Chapter 2: Probability Concepts and Applications · • The probability, P, of any event or state...

1

Chapter 2: Probability Concepts and

Applications

Textbook: pp. 39-80

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Learning Objectives

After completing this chapter, students will be able to:

• Understand the basic foundations of probability analysis.

• Describe statistically dependent and independent events.

• Use Bayes’ theorem to establish posterior probabilities.

• Describe and provide examples of both discrete and

continuous random variables.

• Explain the difference between discrete and continuous

probability distributions.

• Calculate expected values and variances and use the

normal table.

3

Learning Objectives

After completing this chapter, students will be able to:

• Understand the binomial distribution.

• Understand the normal distribution and use the normal

table.

• Understand the F distribution.

• Understand the exponential distribution and its relation to

queuing theory.

• Understand the Poisson distribution and its relation to

queuing theory.

4

• Life is uncertain; we are not sure what the future will

bring

• Probability is a numerical statement about the

likelihood that an event will occur

Introduction

5

• The probability, P, of any event or state of nature

occurring is greater than or equal to 0 and less than or

equal to 1. That is,

0 ≤ P(event) ≤ 1

• A probability of 0 indicates that an event is never

expected to occur. A probability of 1 means that an

event is always expected to occur.

• The sum of the simple probabilities for all possible

outcomes of an activity must equal 1. Regardless of

how probabilities are determined, they must adhere to

these two rules.

Two Basic Rules of Probability

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• Objective Approach

• Relative frequency approach

• Classical or logical method

Types of Probability (1 of 3)

P(event) =Number of occurrences of the event

Total numbers of trials or outcomes

P(head) =

P(head) =

1

2

13

52

Number of ways of getting a head

Number of possible outcomes (head or tail)

Number of chances of drawing a spade

Number of possible outcomes

2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace

7

• Historical demand for white latex paint at = 0, 1, 2, 3, or

4 gallons per day

• Observed frequencies over the past 200 days

Diversey Paint Example (1 of 2)

8

• Historical demand for white latex paint at = 0, 1, 2, 3, or

4 gallons per day

• Observed frequencies over the past 200 days

Diversey Paint Example (2 of 2)

Individual

probabilities are all

between 0 and 1

0 ≤ P (event) ≤ 1

Total of all event

probabilities equals 1

∑ P (event) = 1.00

9

• Objective Approach

o Classical or logical method

• Perform a series of trials


1head =

2

13spade =

52

01

= .25=4

= 25%

Number of ways of getting a headP

Number of possible outcomes head or tail

Number of chances of drawing a spadeP

Number of possible outcomes

( )

10

• Subjective Approach

o Based on the experience and judgment of the

person making the estimate

• Opinion polls

• Judgment of experts

• Delphi method


11

• Events are said to be mutually exclusive if only one of

the events can occur on any one trial

o Tossing a coin will result in either a head or a tail

o Rolling a die will result in only one of six possible

outcomes

Mutually Exclusive and Collectively

Exhaustive Events (1 of 2)

12

• Events are said to be

collectively exhaustive if

the list of outcomes

includes every possible

outcome

• Both heads and tails as

possible outcomes of coin

flips

• All six possible outcomes of

the roll of a die:

Mutually Exclusive and Collectively

Exhaustive Events (2 of 2)

13

Venn Diagrams

A B

Events that are mutually exclusive

Events that are notmutually exclusive

A B

14

• Draw one card from a deck of 52 playing cards

A = event that a 7 is drawn

B = event that a heart is drawn

P (a 7 is drawn) = P(A)= 4/52 = 1/13

P (a heart is drawn) = P(B) = 13/52 = 1/4

• These two events are not mutually exclusive since a 7

of hearts can be drawn

• These two events are not collectively exhaustive

since there are other cards in the deck besides 7s and

hearts

Drawing a Card

15

Differences (mutually exclusive / collectively

exhaustive)

