chapter 4 260110 044531

Chapter 4Chapter 4

Probability & Probability & Counting Rules Counting Rules Probability & Probability &

Counting Rules Counting Rules

Reference: Allan G. Bluman (2004) Elementary Statistics: A Step-by Step Approach. New York : McGraw Hill

ObjectivesObjectives Determine Sample Spaces and find the probability

of an event using classical probability or empirical probability.

Find the probability of compound events using the addition rules and the multiplication rules.

Find the conditional probability of an event. Find the total number of outcomes in a sequence of

events, using the fundamental counting rule. Find the number of ways rr objects can be selected

from nn objects using the permutation rule. Find the number of ways rr objects can be selected

from nn objects without regard to order using the combination rule.

Find the probability of an event, using the counting rule.

Sample Spaces and ProbabilitySample Spaces and Probability

A probability experimentprobability experiment is a process that leads to well-defined results called outcomes.

An outcomeoutcome is the result of a single trial of a probability experiment.

An eventevent consists of a set of outcomes of a probability experiment.

NOTE:NOTE: A tree diagram can be used as a systematic way to find all possible outcomes of a probability experiment.

Tree Diagram for Tossing Two CoinsTree Diagram for Tossing Two Coins

First Toss

H

T

H

T H

T

Second Toss

Sample Spaces -Sample Spaces - Examples

EXPERIMENT SAMPLE SPACE

Toss one coin H, T

Roll a dice 1, 2, 3, 4, 5, 6

Answer a true-false question

True, False

Toss two coins HH, HT, TH, TT

Formula for Classical ProbabilityFormula for Classical Probability

Classical probability assumes that all outcomes in the sample space are equally likely to occur.

That is, equally likelyequally likely events are events that have the same probability of occurring.

Formula for Classical ProbabilityFormula for Classical Probability

( ) = ( )

( )

,

.

The probability of any event E is

number of outcomes in E

total number of outcomes in the sample space

This probability is denoted by

P En E

n S

This probability is called classical probability

and it uses the sample space S

.

.

Classical Probability -Classical Probability - Examples

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:Solution:

(a) Since there are 4 queens and 52 cards, P(queen) = 4/52 = 1/13P(queen) = 4/52 = 1/13.

(b) Since there is only one 6 of clubs, then P(6 of P(6 of clubs) = 1/52clubs) = 1/52.


(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, thus P(3 or diamond) = 16/52 = 4/13P(3 or diamond) = 16/52 = 4/13.


When a single dice is rolled, find the probability of getting a 9.

Solution: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 = 0P(9) = 0/6 = 0. NOTE:NOTE: The sum of the probabilities of all

outcomes in a sample space is one.

Complement of an EventComplement of an Event

T h e c o m p le m e n t o f a n e v e n t E is th e se t o f o u tc o m e s in th e

sa m p le sp a c e th a t a re n o t in c lu d e d in th e o u tc o m e s

o f e v e n t E T h e c o m p le m e n t o f E is d e n o te d b y E E b a r

. ( ) .

E

E

Complement of an Event - Complement of an Event - Example

Find the complement of each event. Rolling a dice and getting a 4. Solution:Solution: Getting a 1, 2, 3, 5, or 6. Selecting a letter of the alphabet and getting

a vowel. Solution:Solution: Getting a consonant.

Complement of an Event -Complement of an Event - Example

Selecting a day of the week and getting a weekday.

Solution:Solution: Getting Saturday or Sunday. Selecting a one-child family and getting a

boy. Solution:Solution: Getting a girl.

Rule for Complementary EventRule for Complementary Event

P E P E

or

P E P E

or

P E P E

( ) ( ) 1

1

1

( ) = ( )

( ) + ( ) = .

Empirical ProbabilityEmpirical Probability

The difference between classical and empirical probabilityempirical 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.

Formula for Empirical ProbabilityFormula for Empirical Probability

Given a frequency distribution

the probability of an event being

in a given class is

P Efrequency for the class

total frequencies in the distribution

f

nThis probability is called the empirical

probability and is based on observation

,

( ) =

.

.

Empirical Probability -Empirical Probability - Example

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.


Type Frequency

ABABO

225221

50 = n


Find the following probabilities for the previous example.

A person has type O blood. Solution:Solution: P(O) = f /n = 21/50. A person has type A or type B blood. Solution:Solution: P(A or B) = 22/50+ 5/50

= 27/50.

The Addition Rules for ProbabilityThe Addition Rules for Probability

Two events are mutually exclusivemutually exclusive if they cannot occur at the same time (i.e. they have no outcomes in common).

The Addition Rules for ProbabilityThe Addition Rules for Probability

A B

A and B are mutually exclusive

Addition Rule 1Addition Rule 1

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

P A or B P A P B ( ) ( ) ( )

Addition Rule 1-Addition Rule 1- Example

At a political rally, there are 20 Republicans (R), 13 Democrats (D), and 6 Independents (I). If a person is selected, find the probability that he or she is either a Democrat or an Independent.

