Chapter 4 Probability

4.1 - 1Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Chapter 4Probability

4-1 Review and Preview

4-2 Basic Concepts of Probability

4-3 Addition Rule

4-4 Multiplication Rule: Basics

4-5 Multiplication Rule: Complements and Conditional Probability

4-6 Counting


Section 4-1Review and Preview


Review

Necessity of sound sampling methods.

Common measures of characteristics of data

Mean

Standard deviation


Preview

Rare Event Rule for Inferential Statistics:

If, under a given assumption, the probability of a particular observed event is extremely small, we conclude that the assumption is probably not correct.

Statisticians use the rare event rule for inferential statistics.


Section 4-2 Basic Concepts of

Probability


Part 1

Basics of Probability


Events and Sample Space

Event

any collection of results or outcomes of a procedure

Simple Event

an outcome or an event that cannot be further broken down into simpler components

Sample Space

for a procedure consists of all possible simple events; that is, the sample space consists of all outcomes that cannot be broken down any further


Example A pair of dice are rolled. The sample space has

36 simple events:

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

where the pairs represent the numbers rolled on each dice.

Which elements of the sample space correspond to the event that the sum of each dice is 4?


Example Which elements of the sample space correspond

to the event that the sum of each dice is 4?

ANSWER:

3,1 2,2 1,3


Notation for Probabilities

P - denotes a probability.

A, B, and C - denote specific events.

P(A) - denotes the probability of event A occurring.


Basic Rules for Computing Probability

Rule 1: Relative Frequency Approximation of Probability

Conduct (or observe) a procedure, and count the number of times event A actually occurs. Based on these actual results, P(A) is approximated as follows:

P(A) = # of times A occurred

# of times procedure was repeated


ExampleProblem 20 on page 149

F = event of a false negative on polygraph test

Thus this is not considered unusual since it is more than 0.001 (see page 146). The test is not highly accurate.

0918.098

9)( FP


Rounding Off Probabilities

When expressing the value of a probability, either give the exact fraction or decimal or round off final decimal results to three significant digits. All digits are significant except for the zeros that are included for proper placement of the decimal point.

Example:

0.1254 has four significant digits

0.0013 has two significant digits



F = event of a selecting a female senator

NOTE: total number of senators=100

Thus this does not agree with the claim that men and women have an equal (50%) chance of being selected as a senator.

160.0100

16

1684

16)(

FP



A = event that Delta airlines passenger is involuntarily bumped from a flight

Thus this is considered unusual since it is less than 0.05 (see directions on page 149). Since probability is very low, getting bumped from a flight on Delta is not a serious problem.

000195.015378

3)( AP


Basic Rules for Computing Probability

Rule 2: Classical Approach to Probability (Requires Equally Likely Outcomes)

Assume that for a given procedure each simple event has an equal chance of occurring.

P(A) = number of ways A can occur

number of different simple events in the sample space


ExampleWhat is the probability of rolling two die and getting a sum of 4?

A = event that sum of the dice is 4

Assume each number is equally likely to be rolled on the die. Rolling a sum of 4 can happen in one of three ways (see previous slide) with 36 simple events so:

0833.036

3)( AP


ExampleWhat is the probability of getting no heads when a “fair” coin is tossed three times? (A fair coin has an equal probability of showing heads or tails when tossed.)

A = event that no heads occurs in three tosses

Sample Space (in order of toss):

TTTTTHTHTTHHHTTHTHHHTHHH ,,,,,,,


ExampleSample space has 8 simple events.

Event A corresponds to TTT only so that:

125.08

1)( AP



Let:

S = event that son inherits disease (xY or Yx)

D = event that daughter inherits disease (xx)



(a) Father: xY Mother: XX

Sample space for a son:

YX YX

Sample space has no simple

events that represent a son that has the disease so:

02

0)( SP



(b) Father: xY Mother: XX

Sample space for a daughter:

xX xX


events that represent a daughter that has the disease so:

02

0)( DP



(c) Father: XY Mother: xX

Sample space for a son:

Yx YX

Sample space has one simple

event that represents a son that has the disease so:

5.02

1)( SP



(d) Father: XY Mother: xX

Sample space for a daughter:

