probability concepts and applications

Probability Concepts and Probability Concepts and ApplicationsApplications

Chapter Outline

2.1 Introduction

2.2 Fundamental Concepts

2.3 Mutually Exclusive and Collectively Exhaustive Events

2.4 Statistically Independent Events

2.5 Statistically Dependent Events

2.6 Revising Probabilities with Bayes’ Theorem

Chapter Outline - continued

2.7 Further Probability Revisions

2.8 Random Variables

2.9 Probability Distributions

2.10 The Binomial Distribution

2.11 The Normal Distribution

2.12 The Exponential Distribution

2.13 The Poisson Distribution

Introduction

• Life is uncertain!• We must deal with risk!

• A probability is a numerical statement about the likelihood that an event will occur

Basic Statements About Probability

1. 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

2. The sum of the simple probabilities for all possible

outcomes of an activity must equal 1.

Example 2.1

• Demand for white latex paint at Diversey Paint

and Supply has always been 0, 1, 2, 3, or 4

gallons per day. (There are no other possible

outcomes; when one outcome occurs, no other

can.) Over the past 200 days, the frequencies of

demand are represented in the following table:

Example 2.1 - continued

Quantity Demanded (Gallons)

0

1

2

3

4

Number of Days

40

80

50

20

10

Total 200

Frequencies of DemandFrequencies of Demand

Example 2.1 - continued

Quant. Freq.

Demand (days)

0 40

1 80

2 50

3 20

4 10

Total days = 200

Probability

(40/200) = 0.20

(80/200) = 0.40

(50/200) = 0.25

(20/200) = 0.10

(10/200) = 0.05Total Prob = 1.00

Probabilities of DemandProbabilities of Demand

Types of Probability

Objective probability:

Determined by experiment or observation:– Probability of heads on coin flip– Probably of spades on drawing card from deck

occurrencesor outcomes ofnumber Total

occursevent timesofNumber )( eventP

Types of Probability

Subjective probability:

Based upon judgement

Determined by:

– judgement of expert

– opinion polls

– Delphi method

– etc.

Mutually Exclusive Events

• Events are said to be mutually

exclusive if only one of the events

can occur on any one trial

Collectively Exhaustive Events

• Events are said to be collectively

exhaustive if the list of outcomes

includes every possible outcome:

heads and tails as possible outcomes

of coin flip

Example 2

Outcome

of Roll

1

2

3

4

5

6

Probability

1/6

1/6

1/6

1/6

1/6

1/6

Total = 1

Rolling a die has six possible outcomes

Example 2a

Outcome

of Roll = 5

Die 1 Die 2

1 4

2 3

3 2

4 1

Probability

1/36

1/36

1/36

1/36

Rolling two dice

results in a total of

five spots

showing. There

are a total of 36

possible outcomes.

Example 3

Draw a spade or a club

Draw a face card or a

number card

Draw an ace or a 3

Draw a club or a nonclub

Draw a 5 or a diamond

Draw a red card or a

diamond

Yes No

Yes Yes

Yes No

Yes Yes

No No

No No

Draw Mutually Collectively Exclusive Exhaustive

Probability : Mutually Exclusive

P(event A or event B) =

P(event A) + P(event B)

or:

P(A or B) = P(A) + P(B)

i.e.,

P(spade or club) = P(spade) + P(club)

= 13/52 + 13/52

= 26/52 = 1/2 = 50%

Probability:Not Mutually Exclusive

P(event A or event B) =

P(event A) + P(event B) -

P(event A and event B both occurring)

or

P(A or B) = P(A)+P(B) - P(A and B)

P(A and B)(Venn Diagram)

P(A) P(B)

P(A and B

)

P(A or B)

+ -

=

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

P(A or B)

Statistical Dependence

• Events are either

– statistically independent (the occurrence of one

event has no effect on the probability of

occurrence of the other) or

– statistically dependent (the occurrence of one

event gives information about the occurrence of

the other)

Which Are Independent?