Solution:Solution: P(D or I) = P(D) + P(I) = 13/39 + 6/39 = 19/39.


A day of the week is selected at random. Find the probability that it is a weekend.

Solution:Solution: P(Saturday or Sunday) = P(Saturday) + P(Sunday) = 1/7 + 1/7 = 2/7.


When two events A and B

are not mutually exclusive, the

probabilityy that A or B will

occur is

P A or B P A P B P A and B

( ) ( ) ( ) ( )


A B

A and B (common portion)


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.

Addition Rule 2 -Addition Rule 2 - Example

STAFF FEMALES MALES TOTAL

NURSES 7 1 8

PHYSICIANS 3 2 5

TOTAL 10 3 13

STAFF FEMALES MALES TOTAL

NURSES 7 1 8

PHYSICIANS 3 2 5

TOTAL 10 3 13


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


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?


Solution:Solution:

P(intoxicated or accident) = P(intoxicated) + P(accident) – P(intoxicated and accident)

= 0.32 + 0.09 – 0.06 = 0.35.

Two events A and B are independentindependent if the fact that A occurs does not affect the probability of B occurring.

Example:Example: Rolling a dice and getting a 6, and then rolling another dice and getting a 3 are independent events.

The Multiplication Rules and Conditional The Multiplication Rules and Conditional ProbabilityProbability

Multiplication Rule 1Multiplication Rule 1


are independent the

probability of both

occurring is

P A and B P A P B

,

( ) ( ) ( ).

Multiplication Rule 1 -Multiplication Rule 1 - Example

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:Solution: Because these two events are independent (why?), P(queen and ace) = (4/52)(4/52) = 16/2704 = 1/169.


A Harris pole found that 46% of Americans 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:Solution: Let S denote stress. Then P(S and S and S) = (0.46)3 = 0.097.


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:Solution: Let T denote a positive test result. Then P(T and T and T and T) = (0.32)4 = 0.010.

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:Example: Having high grades and getting a scholarship are dependent events.


The conditional probabilityconditional probability of 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:NOTE: This does not mean B A.


Multiplication Rule 2Multiplication Rule 2


are dependent the

probability of both

occurring is

P A and B P A P B A

,

( ) ( ) ( | ).

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:Solution: See next slide.

The Multiplication Rules and Conditional The Multiplication Rules and Conditional Probability -Probability - Example

Solution:Solution: Since the events are dependent,

P(D1 and D2) = P(D1)P(D2| D1) = (2/25)(1/24)

= 2/600 = 1/300.


The WW 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.


Solution:Solution: Since the events are dependent,

P(H and A) = P(H)P(A|H) = (0.53)(0.27) = 0.1431.


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.


Tree Diagram for Tree Diagram for Example

P(B1) 1/2

Red

Red

Blue

Blue

Box 1

P(B2) 1/2Box 2

P(R|B1) 2/3

P(B|B1) 1/3

P(R|B2) 1/4

P(B|B2) 3/4

(1/2)(2/3)

(1/2)(1/3)

(1/2)(1/4)

(1/2)(3/4)

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


Conditional Probability -Conditional Probability - FormulaFormula

.

( | ) =( )

( )

The probability that the event B occurs

given that the first event A has occurred can be

found by dividing the probability that both events

occurred by the probability that the first event has

occurred The formula is

P B AP A and B

P A

second

.

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.

Conditional Probability -Conditional Probability - Example

Solution: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.


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.



Gender Yes No Total

Male 32 18 50

Female 8 42 50

Total 40 60 100

Gender Yes No Total

Male 32 18 50

Female 8 42 50

Total 40 60 100

Find the probability that the respondent answered “yes” given that the respondent was a female.

Solution:Solution: Let M = respondent was a male;

F = respondent was a female;

Y = respondent answered “yes”; N = respondent answered “no”.


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.


Tree Diagrams Tree Diagrams

A tree diagramtree diagram is a device used to list all possibilities of a sequence of events in a systematic way.

Tree Diagrams -Tree Diagrams - Example

Suppose a sales person can travel from New York to Pittsburgh by plane, train, or bus, and from Pittsburgh to Cincinnati by bus, boat, or automobile. Display the information using a tree diagram.

Tree Diagrams -Tree Diagrams - Example

Cincinnati

Bus

New York

Pittsburgh

Plane

Train

Bus

Boat

Auto

Bus

Boat

Boat

Bus

Auto

Auto

Plane, Bus

Plane, boat

Plane, autoTrain, bus

Train, boat

Train, autoBus, busBus, boat

Bus, auto

The Multiplication Rule for CountingThe Multiplication Rule for Counting

Multiplication Rule :Multiplication Rule : In a sequence of nn events in which the first one has kk11 possibilities and the second event has kk22 and the third has kk33, and so forth, the total possibilities of the sequence will be kk11kk22kk33kknn.

