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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.

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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.

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• 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

For example, the market for the new Iphone X might be

“good” with a chance of 60% (a probability of 0.6) or “not

good” with a chance of 40% (a probability of 0.4).

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• 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

Types of Probability (1 of 3)

P(event) =Number of occurrences of the event

Total numbers of trials or outcomes

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• 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)

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• 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

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• Objective Approach

o Classical or logical method

• Perform a series of trials

Types of Probability (2 of 3)

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

( )

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

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• Subjective Approach

o When logic and past history are not appropriate,

probability values can be assessed subjectively!

o Based on the experience and judgment of the

person making the estimate

• Opinion polls (e.g. to determine election returns)

• Judgment of experts

• Delphi method (a panel of experts is assembled to

make their predictions of the future)

Types of Probability (3 of 3)

Example: What is the probability that the Chinese

economy will be in a severe depression in 2018?

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• 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)

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• 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)

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Venn Diagrams

A B

Events that are mutually exclusive

Events that are notmutually exclusive

A B

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• 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

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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)

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• 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)

Unions and Intersections of Events (2 of 3)

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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

Unions and Intersections of Events (3 of 3)

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• 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

Probability Rules (2 of 5)

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)

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• 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

Probability Rules (5 of 5)

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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!

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)

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Dependent Events (2 of 3)

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• The conditional probability that the ball drawn is

lettered, given that it is yellow

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

Dependent Events (3 of 3)

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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

Probability Rules (3 of 5)

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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

Probability Rules (4 of 5)

Dependent events

Independent events

Independent events

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• 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)

This means that we can take new or recent

data and then revise and improve upon our

old probability estimates for an event!

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• 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?

Revising Probabilities with Bayes’ Theorem (2 of 7)

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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

Revising Probabilities with Bayes’ Theorem (3 of 7)

We can answer these questions by using the formula for joint

probability under statistical dependence and Bayes’ theorem.

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• A 3 can occur in combination with the state “fair die” or

in combination with the state “loaded die”.

• The sum of their probabilities gives the unconditional

(marginal probability) of a 3 on the toss:

P(3) = 0.083 + 0.300 = 0.383

Revising Probabilities with Bayes’ Theorem (4 of 7)

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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

Revising Probabilities with Bayes’ Theorem (5 of 7)

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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

• These are the revised or posterior probabilities for the

next roll of the die

• We use these to revise our prior probability estimates

Revising Probabilities with Bayes’ Theorem (6 of 7)

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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)

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• Please read Chapter 3!

Homework --- Chapter 2