16

Differences (mutually exclusive / collectively

exhaustive)

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• Intersection – the set of all outcomes that are common to

both events

• Intersection of event A and event B = A and B

= A ∩ B

= AB

• Probability notation

P(Intersection of event A and event B) = P(A and B)

= P(A ∩ B)

= P(AB)

Sometimes called joint probability

Unions and Intersections of Events (1 of 3)

18

• Union – the set of all outcomes that are contained in

either of two events

Union of event A and event B = A or B

• Probability notation

P(Union of event A and event B) = P(A or B)

= P(A ∪ B)


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• In the previous example

• Intersection of event A and event B

(A and B) = the 7 of hearts is drawn

P(A and B) = P(7 of hearts is drawn) = 1/52

• Union of event A and event B

(A or B) = either a 7 or a heart is drawn

P(A or B) = P(any 7 or any heart is drawn) = 16

/52


20

• General rule for union of two events,

additive rule

P(A or B) = P(A) + P(B) – P(A and B)

o Union of two events, a 7 or a heart

P(A or B) = P(A) + P(B) – P(A and B)

= 4/52 + 13/52 – 1/52

= 16/52

Probability Rules (1 of 5)

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• Conditional probability – probability that an event

occurs given another event has already happened

o Probability of a 7 given a heart has been drawn


P(A | B) = = = 1/13

P(AB)

P(B)

1/52

13/52

P(A | B) =P(AB)

P(B)

P(AB) = P(A | B) P(B)

22

Which sets are independent?1. (a) Your education

(b) Your income level

2. (a) Draw a jack of hearts from a full 52-card deck

(b) Draw a jack of clubs from a full 52-card deck

3. (a) Snow in Zhengzhou

(b) Rain in Bangor / Wales


23

• Independent one event has no effect on the other event

P(A | B) = P(A)

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

• For a fair coin tossed twice

A = event that a head is the result of the first toss

B = event that a head is the result of the second toss

P(A) = 0.5 and P(B) = 0.5

P(AB) = P(A)P(B) = 0.5(0.5) = 0.25


24

• A bucket contains 3 black balls and 7 green balls

Draw a ball from the bucket, replace it, and draw a

second ball!

1. The probability of a black ball drawn on first draw is:

P(B) = 0.30

2. The probability of two green balls drawn is:P(GG) = P(G) x P(G) = 0.7 x 0.7 = 0.49

Independent Events (1 of 2)

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• A bucket contains 3 black balls and 7 green balls

Draw a ball from the bucket, replace it, and draw a

second ball!

3. The probability of a black ball drawn on the second

draw if the first draw is green is:

P(B | G) = P(B) = 0.30

4. The probability of a green ball drawn on the second

draw if the first draw is green is:

P(G | G) = P(G) = 0.70

Independent Events (2 of 2)

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• An urn contains the following 10 balls:

o 4 are white (W) and lettered (L)

o 2 are white (W) and numbered (N)

o 3 are yellow (Y) and lettered (L)

o 1 is yellow (Y) and numbered (N)

P(WL) = 4/10 = 0.4 P(YL) = 3/10 = 0.3

P(WN) = 2/10 = 0.2 P(YN) = 1/10 = 0.1

P(W) = 6/10 = 0.6 P(L) = 7/10 = 0.7

P(Y) = 4/10 = 0.4 P(N) = 3/10 = 0.3

Dependent Events (1 of 3)

27


28

• The conditional probability that the ball drawn is

lettered, given that it is yellow

We can verify P(YL) using the joint probability formula:


29

• Bayes’ theorem is used to incorporate additional

information and help create posterior probabilities from

original or prior probabilities

Revising Probabilities with Bayes’ Theorem (1 of 7)

30

• A cup contains two dice identical in appearance but one

is fair (unbiased) and the other is loaded (biased)

o The probability of rolling a 3 on the fair die is 1/6 or 0.166

o The probability of tossing the same number on the loaded

die is 0.60

o We select one by chance, toss it, and get a 3

o What is the probability that the die rolled was fair?

o What is the probability that the loaded die was rolled?


