/*! 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 14 A college job placement center h... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 14

A college job placement center has requests from five students for employment interviews. Three of these students are math majors, and the other two students are statistics majors. Unfortunately, the interviewer has time to talk to only two of the students. These two will be randomly selected from among the five. a. What is the sample space for the chance experiment of selecting two students at random? (Hint: You can think of the students as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E}\). One possible selection of two students is \(\mathrm{A}\) and \(\mathrm{B}\). There are nine other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that both selected students are statistics majors? d. What is the probability that both students are math majors? e. What is the probability that at least one of the students selected is a statistics major? f. What is the probability that the selected students have different majors?

Short Answer

Expert verified
The sample space for the chance experiment of selecting 2 students at random is {AB, AC, AD, AE, BC, BD, BE, CD, CE, DE}. The outcomes are equally likely, with each having a 1/10 chance. The probability of both students being statistics majors is 1/10, both being math majors is 3/10, at least one being a statistics major is 7/10, and the pair having different majors is 3/5.

Step by step solution

01

Sample Space

List all possible unique pairs of students without considering their majors. Label the students as A, B, C, D, and E. There are ten possible pairs: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.
02

a. Sample Space

The sample space for the chance experiment of selecting two students at random is {AB, AC, AD, AE, BC, BD, BE, CD, CE, DE}.
03

b. Equally Likely Outcomes

Yes, the outcomes in the sample space are equally likely. Each pair has a 1/10 chance of being selected.
04

c. Probability of Both Statistics Majors

Let A and B be the statistics majors. We only have 1 pair consisting of both statistics majors, which is AB. As there are 10 possible pairs, the probability of selecting both statistics majors is 1/10.
05

d. Probability of Both Math Majors

Let C, D, and E be the math majors. We have 3 pairs consisting of only math majors: CD, CE, and DE. As there are 10 possible pairs, the probability of selecting both math majors is 3/10.
06

e. Probability of at Least One Statistics Major

To find the probability of selecting at least one statistics major, we can use the complement rule. The complement of selecting at least one statistics major is selecting no statistics majors, which means selecting both math majors. The probability of selecting both math majors is 3/10. Therefore, the probability of selecting at least one statistics major is 1 - 3/10 = 7/10.
07

f. Probability of Different Majors

To find the probability of selecting students with different majors, we can count the number of pairs with different majors. These pairs are AC, AD, AE, BC, BD, and BE. As there are 10 possible pairs, the probability of selecting students with different majors is 6/10 or 3/5.

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.

Sample Space
In probability theory, the concept of a sample space is crucial. It refers to the set of all possible outcomes of a probability experiment. When you conduct a chance experiment, the sample space includes every possible result.
For the exercise given, the task is to choose 2 students out of 5, labeled as A, B, C, D, and E. Here, it shows all potential pairs that can be formed by selecting two students, without focusing on their individual majors.
So the sample space, which represents every unique pairing from the five students, is:
  • {AB, AC, AD, AE, BC, BD, BE, CD, CE, DE}
This list, with its ten pairs, represents all possible outcomes for the selection process. Understanding the sample space is the first step in analyzing any probability problem.
Equally Likely Outcomes
The idea of equally likely outcomes means that every outcome in a sample space has the same probability of occurring. In many problems, especially those involving fair selections or random draws, each outcome often carries the same weight.
For our current exercise, selecting pairs of students is done randomly, indicating that any pair is just as likely to be chosen as any other. This is confirmed by noting that each pair has a probability of \[\frac{1}{10}\]to be selected, which reflects equal likelihood since there are ten pairs total.
Understanding that outcomes are equally likely simplifies calculations and allows you to apply fundamental probability rules accurately.
Complement Rule
The complement rule is a valuable tool in probability theory. It allows you to find the probability of an event by considering the probability of its complement, or the event not happening. If you know the probability of an event, then its complement has a probability of one minus that event's probability.
In our exercise, to find the probability of selecting at least one statistics major, we use this rule. First, calculate the probability of the opposite event, which is selecting only math majors, and then subtract this probability from 1.
Given there are 3 pairs of math majors and 10 total pairs, the probability of choosing both math majors is \[\frac{3}{10}\]. Thus, the probability of selecting at least one statistics major is \[1 - \frac{3}{10} = \frac{7}{10}\].
This method is intuitive and often more straightforward than directly calculating the desired probability, especially when dealing with the "at least" type of things.