The Multiplication Rule for Counting -The Multiplication Rule for Counting - Example

A nurse has three patients to visit. How many different ways can she make her rounds if she visits each patient only once?

The Multiplication Rule for Counting - The Multiplication Rule for Counting - ExampleExample

She can choose from three patients for the first visit and choose from two patients for the second visit, since there are two left. On the third visit, she will see the one patient who is left. Hence, the total number of different possible outcomes is 3 2 1= 6.

The Multiplication Rule for Counting - The Multiplication Rule for Counting - Example

Employees of a large corporation are to be issued special coded identification cards. The card consists of 4 letters of the alphabet. Each letter can be used up to 4 times in the code. How many different ID cards can be issued?

The Multiplication Rules for Counting -The Multiplication Rules for Counting - Example

Since 4 letters are to be used, there are 4 spaces to fill ( _ _ _ _ ). Since there are 26 different letters to select from and each letter can be used up to 4 times, then the total number of identification cards that can be made is 26 2626 26= 456,976.


The digits 0, 1, 2, 3, and 4 are to be used in a 4-digit ID card. How many different cards are possible if repetitions are permitted?

Solution:Solution: Since there are four spaces to fill and five choices for each space, the solution is 5 5 5 5 = 54 = 625.


What if the repetitions were not permitted in the previous example?

Solution:Solution: The first digit can be chosen in five ways. But the second digit can be chosen in only four ways, since there are only four digits left; etc. Thus the solution is 5 4 3 2 = 120.

PermutationsPermutations

Consider the possible arrangements of the letters aa, bb, and cc.

The possible arrangements are: abc, acb, bac, bca, abc, acb, bac, bca, cab, cbacab, cba..

If the order of the arrangement is importantorder of the arrangement is important then we say that each arrangement is a permutation of the three letters. Thus there are six permutations of the three letters.


An arrangement of n distinct objects in a specific order is called a permutationpermutation of the objects.

Note:Note: To determine the number of possibilities mathematically, one can use the multiplication rule to get: 3 2 1 = 6 permutations.


Permutation Rule :Permutation Rule : The arrangement of nn objects in a specific order using r objects at a time is called a permutation of nn objects taken rr objects at a time. It is written as nnPPrr and the formula is given by

nnPPrr = n! / (n – r)! = n! / (n – r)!.

Permutations -Permutations - Example

How many different ways can a chairperson and an assistant chairperson be selected for a research project if there are seven scientists available?

Solution:Solution: Number of ways = 77PP22 = 7! / (7 – 2)! = 7! / (7 – 2)! = 7!/5! = 42= 7!/5! = 42.

Permutations -Permutations - Example

How many different ways can four books be arranged on a shelf if they can be selected from nine books?

Solution:Solution: Number of ways =99PP44 = 9! / (9 – 4)! = 9! / (9 – 4)! = 9!/5! = 3024= 9!/5! = 3024.

CombinationsCombinations

Consider the possible arrangements of the letters aa, bb, and cc.

The possible arrangements are: abc, acb, bac, bca, abc, acb, bac, bca, cab, cbacab, cba..

If the order of the arrangement is not importantorder of the arrangement is not important then we say that each arrangement is the same. We say there is one combination of the three letters.

CombinationsCombinations

Combination Rule :Combination Rule : The number of combinations of of rr objects from nn objects is denoted by nnCCrr and the formula is given b nnCCrr = n! / [(n – r)!r!] = n! / [(n – r)!r!] .

Combinations -Combinations - Example

How many combinations of four objects are there taken two at a time?

Solution:Solution: Number of combinations: 44CC22

= 4! / [(4 – 2)!= 4! / [(4 – 2)! 2!] = 4!/[2!2!] = 62!] = 4!/[2!2!] = 6..


In order to survey the opinions of customers at local malls, a researcher decides to select 5 malls from a total of 12 malls in a specific geographic area. How many different ways can the selection be made?

Solution:Solution: Number of combinations: 1212CC55 = 12! / = 12! /

[(12 – 5)![(12 – 5)! 5!] = 12!/[7!5!] = 7925!] = 12!/[7!5!] = 792..


In a club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different possibilities are there?

Solution:Solution: Number of possibilities: (number of ways of selecting 3 women from 7) (number of ways of selecting 2 men from 5) = 77CC33 55CC22 = (35) = (35)

(10) = 350(10) = 350..


A committee of 5 people must be selected from 5 men and 8 women. How many ways can the selection be made if there are at least 3 women on the committee?


Solution:Solution: The committee can consist of 3 women and 2 men, or 4 women and 1 man, or 5 women. To find the different possibilities, find each separately and then add them:

88CC33 55CC22 + + 88CC44 55CC11 + + 88CC55 55CC00

= (56)(10) + (70)(5) + (56)(1) = (56)(10) + (70)(5) + (56)(1)

= 966= 966..

chapter 4 260110 044531

Education

classical

show positive

tree diagram

tree diagrams

multiplication

sample space

addition rule

addition rule