Xx XX


event that represents a daughter that has the disease so:

02

0)( DP


Problem 18 on page 149

Table 4-1 on page 137 (polygraph data)

Example

Did Not LieDid Not Lie Did LieDid Lie

Positive Test ResultPositive Test Result 1515

(false positive)(false positive)

4242

(true positive)(true positive)

NegativeTest ResultNegativeTest Result 3232

(true negative)(true negative)

99

(false negative)(false negative)


It is helpful to first total the data in the table:

(a) How many responses were lies:

ANSWER: 51

Example

Did Not LieDid Not Lie Did LieDid Lie TOTALSTOTALS



4242


5757



99


4141

TOTALSTOTALS 4747 5151 9898


(b) If one response is randomly selected, what is the probability it is a lie?

L = event of selecting one of the lie responses

(c)

Example

98

51)( LP

520.098

51


Basic Rules for Computing Probability - continued

Rule 3: Subjective Probabilities

P(A), the probability of event A, is

estimated by using knowledge of the

relevant circumstances.



Probability should be high based on experience (it is rare to be delayed because of an accident).

Guess: P=99/100 (99 out of 100 times you will not be delayed because of an accident)

ANSWERS WILL VARY


Example: Classical probability predicts the probability of flipping a (non-biased) coin and it coming up heads is ½=0.5

Ten coin flips will sometimes result in exactly 5 heads and a frequency probability of heads 5/10=0.5; but often you will not get exactly 5 heads in ten flips.


Law of Large Numbers

As a procedure is repeated again and again, the relative frequency probability of an event tends to approach the actual probability.

Example: If we flip a coin 1 million times the frequency probability should be approximately 0.5


Probability Limits

The probability of an event that is certain to occur is 1.

The probability of an impossible event is 0.

For any event A, the probability of A is between 0 and 1 inclusive. That is:

0 P(A) 1

Always express a probability as a fraction or decimal number between 0 and 1.


Possible Values for Probabilities


Complementary Events

The complement of event A, denoted by

A, consists of all outcomes in which the

event A does not occur.


ExampleIf a fair coin is tossed three times and

A = event that exactly one heads occurs

Find the complement of A.


ExampleSample space:

Event A corresponds to HTT, THT, TTH

Therefore, the complement of A are the simple events:

HHH, HHT, HTH, THH, TTT

TTTTTHTHTTHHHTTHTHHHTHHH ,,,,,,,


Part 2

Beyond theBasics of Probability: Odds


Odds

The actual odds in favor of event A occurring are the ratio P(A)/ P(A), usually expressed in the form of a:b (or “a to b”), where a and b are integers having no common factors.

The actual odds against event A occurring are the ratio P(A)/P(A), which is the reciprocal of the actual odds in favor of the event. If the odds in favor of A are a:b, then the odds against A are b:a.

The payoff odds against event A occurring are the ratio of the net profit (if you win) to the amount bet.

payoff odds against event A = (net profit) : (amount bet)


Example

Problem 38, page 149

W = simple event that you win due to an odd number

Sample Space00, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36

(a) There are 18 odd numbers so that

P(W) = 18/38


Example


There are 20 events that correspond to winning from a number that is not odd (i.e. you do not win due to an odd number) so:

(b) Odds against winning are

9:10or 9/1018

20

38/18

38/20

)(

)(

WP

WP

38

20)( WP


Example


(c) Payoff odds against winning are 1:1 That is, $1 net profit for every $1 bet

Thus, if you bet $18 and win, your net profit is $18 which can be found by solving the proportion:

The casino returns $18+$18=$36 to you.

1

1

18

profitnet


Example


(d) Actual odds against winning are 10:9 That is, $10 net profit for every $9 bet

Thus, if you bet $18 and win, your net profit is $20 which can be found by solving the proportion:

The casino returns $18+$20=$38 to you.

9

10

18

profitnet


Recap

In this section we have discussed:

Rare event rule for inferential statistics.

Probability rules.

Law of large numbers.

Complementary events.

Rounding off probabilities.

Odds.


Section 4-3 Addition Rule


Key Concept

This section presents the addition rule as a device for finding probabilities that can be expressed as P(A or B), the probability that either event A occurs or event B occurs (or they both occur) as the single outcome of the procedure.