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• The probability of the die being fair or loaded is

P(fair) = 0.50 P(loaded) = 0.50

and that

P(3 | fair) = 0.166 P(3 | loaded) = 0.60

• The probabilities of P(3 and fair) and P(3 and loaded)

are

P(3 and fair) = P(3 | fair) x P(fair) = (0.166)(0.50) = 0.083

P(3 and loaded) = P(3 | loaded) x P(loaded) = (0.60)(0.50)

= 0.300


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• The probability of the die being fair or loaded is

P(fair) = 0.50 P(loaded) = 0.50

and that

P(3 | fair) = 0.166 P(3 | loaded) = 0.60

• The probabilities of P(3 and fair) and P(3 and loaded)

are

P(3 and fair) = P(3 | fair) x P(fair) = (0.166)(0.50) = 0.083

P(3 and loaded) = P(3 | loaded) x P(loaded) = (0.60)(0.50)

= 0.300

The sum of these probabilities gives us the unconditional

probability of tossing a 3

P(3) = 0.083 + 0.300 = 0.383


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• If a 3 does occur, the probability that the die rolled was

the fair one is

• The probability that the die was loaded is


34

• If a 3 does occur, the probability that the die rolled was

the fair one is

• The probability that the die was loaded is

• These are the revised or posterior probabilities for the

next roll of the die

• We use these to revise our prior probability estimates


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• Event B has occurred

• A 3 is rolled


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• We can compute revised probabilities more directly by

using

where

A’ = the complement of the event A;

for example, if A is the event “fair die”, then A’ is “loaded

die”

General Form of Bayes’ Theorem (1 of 2)

P(A | B) =P(B | A)P(A)

P(B | A)P(A)+P(B | ¢A )P( ¢A )

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• Conditional probability

From the previous example

Replace A with “fair die”, A’ with “loaded die”, B with “3

rolled”

General Form of Bayes’ Theorem (2 of 2)

38

• Additional information from a second experiment

o If you can afford it, perform experiments several

times

o We roll the die again and again get a 3

Further Probability Revisions (1 of 4)

39


• If you can afford it, perform experiments several times

• We roll the die again and again get a 3


40


• If you can afford it, perform experiments several times

• We roll the die again and again get a 3


41

• After the first roll of the die

probability the die is fair = 0.22

probability the die is loaded = 0.78

• After the second roll of the die

probability the die is fair = 0.067

probability the die is loaded = 0.933


42

• A random variable assigns a real number to every

possible outcome or event in an experiment

X = number of refrigerators sold during the day

• Discrete random variables can assume only a finite or

limited set of values

• Continuous random variables can assume any one of

an infinite set of values

Random Variables (1 of 3)

43

Random Variables --- that are Numbers (2 of 3)

44

Random Variables --- that are not Numbers (3 of 3)

45

• For discrete random variables, probability value

assigned to each event

o Statistics class of 100 students

o Quiz with five problems with 1 point for each correct

answer

o Lowest score = 1, highest score = 5

• Examples follows the three rules

o Events are mutually exclusive and collectively exhaustive

o Individual probability values between 0 and 1

o Total probability sums to 1

Probability Distributions (1 of 4)

46

• Developed using relative frequency approach

Probability Distributions --- Quiz Scores (2 of 4)

47

Probability Distributions --- Class Distribution(3 of 4)

48

Probability Distributions --- Class Distribution (4 of 4)

• Central tendency of the

distribution is the mean

or expected value

• Amount of variability is

the variance

49

• Expected value is a measure of the central tendency of

the distribution

where

Expected Value of a Discrete Probability

Distribution (1 of 2)

50

• For the quiz scores

Expected Value of a Discrete Probability

Distribution (2 of 2)