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

A deck of 52 cards is mixed well, and 5 cards are dealt. a. It can be shown that (disregarding the order in which the cards are dealt) there are 2,598,960 possible hands, of which only 1287 are hands consisting entirely of spades. What is the probability that a hand will consist entirely of spades? What is the probability that a hand will consist entirely of a single suit? b. It can be shown that exactly 63,206 of the possible hands contain only spades and clubs, with both suits represented. What is the probability that a hand consists entirely of spades and clubs with both suits represented?

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.64\) and \(P(F)=0.17\) i. What is the value of \(P(E \cap F)\) ? ii. What is the value of \(P(E \cup F)\) ? b. Suppose that \(A\) and \(B\) are events with \(P(A)=0.3, P(B)=0.5\), and \(P(A \cap B)=0.15 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell? c. Suppose that \(A\) and \(B\) are events with \(P(A)=0.65\) and \(P(B)=0.57 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

Lyme disease is transmitted by infected ticks. Several tests are available for people with symptoms of Lyme disease. One of these tests is the EIA/IFA test. The paper "Lyme Disease Testing by Large Commercial Laboratories in the United States" (Clinical Infectious Disease [2014]: \(676-681\) ) found that \(11.4 \%\) of those tested actually had Lyme disease. Consider the following events: \(+\) represents a positive result on the blood test \- represents a negative result on the blood test \(L\) represents the event that the patient actually has Lyme disease \(L^{C}\) represents the event that the patient actually does not have Lyme disease The following probabilities are based on percentages given in the paper: $$ \begin{array}{r} P(L)=0.114 \\ P\left(L^{C}\right)=0.886 \end{array} $$ $$ \begin{array}{c} P(+\mid L)=0.933 \\ P(-\mid L)=0.067 \\ P\left(+\mid L^{C}\right)=0.039 \\ P\left(-\mid L^{C}\right)=0.961 \end{array} $$ a. For each of the given probabilities, write a sentence giving an interpretation of the probability in the context of this problem. b. Use the given probabilities to construct a hypothetical 1000 table with columns corresponding to whether or not a person has Lyme disease and rows corresponding to whether the blood test is positive or negative. c. Notice the form of the known conditional probabilities; for example, \(P(+\mid L)\) is the probability of a positive test given that a person selected at random from the population actually has Lyme disease. Of more interest is the probability that a person has Lyme disease, given that the test result is positive. Use information from the table constructed in Part (b) to calculate this probability.

In an article that appears on the website of the American Statistical Association, Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in 100 . And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given say 99 in 100 -are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty \(T C=\) event that the test result is clean \(D=\) event that the person tested is actually dirty \(C=\) event that the person tested is actually clean a. Using the information in the quote, what are the values of $$ \text { i. } P(T D \mid D) $$ iii. \(P(C)\) $$ \text { ii. } P(T D \mid C) $$ iv. \(P(D)\) b. Use the probabilities from Part (a) to construct a hypothetical 1000 table. c. What is the value of \(P(T D)\) ? d. Use the information in the table to calculate the probability that a person is clean given that the test result is dirty, \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

The article "A Crash Course in Probability" from The Economist included the following information: The chance of being involved in an airplane crash when flying on an Airbus 330 from London to New York City on Virgin Atlantic Airlines is 1 in \(5,371,369 .\) This was interpreted as meaning that you "would expect to go down if you took this flight every day for 14,716 years." The article also states that a person could "expect to fly on the route for 14,716 years before plummeting into the Atlantic." Comment on why these statements are misleading.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Mechanics 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.