The key word in this section is “or.” It is the inclusive or, which means either one or the other or both.


Compound Event any event combining 2 or more simple events

Compound Event

Notation

P(A or B) = P (in a single trial, event A occurs or event B occurs or they both occur)


When finding the probability that event A occurs or event B occurs, find the total number of ways A can occur and the number of ways B can occur, but find that total in such a way that no outcome is counted more than once.

General Rule for a Compound Event


A random survey of members of the class of 2005 finds the following:

What is the probability the student did not graduate or was a man?

Example

Number of Number of students who students who graduatedgraduated

Number of students Number of students who did not graduatewho did not graduate

TOTALSTOTALS

WomenWomen 672672 2222 694694

MenMen 582582 1919 601601

TOTALSTOTALS 12541254 4141 12951295


Method 1: directly add up those who did not graduate and those who are men (without counting men twice):

22 + 19 + 582 = 623

Method 2: add up the total number who did not graduate and the total number of men, then subtract the double count of men:

41 + 601 - 19 = 623

Example


The probability the student did not graduate or was a man :

623/1295 = 0.481

Example


Compound Event

Formal Addition Rule

P(A or B) = P(A) + P(B) – P(A and B)

where P(A and B) denotes the probability that A and B both occur at the same time as an outcome in a trial of a procedure.


N = event that student did not graduate

M = event that student was a man

P(N) = 41/1295 = 0.0317

P(M) = 601/1295 = 0.464

P(N and M) = 19/1295 = 0.0147

P(N or M) = P(N) + P(M) - P(N and M)

= 0.0317 + 0.464 - 0.0147

= 0.481

Previous Example



What is the probability the subject had a negative test result or lied?

Example




4242


5757



99


4141



N = event that there is a negative test result

L = event that the subject lied

P(N) = 41/98 = 0.418

P(L) = 51/98 = 0.520

P(N and L) = 9/98 = 0.0918

P(N or L) = P(N) + P(L) - P(N and L)

= 0.418 + 0.520 - 0.0918

= 0.846

Example (cont.)


Problem 30 on page 158N=event that person refuses to respond

F=event that person’s age is 60 or older

Note: total number of people in the study is 1205

total number who refused to respond is 156 so:

P(N)=156/1205

total number who are 60 or older is 251 so:

P(F)=251/1205

total number who refuse to respond and 60 or older is 49 so:

P(N and F)=49/1205

P(N OR F) = 156/1205 + 251/1205 – 49/1205 = 358/1205 = 0.297

Example


Disjoint or Mutually ExclusiveEvents A and B are disjoint (or mutually exclusive) if they cannot occur at the same time. (That is, disjoint events do not overlap.)

Previous Example:

G = students who graduated

N = students who did not graduate

N and G are disjoint because no student who did not graduate also graduated


Disjoint or Mutually ExclusiveEvents A and B are not disjoint if they overlap.

Previous Example:

M = male students

G = students who graduated

M and G are not disjoint because some students who graduated are also male


ExampleProblem 10 and 12 on page 157

10.These are disjoint since a subject treated with Lipitor cannot be a subject given no medication.

12. These are not disjoint since it is possible for a homeless person to be a college graduate.


Venn DiagramA Venn diagram is a way to picture how sets overlap.

Venn Diagram for Events That Are Not Disjoint and overlap.

Venn Diagram for Disjoint Events which do not overlap.


Complementary Events

It is impossible for an event and its complement to occur at the same time.

That is, for any event A,

are disjoint

AA and


Probability Rule of Complementary Events

P(A) + P(A) = 1

= 1 – P(A)

P(A) = 1 – P(A)

P(A)


Venn Diagram for the Complement of Event A



is the probability that a screened driver is not intoxicated

)(IP

991.000888.01)(1)( IPIP


Recap


Compound events.

Formal addition rule.

Intuitive addition rule.

Disjoint events.

Complementary events.


Section 4-4 Multiplication Rule:

Basics


Tree Diagrams

A tree diagram is a picture of the possible outcomes of a procedure, shown as line segments emanating from one starting point. These diagrams are sometimes helpful in determining the number of possible outcomes in a sample space, if the number of possibilities is not too large.