51

where

Xi = random variable’s possible values

E(Xi) = expected value of the random variable

[Xi – E(X)] = difference between each value of the

random variable and the expected

value

E(X) = probability of each possible value of

the random variable

Variance of a Discrete Probability Distribution(1 of 4)

52

• For quiz scores

Variance = (5 – 2.9)2(0.1) + (4 – 2.9)2(0.2) + (3 – 2.9)2(0.3)

+ (2 – 2.9)2(0.3) + (1 – 2.9)2(0.1)

= (2.1)2(0.1) + (1.1)2(0.2) + (0.1)2(0.3)

+ (–0.9)2(0.3) + (–1.9)2(0.1)

= 0.441 + 0.242 + 0.003 + 0.243 + 0.361

= 1.29


53

• Standard deviation is the square root of the variance

where


s = Variance = s 2

= square root

s = standard deviation

For this example:

s = Variance

= 1.29 =1.14

56

• The fundamental rules for continuous random variables

must be modified

o The sum of the probability values must still equal 1

o The probability of each individual value of the random

variable occurring must equal 0 or the sum would be

infinitely large

• The probability distribution is defined by a continuous

mathematical function called the probability density

function or just the probability function represented

by f (X)

Probability Distribution of a Continuous

Random Variable (1 of 3)

57

Graph of Sample Density Function:



58

• For any continuous distribution, the probability does not

change if a single point is added to the range of values

that is being considered.

• The following probabilities are all exactly the same:

P(5.22 < X < 5.26) = P(5.22 < X ≤ 5.26) = P(5.22 ≤ X < 5.26)

= P(5.22 ≤ X ≤ 5.26)



59

• Many business experiments can be characterised by

the Bernoulli process

• The Bernoulli process is described by the binomial

probability distribution

o Each trial has only two possible outcomes

o The probability of each outcome stays the same

from one trial to the next

o The trials are statistically independent

o The number of trials is a positive integer

The Binomial Distribution (1 of 3)

60

• The binomial distribution is used to find the probability of

a specific number of successes in n trials

We need to know

n = number of trials

p = the probability of success on any single trial

We let

r = number of successes

q = 1 – p = the probability of a failure


61

• The binomial formula is

Probability of r success in n trials

The symbol ! means factorial, and n! = n(n − 1)(n − 2)…(1)

4! = (4)(3)(2)(1) = 24

Also, 1! = 1 and 0! = 0 by definition


!

!( )!

r n rnp q

r n r

n = number of trials p = the probability of success on any single trial

r = number of successes q = 1 – p = the probability of a failure

62

Find the probability of getting 4 heads in 5 tosses of a coin

n = 5, r = 4, p = 0.5, and q = 1 − 0.5 = 0.5

P(4 successes in 5 trials)

Solving Problems with the Binomial Formula (1 of 3)

4 5-45!= 0.5 0.5

4!(5-4)!

5(4)(3)(2)(1)= (0.0625)(0.5) = 0.15625

4(3)(2)(1)1!



63

Binomial Distribution for n = 5, p = 0.50




64

Binomial Distribution for n = 5, p = 0.50


65

MSA Electronics is experimenting with the manufacture of

a new transistor:

o Every hour a random sample of 5 transistors is taken

o The probability of one transistor being defective is

0.15

o What is the probability of finding 3, 4, or 5 defective?

n = 5, p = 0.15, and r = 3, 4, or 5

Solving Problems with Binomial Tables (1 of 4)

66

A Sample Table for the Binomial Distribution:

We find the three probabilities in the table for

n = 5, p = 0.15, and r = 3, 4, and 5 and add them together


67


A Sample Table for the Binomial Distribution:

We find the three probabilities in the table for

n = 5, p = 0.15, and r = 3, 4, and 5 and add them together

P(3 or more defects) = P(3) + P(4) + P(5)

= 0.0244 + 0.0022 + 0.0001

= 0.0267

68

Expected value is:

Expected value (mean) = np

Variance = np(1 − p)

