/*! 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} Q1E Four universities鈥�1, 2, 3, and... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q1E (page 56)

Four universities鈥�1, 2, 3, and 4鈥攁re participating in a holiday basketball tournament. In the first round, 1 will play 2 and 3 will play 4. Then the two winners will play for the championship, and the two losers will also play. One possible outcome can be denoted by 1324 (1 beats 2 and 3 beats 4 in first-round games, and then 1 beats 3 and 2 beats 4).

a. List all outcomes in S.

b. Let A denote the event that 1 wins the tournament. List outcomes in A.

c. Let Bdenote the event that 2 gets into the championship game. List outcomes in B.

d. What are the outcomes in A\( \cup \)B and in A\( \cap \)B? What are the outcomes in A鈥�?

Short Answer

Expert verified

a. The outcomes are:

\(S = \left\{ {1324,3124,1342,3142,1423,1432,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

b. The possible outcomes of A are:

\(A = \left\{ {1324,1342,1423,1432} \right\}\)

c. The possible outcomes of B are:

\(B = \left\{ {2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

d. The possible outcomes are,

\(A \cup B = \left\{ {1324,1342,1423,1432,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

\(A \cap B = \phi \)

\(A' = \left\{ {3124,3142,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

Step by step solution

01

Given information

The number of universities is 4; they are 1, 2, 3, and 4.

In the first-round, 1 will play 2 and 3 will play 4. Then two winners and two losers will play respectively.

The possible outcome 1324 implies that 1 beat 2 and 3 beat 4 in the first round and then 1 beat 3 and 2 beat 4.

02

Providing the elements of a sample space

a.

The sample space for the provided scenario is as follows,

\(S = \left\{ {1324,3124,1342,3142,1423,1432,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

03

Providing the possible outcomes of A

b.

A represents the event that 1 wins the tournament.

Referring to the sample space of part a,

\(S = \left\{ {1324,3124,1342,3142,1423,1432,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

The outcomes in A will be the ones where 1 is at the beginning.

The possible outcomes of A is represented as,

\(A = \left\{ {1324,1342,1423,1432} \right\}\)

04

Providing the possible outcomes of B

c.

B represents the event that 2 gets into the championship game.

Referring to the sample space of part a,

\(S = \left\{ {1324,3124,1342,3142,1423,1432,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

The outcomes in B will be the ones where 2 is at any of the first two places.

The possible outcomes of B is represented as,

\(B = \left\{ {2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

05

List of the possible outcomes of union and intersection of events.

Referring to part b,

\(A = \left\{ {1324,1342,1423,1432} \right\}\)

Referring to part c,

\(B = \left\{ {2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

A union of two events A and B consists of all the outcomes that are either in A or B or in both events.

The possible outcomes in\(A \cup B\)is given as,

\(A \cup B = \left\{ {1324,1342,1423,1432,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

The intersection of two events A and B consists of all the outcomes that are present in both events.

The possible outcomes in\(A \cap B\)is given as,

\(A \cap B = \phi \)

This implies that are no possible outcomes in\(A \cap B\).

The complementary of an event A consists of all the outcomes that are not contained in A.

The possible outcomes in\(A'\)are given by,

\(A' = \left\{ {3124,3142,4123,4132,2314,2341,3214,3241,2413,2431,4213,4231} \right\}\)

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!

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 composer Beethoven wrote \({\rm{9}}\) symphonies, \({\rm{9}}\) piano concertos (music for piano and orchestra), and \({\rm{32}}\) piano sonatas (music for solo piano).

a. How many ways are there to play first a Beethoven symphony and then a Beethoven piano concerto?

b. The manager of a radio station decides that on each successive evening (\({\rm{7}}\) days per week), a Beethoven symphony will be played followed by a Beethoven piano concerto followed by a Beethoven piano sonata. For how many years could this policy be continued before exactly the same program would have to be repeated?