Tree Diagrams

This figure summarizes the possible outcomes for a true/false question followed by a multiple choice question.

Note that there are 10 possible combinations.


Example: Tree Diagrams

A bag contains three different colored marbles: red, blue, and green. Suppose two marbles are drawn from the bag and after the first marble is drawn, it is put back into the bag before the second marble is drawn. Construct a tree diagram that depicts all possible outcomes.




Example: Computing Probability withTree Diagrams

Use the previous example of drawing two marbles with replacement to compute the probability of drawing a red marble on the first draw and a blue marble on the second draw.



P(Y) = number of ways Y can occur


9

1)( YP

Y = event of drawing red first then blue



Consider each event separately:

3

1)( RP

R = event of drawing red on first draw

B = event of drawing blue on second draw

3

1)( BP


Notation

P(A and B) =

P(event A occurs in a first trial and

event B occurs in a second trial)


Key Concept

The basic multiplication rule is used for finding P(A and B), the probability that event A occurs in a first trial and event B occurs in a second trial.



The probability of drawing red on first draw and drawing blue on second draw is the product of the individual probabilities:

9

1

3

1

3

1) and ( BRP


Key Concept

NEXT EXAMPLE SHOWS:

If the outcome of the first event A somehow affects the probability of the second event B, it is important to adjust the probability of B to reflect the occurrence of event A.

This is Conditional Probability



A bag contains three different colored marbles: red, blue, and green. Suppose two marbles are drawn from the bag without replacing the first marble after it is drawn. Construct a tree diagram that depicts all possible outcomes.





Use the previous example to compute the probability of drawing a red marble on the first draw and a blue marble on the second draw.



P(Y) = number of ways Y can occur


6

1)( YP

Y = event of drawing red first then blue



Multiplication Method

3

1)( RP

R = event of drawing red on first draw

B = event of drawing blue on second draw (given that there are now only two marbles in the bag)

2

1)( BP



The probability of drawing red on first draw and drawing blue on second draw is the product of the individual probabilities:

6

1

2

1

3

1) and ( BRP


Conditional ProbabilityKey Point

Without replacement, we must adjust the probability of the second event to reflect the outcome of the first event.


Conditional ProbabilityImportant Principle

The probability for the second event B should take into account the fact that the first event A has already occurred.


Notation for Conditional Probability

P(B|A) represents the probability of event B occurring after it is assumed that event A has already occurred (read B|A as “B given A.”)


Multiplication Rule and Conditional Probability

P(A and B) = P(A) • P(B|A)


Intuitive Multiplication Rule

When finding the probability that event A occurs in one trial and event B occurs in the next trial, multiply the probability of event A by the probability of event B, but be sure that the probability of event B takes into account the previous occurrence of event A.



If three are selected without replacement, what is probability they all had false positive test results?

Example




4242


5757



99


4141




On first selection there are 98 subjects, 15 of which are false positive:

Example

98

15)( 1 FP

selectionth on positive false a is thereevent that nFn



On second selection there are 97 subjects (without replacement)

If the first selection was false positive, there are 14 false positives left:

Example

97

14)|( 12 FFP



On third selection there are 96 subjects (without replacement)

If the first and second selections were false positive, there are 13 false positives left:

Example

96

13) and |( 213 FFFP



ANSWER:

This is considered unusual since probability is less than 0.05

Example

) and |()|()() and and ( 213121321 FFFPFFPFPFFFP

00299.096

13

97

14

98

15


Dependent and Independent

Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other. (Several events are similarly independent if the occurrence of any does not affect the probabilities of the occurrence of the others.) If A and B are not independent, they are said to be dependent.



Finding that your calculator works

Finding that your computer works

Assuming your calculator and your computer are not both running off the same source of power, these are independent events.

Examples


Previous example of drawing two marbles with replacement

Since the probability of drawing the second marble is not affected by drawing the first marble, these are independent events.

Examples


Dependent Events

Two events are dependent if the occurrence of one of them affects the probability of the occurrence of the other, but this does not necessarily mean that one of the events is a cause of the other.


Previous example of drawing two marbles without replacement

Since the probability of drawing the second marble is affected by drawing the first marble, these are dependent events.