For the MSA example:

Expected value = np 5(0.15) = 0.75

Variance = np(1 − p) = 5(0.15)(0.85) = 0.6375


71

• One of the most popular and useful continuous

probability distributions

o The probability density function

o Completely specified by the mean, μ, and the

standard deviation, σ

The Normal Distribution (1 of 4)

2

2

( )

21

( ) =2

x

f X e

72

Normal Distribution with Different Values for μ


73

Normal Distribution with Different Values for σ


74

• Symmetrical with the midpoint representing the mean

• Shifting the mean does not change the shape

• Values on the X axis measured in the number of

standard deviations away from the mean

• As standard deviation becomes larger, curve flattens

• As standard deviation becomes smaller, curve becomes

steeper


75

Step 1:

• Convert the normal distribution into a standard normal

distribution:

o Mean of 0 and a standard deviation of 1

o The new standard random variable is Z

where

X = value of the random variable we want to measure

μ = mean of the distribution

σ = standard deviation of the distribution

Z = number of standard deviations from X to the mean, μ

Using the Standard Normal Table (1 of 4)

=X

Z

76

For μ = 100, σ = 15, find the probability that X is less than

130


130 100= =

15

30= = 2 std dev

15

XZ

Normal Distribution Showing the

Relationship Between Z Values and X

Values:

77

Step 2:

• Look up the probability from a table of normal curve

areas

• Use Appendix A or Table 2.10

• Column on the left is Z value

• Row at the top has second decimal places for Z values


78

Standardised Normal Distribution Function:


79

Builds three- and four-unit

apartment buildings:

o Total construction time follows

a normal distribution

o For triplexes, μ = 100 days and

σ = 20 days

o Contract calls for completion in

125 days

o Late completion will incur a

severe penalty fee

o Probability of completing in 125

days?

Haynes Construction Company (1 of 6)

Normal Distribution for

Haynes Construction

80

• Compute Z


125 – 100= =

20

25= =1.25

20

XZ

From Appendix A,

for Z = 1.25

area = 0.89435

Normal Distribution for

Haynes Construction

The probability is about 0.89

that Haynes will not violate the contract!

81

If finished in 75 days or less, bonus = $5,000

o Probability of bonus?

o Because the distribution is symmetrical, equivalent

to Z = 1.25 … so area = 0.89435


75 – 100

20

–25–1.25

20

XZ

Probability That Haynes Will

Receive the Bonus by Finishing in

75 Days or Less

82

If finished in 75 days or less, bonus = $5,000

o Probability of bonus?

o Because the distribution is symmetrical, equivalent

to Z = 1.25 … so area = 0.89435


75 – 100

20

–25–1.25

20

XZ


Receive the Bonus by Finishing in

75 Days or Less

P(X > 125) = 1.0 − P(X ≤ 125)

= 1.0 − 0.89435 = 0.10565

The probability of completing the contract in 75 days or

less is about 11%

83

Probability of completing between 110 and 125 days?

P(110 < X < 125) = P(X ≤ 125) − P(X < 110)

P(X ≤ 125) = 0.89435


110 – 100

20

100.5

20

XZ


Complete in 110 to 125 Days

84

Probability of completing between 110 and 125 days?

P(110 < X < 125) = P(X ≤ 125) − P(X < 110)

P(X ≤ 125) = 0.89435


110 – 100

20

100.5

20

XZ


Complete in 110 to 125 Days

P(110 ≤ X < 125) = 0.89435 − 0.69146

= 0.20289

The probability of completing between 110 and 125 days

is about 20%

85

Standardised Normal Distribution Function

Standard Normal Distribution

87

For a normally distributed random variable with mean μ

and standard deviation σ

o Approximately 68% of values will be within ±1σ of the

mean

o Approximately 95% of values will be within ±2σ of the

mean

o Almost all (99.7%) of values will be within ±3σ of the

mean

The Empirical Rule (1 of 2)

