and has long hair. d. The probability that a student is male, given that the student has long hair. e. The probability that a student has long hair, given that the student is male. f. Of all the female students, the probability that a student has short hair. g. Of all students with long hair, the probability that a student is female. h. The probability that a student is female or has long hair. i. The probability that a randomly selected student is a male student with short hair. j. The probability that a student is female.

Use the following information to answer the next six exercises. There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 in Oceania (Pacific Ocean region). Let A = the event that a country is in Asia. Let E = the event that a country is in Europe. Let F = the event that a country is in Africa. Let N = the event that a country is in North America. Let O = the event that a country is in Oceania. Let S = the event that a country is in South America. Find P(N).

A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. Randomly select one book Let F = event that book is fiction Let N = event that book is nonfiction What is the sample space?

What is the probability of drawing a red card in a standard deck of 52 cards?

The following table of data obtained from www.baseball-almanac.com shows hit information for four players. Suppose that one hit from the table is randomly selected. $$\begin{array}{|l|l|l|l|l|}\hline \text { Name } & {\text { single }} & {\text { Double }} & {\text { Triple }} & {\text { Home Run }} & {\text { Total Hits }} \\ \hline \text { Babe Ruth } & {1,517} & {506} & {136} & {714} & {2,873} \\ \hline \text { Jackie Robinson } & {1,054} & {273} & {54} & {137} & {1,518} \\ \hline \text { Ty Cobb } & {3,603} & {174} & {295} & {114} & {4,189} \\ \hline \text { Hank Aaron } & {2,294} & {624} & {98} & {755} & {3,771} \\ \hline\end{array}$$ Are "the hit being made by Hank Aaron" and "the hit being a double" independent events? a. Yes, because P(hit by Hank Aaron|hit is a double) = P(hit by Hank Aaron) b. No, because P(hit by Hank Aaron|hit is a double) ? P(hit is a double) c. No, because P(hit is by Hank Aaron|hit is a double) ? P(hit by Hank Aaron) d. Yes, because P(hit is by Hank Aaron|hit is a double) = P(hit is a double)
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