Examples


Multiplication Rule for Independent Events

Note that if A and B are independent events, then P(B|A)=P(B) and the multiplication rule is then:

P(A and B) = P(A) • P(B)


Applying the Multiplication Rule

4.1 - 100

Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.

Applying the Multiplication Rule

4.1 - 101


Caution

When applying the multiplication rule, always consider whether the events are independent or dependent, and adjust the calculations accordingly.

4.1 - 102


Multiplication Rule for Several Events

In general, the probability of any sequence of independent events is simply the product of their corresponding probabilities.

4.1 - 103


Example

Page 169, problem 26

These events are independent since the probability of getting a girl on any try is not affected by the occurence of getting a girl on a previous try.

th tryon birth girl a is thereevent that nGn

2

1)( nGP

4.1 - 104


Example


ten factors of 1/2

10

2

1

2

1...

2

1

2

1

)(...)()(

) and ... and and (

1021

1021

GPGPGP

GGGP

4.1 - 105


ExamplePage 169, problem 26

Calculator: use “hat key” to evaluate powers: 000977.05.0

2

1 1010

10 ^ 5.0)5.0( 10

4.1 - 106



Since:

We see that getting 10 girls by chance alone is unusual and conclude that the gender selection method is effective.

Calculator: use “hat key” to evaluate powers:

05.0000977.0

4.1 - 107


Treating Dependent Events as Independent

Some calculations are cumbersome, but they can be made manageable by using the common practice of treating events as independent when small samples are drawn from large populations. In such cases, it is rare to select the same item twice (sample with replacement).

4.1 - 108


The 5% Guideline for Cumbersome Calculations

If a sample size is no more than 5% of the size of the population, treat the selections as being independent (even if the selections are made without replacement, so they are technically dependent).

4.1 - 109


Example


a) If we select without replacement, then randomly selecting an ignition system are not independent. But since 3/200 = 0.015 =1.5%, we could use the 5% guideline and regard these events as independent.

4.1 - 110


Example


b) If these events are not independent (dependent) then:

) and |()|()() and and ( 213121321 GGGPGGPGPGGGP

926.0198

193

199

194

200

195

4.1 - 111


Example


c) If these events are independent then:

)()()() and and ( 321321 GPGPGPGGGP 3

200

195

200

195

200

195

200

195

927.0975.0 3

4.1 - 112


Example


d) The answer from part (b) is exact so it is better.

4.1 - 113


Summary of Fundamentals

In the addition rule, the word “or” in P(A or B) suggests addition. Add P(A) and P(B), being careful to add in such a way that every outcome is counted only once.

In the multiplication rule, the word “and” in P(A and B) suggests multiplication. Multiply P(A) and P(B), but be sure that the probability of event B takes into account the previous occurrence of event A.

4.1 - 114


Recap


Notation for P(A and B).

Notation for conditional probability.

Independent events.

Formal and intuitive multiplication rules.

Tree diagrams.

4.1 - 115


Section 4-5 Multiplication Rule:Complements and

Conditional Probability

4.1 - 116


Key Concepts

Probability of “at least one”:Find the probability that among several trials, we get at least one of some specified event.

Conditional probability:Find the probability of an event when we have additional information that some other event has already occurred.

4.1 - 117


Complements: The Probability of “At Least One”

The complement of getting at least one item of a particular type is that you get no items of that type.

“At least one” is equivalent to “one or more.”

4.1 - 118


Example

Page 175, problems 6

If not all 6 are free from defects, that means that at least one of them is defective.

4.1 - 119


Example

Page 175, problems 8

If it is not true that at least one of the five accepts an invitation, then all five did not accept the invitation.

4.1 - 120


Finding the Probability of “At Least One”

To find the probability of at least one of something, calculate the probability of none, then subtract that result from 1. That is,

P(at least one) = 1 – P(none).

4.1 - 121


Example


P(at least one girl) = 1 – P(all boys)

)(...)()(1

) and ... and and (1)boys all(1

821

821

BPBPBP

BBBPP

th tryon birth boy a is thereevent that nBn

996.0

00391.012

11

8

4.1 - 122


Example


The probability of having 8 children and none of them are girls (all boys) is 0.00391 which means this is a rare event.