88

Approximate Probabilities from

the Empirical Rule´:

The Empirical Rule (2 of 2)

89

It is a continuous probability distribution:

o The F statistic is the ratio of two sample

variances

o F distributions have two sets of degrees of freedom

o Degrees of freedom are based on sample size and

used to calculate the numerator and denominator

df1 = degrees of freedom for the numerator

df2 = degrees of freedom for the denominator

o The probabilities of large values of F are very small

The F Distribution (1 of 4)

90


91

Consider the example:

df1 = 5

df2 = 6

= 0.05

From Appendix D, we get

F, df1, df2= F0.05, 5, 6 = 4.39

This means

P(F > 4.39) = 0.05

The probability is only 0.05 F will exceed 4.39


92

F Value for 0.05 Probability with 5 and 6 Degrees of

Freedom:


94

• Also called the negative exponential distribution

o A continuous distribution often used in queuing

models

o Probability function given by

where

X = random variable (service times)

μ = average number of units the service facility can

handle in a specific period of time

e = 2.718 (the base of natural logarithms)

The Exponential Distribution (1 of 2)

( ) xf X e

95

The Exponential Distribution (2 of 2)

96

• Installs new mufflers on automobiles and small trucks

o Can install 3 new mufflers per hour

o Service time is exponentially distributed

• What is the probability that the time to install a new

muffler would be ½ hour or less?

X = Exponentially distributed service time

μ = average number of units the served per time period = 3

per hour

t = ½ hour = 0.5 hour

P(X ≤ 0.5) = 1 − e−3(0.5) = 1 − e −1.5 = 1 = 0.2231 = 0.7769

Arnold’s Muffler Shop (1 of 3)

97

Probability That the Mechanic Will Install a Muffler in 0.5

Hour:


98

Similarly:

And:


13

13

23

23

11 1 1 0.3679 0.6321

3

21 1 1 0.1353 0.8647

3

P X e e

P X e e

P(X > 0.5) = 1 − P(X ≤ 0.5) = 1 − 0.7769 = 0.2231

100

• A discrete probability distribution

o Often used in queuing models to describe arrival

rates over time

o Probability function given by:

where

P(X) = probability of exactly X arrivals or occurrences

= average number of arrivals per unit of time (the mean arrival rate)

e = 2.718, the base of natural logarithms

X = number of occurrences (0, 1, 2, 3, …)

The Poisson Distribution (1 of 3)

( )!

xeP X

X

101

From Appendix C for λ = 2


2 0

2 1 2

2 2 2

( )!

2 (0.1353)1(0) 0.1353 14%

0! 1

2 2 0.1353(2)(1) 0.2706 27%

1! 1 1

2 4 0.1353(4)(2) 0.2706 27%

2! 2(1) 2

xeP X

X

eP

e eP

e eP

102

Sample Poisson Distributions with λ = 2 and λ = 4


104

• Please read Chapter 3!

Homework --- Chapter 2

105

If the sale of ice cream and pizza are independent, then as

ice cream sales decrease by 60 percent during the winter

months, pizza sales will

A. increase by 60 percent.

B. increase by 40 percent.

C. decrease by 40 percent.

D. be unrelated.

Multiple Choice Question

106

If P(A) = 0.3, P(B) = 0.2, P(A and B) = 0.0, what can be

said about events A and B?

A. They are independent.

B. They are mutually exclusive.

C. They are posterior probabilities.

D. They are collectively exhaustive.


107

Disco Fever is randomly found in one half of one percent

of the general population. Testing a swatch of clothing for

the presence of polyester is 99% effective in detecting the

presence of this disease. The test also yields a false-

positive in 4% of the cases where the disease is not

present. What is the probability that the test result is

negative if the disease is absent?

A. 0.95

B. 0.96

C. 0.99

D. 0.995


Chapter 2: Probability Concepts and Applications · • The probability, P, of any event or state...

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