• (a) Your education

(b) Your income level

• (a) Draw a Jack of Hearts from a full 52 card deck

(b) Draw a Jack of Clubs from a full 52 card deck

Probabilities - Independent Events

• Marginal probability: the probability of an event occurring:

[P(A)]• Joint probability: the probability of multiple,

independent events, occurring at the same time

P(AB) = P(A)*P(B)

• Conditional probability (for independent events): – the probability of event B given that event A has

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

– or, the probability of event A given that event B has

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

Probability(A|B) Independent Events

P(B

)

P(A

)

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

Statistically Independent Events

1. P(black ball drawn on first draw)

2. P(two green balls drawn)

A bucket contains 3 black balls, and 7 green balls. We draw a ball from the bucket, replace it, and draw a second ball

Statistically Independent Events - continued

1. P(black ball drawn on second draw, first draw was

green)

2. P(green ball drawn on second draw, first draw was

green)

Probabilities - Dependent Events

• Marginal probability: probability of an event

occurring P(A)

• Conditional probability (for dependent events): – the probability of event B given that event A has

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

– the probability of event A given that event B has

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

Probability(A|B)

/

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

P(AB) P(B)P(A)

Probability(B|A)

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

/

P(AB)P(B) P(A)

Statistically Dependent Events

Assume that we have an urn containing 10 balls of the following descriptions:

•4 are white (W) and lettered (L)

•2 are white (W) and numbered N

•3 are yellow (Y) and lettered (L)

•1 is yellow (Y) and numbered (N)

Then:

• P(WL) = 4/10 = 0.40

• P(WN) = 2/10 = 0.20

• P(W) = 6/10 = 0.60

• P(YL) = 3/10 = 0.3

• P(YN) = 1/10 = 0.1

• P(Y) = 4/10 = 0.4

Statistically Dependent Events - Continued

Then:– P(L|Y) = P(YL)/P(Y)

= 0.3/0.4 = 0.75

– P(Y|L) = P(YL)/P(L)

= 0.3/0.7 = 0.43

– P(W|L) = P(WL)/P(L)

= 0.4/0.7 = 0.57

Joint Probabilities, Dependent Events

Your stockbroker informs you that if the stock market

reaches the 10,500 point level by January, there is a 70%

probability the Tubeless Electronics will go up in value.

Your own feeling is that there is only a 40% chance of the

market reaching 10,500 by January.

What is the probability that both the stock market will reach

10,500 points, and the price of Tubeless will go up in

value?

Joint Probabilities, Dependent Events - continued

Then:

P(MT) =P(T|M)P(M)

= (0.70)(0.40)

= 0.28

Let M represent the event of

the stock market reaching

the 10,500 point level, and T

represent the event that

Tubeless goes up.

Revising Probabilities: Bayes’ Theorem

Bayes’ theorem can be used to calculate revised or posterior probabilities

Prior Probabilities

Bayes’ Process

Posterior Probabilities

New Information

General Form of Bayes’ Theorem

Aevent theof complementA where

)()|()()|(

)()|()|(

)(

)()|(

APABPAPABP

APABPBAP

orBP

ABPBAP

die.unfair"" is Aevent then the

die,fair"" isA event theifexample,For

Posterior Probabilities

A cup contains two dice identical in appearance.

One, however, is fair (unbiased), the other is loaded (biased).

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

The probability of tossing the same number on the loaded die is 0.60.

We have no idea which die is which, but we select one by chance, and toss it. The result is a 3.

What is the probability that the die rolled was fair?

Posterior Probabilities Continued

• We know that:P(fair) = 0.50 P(loaded) = 0.50

• And:

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

• Then:P(3 and fair) = P(3|fair)P(fair)

= (0.166)(0.50)

= 0.083

P(3 and loaded) = P(3|loaded)P(loaded)

= (0.60)(0.50)

= 0.300

Posterior Probabilities Continued

• 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 or marginal probability

of a 3 on a toss:

P(3) = 0.083 + 0.0300 = 0.383.

• Then, the probability that the die rolled was the fair one is

given by:

0.22 0.383

0.083

P(3)

3) andP(Fair 3)|P(Fair

Further Probability Revisions

• To obtain further information as to whether the die

just rolled is fair or loaded, let’s roll it again.

• Again we get a 3.

Given that we have now rolled two 3s, what is the

probability that the die rolled is fair?