4.1 - 123


Example


P(at least one working calculator)

= 1 – P(all calculators fail)

Note: these are independent events and

fails calculatorth event that nFn

04.096.01)( nFP

P(a calculator fails) = 1 – P(a calculator does not fail)

4.1 - 124


Example


998.0

0016.0104.01

)()(1

) and (1)fail scalculator all(1

2

21

21

FPFP

FFPP

4.1 - 125



With one calculator,

P(working calculator) = 0.96

With two calculators,

P(at least one working calculator)

= 0.998

The increase in chance of a working calculator might be worth the effort.

4.1 - 126


Conditional ProbabilityA conditional probability of an event is a probability obtained with the additional information that some other event has already occurred.

P(B|A) denotes the conditional probability of event B occurring, given that event A has already occurred.

4.1 - 127


Intuitive Approach to Conditional Probability

The conditional probability of B given A can be found by assuming that event A has occurred, and then calculating the probability that event B will occur.

4.1 - 128



Table 4-1 on page 137 (polygraph data)

Example

Did Not LieDid Not Lie Did LieDid Lie



4242




99


4.1 - 129


Example


(a)There are 47 subjects who did not lie, 32 of which had a negative test result.

681.047

32lie)not didsubject |result test negative( P

4.1 - 130


Conditional Probability Formula

Conditional probability of event B occurring, given that event A has already occurred

P(B A) = P(A and B)

P(A)

4.1 - 131


Example


(a)Using the formula

681.0 98/47

98/32

lie)not didsubject (

result) test negative had and lienot didsubject (

lie)not didsubject |result test negative(

P

P

P

4.1 - 132


Confusion of the Inverse

To incorrectly believe that P(A|B) and P(B|A) are the same, or to incorrectly use one value for the other, is often called confusion of the inverse.

4.1 - 133


Example


(b) Using the formula

780.0 98/41

98/32

result) test negative(

lie)not didsubject andresult test negative(

result) test negative|lienot didsubject (

P

P

P

4.1 - 134


Example


(c) The results from parts (a) and (b) are not equal.

4.1 - 135


Recap


Concept of “at least one.”

Conditional probability.

Intuitive approach to conditional probability.

4.1 - 136


Section 4-6Counting

4.1 - 137


Key Concept

In many probability problems, the big obstacle is finding the total number of outcomes, and this section presents several methods for finding such numbers without directly listing and counting the possibilities.

4.1 - 138


Fundamental Counting Rule

For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of m n ways.

4.1 - 139


Example

A coin is flipped and then a die is rolled. What are the total number of outcomes?

4.1 - 140


Example


4.1 - 141


Example


ANSWER: There are 2 outcomes for the coin flip and 6 outcomes for the die roll. Total number of outcomes:

1262

4.1 - 142


Example

Page 185, Problem 32 (a)

If 20 newborn babies are randomly selected, how many different gender sequences are possible?

4.1 - 143


Example


ANSWER:

This is a sequence of 20 events each of which has two possible outcomes (boy or girl). By the fundamental counting rule, total number of gender sequences will be:

576,048,122222 20

4.1 - 144


Example

Page 185, Problem 28

A safe combination consists of four numbers between 0 and 99. If four numbers are randomly selected, what is the probability of getting the correct combination on the first attempt?

4.1 - 145


ExamplePage 185, Problem 28

ANSWER:

Total number of possible combinations is

Since there is only one correct combination:

It is not feasible to try opening the safe this way.

000,000,100100100100100100 4

00000001.0000,000,100

1n)combinatiocorrect ( P

4.1 - 146


Notation

The factorial symbol ! denotes the product of decreasing positive whole numbers.

For example,

4! 4 3 2 1 24.

By special definition, 0! = 1.

4.1 - 147


A collection of n different items can be arranged in order n! different ways. (This factorial rule reflects the fact that the first item may be selected in n different ways, the second item may be selected in n – 1 ways, and so on.)

Factorial Rule

4.1 - 148


Example

Page 183, Problem 6

Find the number of different ways that the nine players on a baseball team can line up for the National Anthem by evaluating 9 factorial.

