- 126 Views
- Uploaded on
- Presentation posted in: General

Chapter 3

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Chapter 3

3-1

Probability

3-2

- 3-1 Introduction
- 3-2 Fundamentals
- 3-3 Addition Rules for Probability
- 3-4 Multiplication Rules: Basics
- 3-5 Multiplication Rules: Beyond the Basics

3-3

- Determine Sample Spaces and find the probability of an event using classical probability.
- Find the probability of an event using empirical probability.
- Find the probability of compound events using the addition rules.

3-4

- Find the probability of compound events using the multiplication rules.
- Find the conditional probability of an event.

3-5

- Aprobability experimentis a process that leads to well-defined results called outcomes.
- Anoutcomeis the result of a single trial of a probability experiment.
- NOTE: A tree diagram can be used as a systematic way to find all possible outcomes of a probability experiment.

H

H

T

Second Toss

H

T

T

3-6

First Toss

3-7

3-8

- Classical probability assumes that all outcomes in the sample space are equally likely to occur.
- That is, equally likely events are events that have the same probability of occurring.

3-9

3-10

- For a card drawn from an ordinary deck, find the probability of getting (a) a queen (b) a 6 of clubs (c) a 3 or a diamond.
- Solution:(a) Since there are 4 queens and 52 cards,P(queen) = 4/52 = 1/13.
- (b) Since there is only one 6 of clubs, thenP(6 of clubs) = 1/52.

3-11

- (c) There are four 3s and 13 diamonds, but the 3 of diamonds is counted twice in the listing. Hence there are only 16 possibilities of drawing a 3 or a diamond, thusP(3 or diamond) = 16/52 = 4/13.

3-12

- When a single die is rolled, find the probability of getting a 9.
- Solution:Since the sample space is 1, 2, 3, 4, 5, and 6, it is impossible to get a 9. Hence,P(9) = 0/6 = 0.
- NOTE:The sum of the probabilities of all outcomes in a sample space is one.

3-13

E

3-14

- Find the complement of each event.
- Rolling a die and getting a 4.
- Solution:Getting a 1, 2, 3, 5, or 6.
- Selecting a letter of the alphabet and getting a vowel.
- Solution:Getting a consonant (assume y is a consonant).

3-15

- Selecting a day of the week and getting a weekday.
- Solution:Getting Saturday or Sunday.
- Selecting a one-child family and getting a boy.
- Solution:Getting a girl.

3-16

P

(

E

)

1

P

(

E

)

or

P

(

E

)

=

1

P

(

E

)

or

P

(

E

)

+

P

(

E

)

=

1

.

3-17

- The difference between classical and empirical probability is that classical probability assumes that certain outcomes are equally likely while empirical probability relies on actual experience to determine the probability of an outcome.

3-18

3-19

- In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had AB blood. Set up a frequency distribution.

3-20

Type

Frequency

A

B

AB

O

22

5

2

21

50 = n

3-21

- Find the following probabilities for the previous example.
- A person has type O blood.
- Solution:P(O) = f/n = 21/50.
- A person has type A or type B blood.
- Solution:P(A or B) = 22/50+ 5/50 = 27/50.

3-22

- Two events aremutually exclusiveif they cannot occur at the same time (i.e., they have no outcomes in common).

3-23

A and B are mutually exclusive

A

B

=

+

P

(

A

or

B

)

P

(

A

)

P

(

B

)

3-24

When two events A and B are mutually exclusive, the probabilitythat A or B will occur is

3-25

- At a political rally, there are 20 Liberals (L), 13 Conservatives (C), and 6 NDPs (N). If a person is selected, find the probability that he or she is either a Conservative or an NDP.
- Solution:P(C or N) = P(C) + P(N) = 13/39 + 6/39 = 19/39.

3-26

- A day of the week is selected at random. Find the probability that it is a weekend.
- Solution:P(Saturday or Sunday) = P(Saturday) + P(Sunday) = 1/7 + 1/7 = 2/7.

3-27

When two events A and B

are not mutually exclusive, the

probability

y

that

A

or

B

will

occur

is

P

(

A

or

B

)

P

(

A

)

P

(

B

)

P

(

A

and

B

)

3-28

A and B

(common portion)

A

B

3-29

- In a hospital unit there are eight nurses and five physicians. Seven nurses and three physicians are females. If a staff person is selected, find the probability that the subject is a nurse or a male.
- The next slide has the data.