A certain system can experience three different types of defects. Let \({{\rm{A}}_{\rm{1}}}\)\({\rm{(i = 1,2,3)}}\)denote the event that the system has a defect of type Suppose that

\(\begin{aligned}P\left( {{A_1}} \right) &= .12\;\;\;P\left( {{A_2}} \right) = .07\;\;\;P\left( {{A_3}}\right) = .05 \hfill \\P\left( {{A_1}E {A_2}} \right) = .13\;\;\;P\left( {{A_1}E {A_3}}\right) &= .14 \hfill \\P\left( {{A_2}E {A_2}} \right) = .10\;\;\;P\left( {{A_1}{C}{A_3}{C}{A_7}} \right) &= .01 \hfill \\\end{aligned} \)

a. What is the probability that the system does not have a type 1 defect?

b. What is the probability that the system has both type 1 and type 2 defects?

c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect?

d. What is the probability that the system has at most two of these defects?

A computer consulting firm presently has bids out on three projects. Let \({A_i}\) = {awarded project i}, for i=1, 2, 3, and suppose that\(P\left( {{A_1}} \right) = .22,\;P\left( {{A_2}} \right) = .25,\;P\left( {{A_3}} \right) = .28,\;P\left( {{A_1} \cap {A_2}} \right) = .11,\;P\left( {{A_1} \cap {A_3}} \right) = .05,\)\(P\left( {{A_2} \cap {A_3}} \right) = .07\),\(P\left( {{A_1} \cap {A_2} \cap {A_3}} \right) = .01\). Express in words each of the following events, and compute the probability of each event:

a. {\({A_1} \cup {A_2}\)}

b. \(A_1' \cap A_2'\)(Hint: \(\left( {{A_1} \cup {A_2}} \right)' = A_1' \cap A_2'\) )

c.\({A_1} \cup {A_2} \cup {A_3}\)

d. \(A_1' \cap A_2' \cap A_3'\)

e. \(A_1' \cap A_2' \cap {A_3}\)

f. \(\left( {A_1' \cap A_2'} \right) \cup {A_3}\)

The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered1, 2,鈥�, 6, then one outcome consists of computers 1 and2, another consists of computers 1 and 3, and so on).

a. What is the probability that both selected setups are for laptop computers?

b. What is the probability that both selected setups are desktop machines?

c. What is the probability that at least one selected setup is for a desktop computer?

d. What is the probability that at least one computer of each type is chosen for setup?

For customers purchasing a refrigerator at a certain appliance store, let A be the event that the refrigerator was manufactured in the U.S., B be the event that the refrigerator had an icemaker, and C be the event that the customer purchased an extended warranty. Relevant probabilities are

\(\begin{aligned}{\rm{P(A) = }}{\rm{.75}}\;\;\;{\rm{P(B}}\mid {\rm{A) = }}{\rm{.9}}\;\;\;{\rm{P}}\left( {{\rm{B}}\mid {{\rm{A}}^{\rm{\cent}}}} \right){\rm{ = }}{\rm{.8}}\\{\rm{P(C}}\mid {\rm{A\c{C}B) = }}{\rm{.8}}\;\;\;{\rm{P}}\left( {{\rm{C}}\mid {\rm{A\c{C}}}{{\rm{B}}^{\rm{\cent}}}} \right){\rm{ = }}{\rm{.6}}\\{\rm{P}}\left( {{\rm{C}}\mid {{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}B}}} \right){\rm{ = }}{\rm{.7}}\;\;\;{\rm{P}}\left( {{\rm{C}}\mid {{\rm{A}}^{\rm{\cent}}}{\rm{\c{C}}}{{\rm{B}}^{\rm{\cent}}}} \right){\rm{ = }}{\rm{.3}}\end{aligned}\)

Construct a tree diagram consisting of first-, second-, and third-generation branches, and place an event label and appropriate probability next to each branch. b. Compute \({\rm{P(A\c{C}B\c{C}C)}}\). c. Compute \({\rm{P(B\c{C}C)}}\). d. Compute \({\rm{P(C)}}\). e. Compute \({\rm{P(A}}\mid {\rm{B\c{C}C)}}\), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

Read Explanation

Geometry

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.