880,362123456789!9

4.1 - 149


Factorial on Calculator

Calculator

9 MATH PRB 4:!

To get:

Then Enter gives the result

!9

4.1 - 150


Permutations Rule(when items are all different)

(n - r)!n rP = n!

If the preceding requirements are satisfied, the number of permutations (or sequences) of r items selected from n available items (without replacement) is

Requirements:

1. There are n different items available. (This rule does not apply if some of the items are identical to others.)

2. We select r of the n items (without replacement).

3. We consider rearrangements of the same items to be different sequences. (The permutation of ABC is different from CBA and is counted separately.)

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Example

A bag contains 4 colored marbles: red, blue, green, yellow. If we select 3 of the four marbles from the bag without replacement, how many different color orders are there?

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Example

ANSWER:

241

1234

!1

!4

)!34(

!434

P

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Example


In horse racing, a trifecta is a bet that the first three finishers in a race are selected, and they are selected in the correct order. Find the number of different possible trifecta bets in a race with ten horses by evaluating

310P

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Example


ANSWER:

7201

8910

1234567

12345678910

!7

!10

)!310(

!10310

P

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Permutations on Calculator

Calculator

10 MATH PRB 2:nPr 3

To get:


310P

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Permutations Rule(when some items are identical to others)

n1! . n2! .. . . . . . . nk! n!

If the preceding requirements are satisfied, and if there are n1 alike, n2 alike, . . . nk alike, the number of permutations (or sequences) of all items selected without replacement is

Requirements:

1. There are n items available, and some items are identical to others.

2. We select all of the n items (without replacement).

3. We consider rearrangements of distinct items to be different sequences.

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Example


Find the number of different ways the letters AGGYB can be arranged.

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Example


ANSWER: there are two letters of the five that are alike. Then the total number of arrangements will be

6012

12345

!1!1!1!2

!5

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Example

Page 185, Problem 32 (b)

How many different ways can 10 girls and 10 boys be arranged in a sequence?

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Example


ANSWER: there are twenty total children in the arrangement. There are two groups of 10 that are alike. This gives a total number of arrangements:

756,184!10!10

!20

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Example

Page 185, Problem 32 (c)

What is the probability of getting 10 girls and 10 boys when twenty babies are born?

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Example


ANSWER: as found on previous slides

The total number of boy/girl arrangements of 20 newborns is:

The total number of ways 10 girls and 10 boys can be arranged in a sequence is

576,048,1220

756,184!10!10

!20

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Example


ANSWER:

176.0576,048,1

756,184

born) are babies 20 when boys 10 and girls 10(

P

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(n - r )! r!n!

nCr =

Combinations Rule

If the preceding requirements are satisfied, the number of combinations of r items selected from n different items is

Requirements:

1. There are n different items available.

2. We select r of the n items (without replacement).

3. We consider rearrangements of the same items to be the same. (The combination of ABC is the same as CBA.)

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Example

Page 183, Problem 8

Find the number of different possible 5 card poker hands by evaluating

552C

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Example

Page 183, Problem 8

ANSWER:

960,598,2 !5

4849505152

!5!47

!52

!5)!552(

!52552

C

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Combinations on Calculator

Calculator

52 MATH PRB 3:nCr 5

To get:


552C

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Example


What is the probability of winning a lottery with one ticket if you select five winning numbers from 1,2,…,31.

Note: numbers selected are different and order does not matter.

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Example


ANSWER:

Total number of possible combinations is:

Since only one combination wins:

911,169!5!26

!31531

C

00000589.0911,169

1winning)( P

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Example


What is the probability of winning a lottery with one ticket if you select five winning numbers from 1,2,…,31.

Note: assume now that you must select the numbers in the same order they were drawn so that order does matter.

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Example


ANSWER:

Total number of possible combinations is:

Since only one permutation wins:

320,389,20!26

!31531 P

0000000490.0320,389,20

1winning)( P

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When different orderings of the same items are to be counted separately, we have a permutation problem, but when different orderings are not to be counted separately, we have a combination problem.

Permutations versus Combinations

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Recap

In this section we have discussed: The fundamental counting rule.

The permutations rule (when items are all different).

The permutations rule (when some items are identical to others).

The combinations rule.

The factorial rule.

Documents

Chapter 4 Probability