Further Probability Revisions - continued

P(fair) = 0.50, P(loaded) = 0.50 as before

P(3,3|fair) = (0.166)(0.166) = 0.027

P(3,3|loaded) = (0.60)(0.60) = 0.36

P(3,3 and fair) = P(3,3|fair)P(fair)

= (0.027)(0.05)

= 0.013

P(3,3 and loaded) = P(3,3|loaded)P(loaded)

= (0.36)(0.5)

= 0.18

P(3,3) = 0.013 + 0.18 = 0.193

Further Probability Revisions - continued

933.00.193

0.18

P(3,3)

Loaded) and P(3,3 3,3)|P(Loaded

067.00.193

0.013

P(3,3)

Fair) and P(3,33,3)|P(Fair

To give the final comparison:

P(fair|3) = 0.22

P(loaded|3) = 0.78

P(fair|3,3) = 0.067

P(loaded|3,3) = 0.933

Further Probability Revisions - continued

Random Variables

• Discrete random variable - can assume only a

finite or limited set of values- i.e., the number of

automobiles sold in a year

• Continuous random variable - can assume any

one of an infinite set of values - i.e., temperature,

product lifetime

Random Variables (Numeric)

Experiment Outcome Random Variable Range of Random Variable

Stock 50 Xmas trees

Number of trees sold

X = number of trees sold

0,1,2,, 50

Inspect 600 items

Number acceptable

Y = number acceptable

0,1,2,…, 600

Send out 5,000 sales letters

Number of people e responding

Z = number of people responding

0,1,2,…, 5,000

Build an apartment building

%completed after 4 months

R = %completed after 4 months

0R100

Test the lifetime of a light bulb (minutes)

Time bulb lasts - up to 80,000 minutes

S = time bulb burns

0S80,000

Random Variables (Non-numeric)

Experiment Outcome Random

Variable

Range of

Random

Variable

Students respond to a questionnaire

Strongly agree (SA)Agree (A)Neutral (N)Disagree (D)Strongly Disagree (SD)

X = 5 if SA4 if A3 if N2 if D1 if SD

1,2,3,4,5

One machine is inspected

DefectiveNot defective

Y = 0 if defective1 if not defective

0,1

Consumers respond to how they like a product

GoodAveragePoor

Z = 3 if good2 if average1 if poor

1,2,3

Probability Distributions

Table 2.4

Outcome X Number Responding

P(X)

SA 5 10 0.10

A 4 20 0.20

N 3 30 0.30

D 2 30 0.30

SD 1 10 0.10

D

0.00

0.05

0.10

0.15

0.20

0.25

0.30

1 2 3 4 5

Figure 2.5Probability Function

Expected Value of a Discrete Probability Distribution

n

iii )X(PX)X(E

2.9

)1.0)(1()3.0)(2(

)3.0)(3()2.0)(4()1.0)(5(

)()(

)()()(

)()(

5544

332211

5

1

XPXXPX

XPXXPXXPX

XPXXE ii

i

Variance of a Discrete Probability Distribution

i

n

ii XPXEX

1

22

29.1

0.3610.2430.0030.242- 0.44

)1.0()9.21(

)3.0(2.9)-(2 3.09.23

2.09.241.09.25

2

22

222

Binomial Distribution

Assumptions:

1. Trials follow Bernoulli process – two possible outcomes

2. Probabilities stay the same from one trial to the next

3. Trials are statistically independent

4. Number of trials is a positive integer

Binomial Distribution

rnr qp r)!-(nr!

n!

Probability of r successes in n trials

n = number of trials

r = number of successes

p = probability of success

q = probability of failure

Binomial Distribution

)p(np

np

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

1 2 3 4 5 6

(r) Number of Successes

P(r

)

N = 5, p = 0.50

Binomial Distribution

Probability Distribution Continuous Random Variable

Probability density function - f(X)

5 5.05 5.1 5.15 5.2 5.25 5.3 5.35 5.4

Normal Distribution

2

2)(2/1

2

1)(

X

eXf

Normal Distribution for Different Values of

0

30 40 50 60 70

=50 =60=40

0 0.5 1 1.5 2

Normal Distribution for Different Values of

=0.1

=0.2=0.3

= 1

Three Common Areas Under the Curve

• Three Normal distributions with different areas

Three Common Areas Under the Curve

Three Normal

distributions

with different

areas

The Relationship Between Z and X

55 70 85 100 115 130 145

-3 -2 -1 0 1 2 3

x

Z=100

=15

Haynes Construction Company Example Fig. 2.12

Haynes Construction Company ExampleFig. 2.13

Haynes Construction Company Example Fig. 2.14