3-30

3-31

- Solution:P(nurse or male) = P(nurse) + P(male) – P(male nurse) = 8/13 + 3/13 – 1/13 = 10/13.

3-32

- On New Year’s Eve, the probability that a person driving while intoxicated is 0.32, the probability of a person having a driving accident is 0.09, and the probability of a person having a driving accident while intoxicated is 0.06. What is the probability of a person driving while intoxicated or having a driving accident?

3-33

- Solution:P(intoxicated or accident) = P(intoxicated) + P(accident) – P(intoxicated and accident) = 0.32 + 0.09 – 0.06 = 0.35.

3-34

- Two events A and B areindependentif the fact that A occurs does not affect the probability of B occurring.
- Example:Rolling a die and getting a 6, and then rolling another die and getting a 3 are independent events.

3-35

3-36

- A card is drawn from a deck and replaced; then a second card is drawn. Find the probability of getting a queen and then an ace.
- Solution:Because these two events are independent (why?), P(queen and ace) = (4/52)(4/52) = 16/2704 = 1/169.

3-37

- A Decima pole found that 46% of Canadians say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer stress at least once a week.
- Solution:Let S denote stress. Then P(S and S and S) = (0.46)3 = 0.097.

3-38

- The probability that a specific medical test will show positive is 0.32. If four people are tested, find the probability that all four will show positive.
- Solution:Let T denote a positive test result. Then P(T and T and T and T) = (0.32)4 = 0.010.

3-39

- When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed, the events are said to be dependent.
- Example:Having high grades and getting a scholarship are dependent events.

3-40

- Theconditional probabilityof an event B in relationship to an event A is the probability that an event B occurs after event A has already occurred.
- The notation for the conditional probability of B given A is P(B|A).
- NOTE:This does not mean BA.

3-41

3-42

- In a shipment of 25 microwave ovens, two are defective. If two ovens are randomly selected and tested, find the probability that both are defective if the first one is not replaced after it has been tested.
- Solution:See next slide.

3-43

- Solution:Since the events are dependent, P(D1 and D2) = P(D1)P(D2| D1) = (2/25)(1/24) = 2/600 = 1/300.

3-44

- The KW Insurance Company found that 53% of the residents of a city had homeowner’s insurance with its company. Of these clients, 27% also had automobile insurance with the company. If a resident is selected at random, find the probability that the resident has both homeowner’s and automobile insurance.

3-45

- Solution:Since the events are dependent, P(H and A) = P(H)P(A|H) = (0.53)(0.27) = 0.1431.

3-46

- Box 1 contains two red balls and one blue ball. Box 2 contains three blue balls and one red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the probability of selecting a red ball.

P(R|B1) 2/3

Red

(1/2)(2/3)

P(B1) 1/2

Box 1

Blue

(1/2)(1/3)

P(B|B1) 1/3

P(R|B2) 1/4

Box 2

Red

(1/2)(1/4)

P(B2) 1/2

Blue

(1/2)(3/4)

P(B|B2) 3/4

3-47

3-48

- Solution:P(red) = (1/2)(2/3) + (1/2)(1/4) = 2/6 + 1/8 = 8/24 + 3/24 = 11/24.

3-49

3-50

- The probability that Sam parks in a no-parking zone and gets a parking ticket is 0.06, and the probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.2. On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a ticket.

3-51

- Solution:Let N = parking in a no-parking zone and T = getting a ticket.
- Then P(T|N) = [P(N and T) ]/P(N) = 0.06/0.2 = 0.30.

3-52

- A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results are shown in the table on the next slide.

3-53

3-54

- Find the probability that the respondent answered “yes” given that the respondent was a female.
- Solution:Let M = respondent was a male; F = respondent was a female; Y = respondent answered “yes”; N = respondent answered “no”.

3-55

- P(Y|F) = [P( F and Y) ]/P(F) = [8/100]/[50/100] = 4/25.
- Find the probability that the respondent was a male, given that the respondent answered “no”.
- Solution: P(M|N) = [P(N and M)]/P(N) = [18/100]/[60/100] = 3/10.