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Chapter 2

To accompanyQuant i tat ive Analysis for Management , Eleventh Edit ion , Global Edit ion by Render, Stair, and HannaPower Point slides created by Brian Peterson

Probab i li ty Concepts and

Appl icat ions



Copyright © 2012 Pearson Education 2-2

In t roduct ion

Life is uncertain; we are not surewhat the future will bring.

Probabi l i ty is a numericalstatement about the likelihoodthat an event will occur.




Fundamental Concepts

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 probabilitiesfor all possible outcomes of anactivity must equal 1.




Diversey Paint Examp le

Demand for white latex paint at Diversey Paintand Supply has always been either 0, 1, 2, 3, or 4gallons per day.

Over the past 200 days, the owner has observed

the following frequencies of demand:QUANTITY

DEMANDEDNUMBER OF DAYS PROBABILITY

0 40 0.20 (= 40/200)

1 80 0.40 (= 80/200)

2 50 0.25 (= 50/200)

3 20 0.10 (= 20/200)

4 10 0.05 (= 10/200)

Total 200 Total 1.00 (= 200/200)




Diversey Paint Examp le

Demand for white latex paint at Diversey Paintand Supply has always been either 0, 1, 2, 3, or 4gallons per day

Over the past 200 days, the owner has observed

the following frequencies of demandQUANTITY

DEMANDEDNUMBER OF DAYS PROBABILITY

0 40 0.20 (= 40/200)

1 80 0.40 (= 80/200)

2 50 0.25 (= 50/200)

3 20 0.10 (= 20/200)

4 10 0.05 (= 10/200)

Total 200 Total 1.00 (= 200/200)

Notice the individual probabilitiesare all between 0 and 1

0 ≤ P (event) ≤ 1

And the total of all eventprobabilities equals 1

∑ P (event) = 1.00




Determining ob ject ive probabi l i ty :

Relative frequency Typically based on historical data

Types o f Probab i l ity

P (event) =Number of occurrences of the event

Total number of trials or outcomes

Classical or logical method

Logically determine probabilities withouttrials

P (head) =1

2

Number of ways of getting a head

Number of possible outcomes (head or tail)


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Types o f Probabi l i ty

Subject ive probabi l i ty is based onthe experience and judgment of theperson making the estimate.

Opinion polls

Judgment of experts

Delphi method


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Mutual ly Exc lus ive Even ts

Events are said to be mutual ly

exclus ive if only one of the events canoccur on any one trial.

Tossing a coin will resultin either a head or a tail.

Rolling a die will result inonly one of six possibleoutcomes.




Col lect ively Exhaus t ive Events

Events are said to be collectivelyexhaustive if the list of outcomesincludes every possible outcome.

Both heads andtails as possibleoutcomes of coin flips.

All six possibleoutcomesof the rollof a die.

OUTCOMEOF ROLL

PROBABILITY

1 1 /6

2 1 /6

31

/64 1 /6

5 1 /6

6 1 /6

Total 1




Stat ist ical ly Independen t Even ts

Events may be either independent or dependent. For independent events, the occurrence

of one event has no effect on theprobability of occurrence of the secondevent.




Which Sets o f Events A re 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) Chicago Cubs win the National League pennant

(b) Chicago Cubs win the World Series

4. (a) Snow in Santiago, Chile

(b) Rain in Tel Aviv, Israel

Dependent events

Dependent

events

Independent

events

Ind ependent events




Three Types o f Probab i l i t ies

Marginal (or s imple ) probability is just theprobability of a single event occurring.

P (A)

Jo in t probability is the probability of two or more

events occurring and is equal to the product of their marginal probabilities for independentevents.

P (AB ) = P (A) x P (B )

Condi t ional probability is the probability of eventB 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)




The Binom ial Distr ibut ion

Many business experiments can becharacterized by the Bernoulli process.

The Bernoulli process is described by the

binomial probability distribution.1. Each trial has only two possible outcomes.

2. The probability of each outcome stays thesame from one trial to the next.

3. The trials are statistically independent.4. The number of trials is a positive integer.





The binomial distribution is used to find theprobability of a specific number of successesin n trials.

We need to know:n = number of trials

p = the probability of success on anysingle trial

We letr = number of successes

q = 1 – p = the probability of a failure





The binomial formula is:

r n r q p

r n r

n n r

)!(!

!trialsinsuccessesof yProbabilit

The symbol ! means factorial, and

n ! = n (n – 1)(n – 2)…(1)

For example

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

By definition

1! = 1 and 0! = 1





NUMBER OFHEADS (r ) Probability = (0.5)r (0.5)5 – r

5!r !(5 – r )!

0 0.03125 = (0.5)0(0.5)5 – 0

1 0.15625 = (0.5)1(0.5)5 – 1

2 0.31250 = (0.5)2(0.5)5 – 2

3 0.31250 = (0.5)3

(0.5)5 – 3

4 0.15625 = (0.5)4(0.5)5 – 4

5 0.03125 = (0.5)5(0.5)5 – 5

5!

0!(5 – 0)!5!

1!(5 – 1)!5!

2!(5 – 2)!5!

3!(5 – 3)!5!

4!(5 – 4)!

5!5!(5 – 5)!

Table 2.7

Binomial Distribution for n = 5 and p = 0.50.




Solv ing Problems w i th the

Binom ial Formula

We want to find the probability of 4 heads in 5 tosses.

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

Thus454 5.05.0

)!45(!4

!5)trials5insuccesses4(

P

156250500625011234

12345

.).)(.()!)()()((

))()()((




Solv ing Problems w i th the

Binom ial Formula

P r o b a b i l i t y

P ( r

)

| | | | | | |1 2 3 4 5 6

Values of r (number of successes)

0.4 –

0.3 –

0.2 –

0.1 –

0 –

Figure 2.7

Binomial Probability Distribution for n = 5 and p = 0.50.




Solv ing Problems w ith

B inom ial Tables

MSA Electronics is experimenting with themanufacture of a new transistor.

Every hour a random sample of 5 transistors istaken.

The probability of one transistor being defectiveis 0.15.

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

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

We could use the formula to solve this problem,but using the table is easier.





B inom ial Tables

P

n r 0.05 0.10 0.15

5 0 0.7738 0.5905 0.4437

1 0.2036 0.3281 0.3915

2 0.0214 0.0729 0.13823 0.0011 0.0081 0.0244

4 0.0000 0.0005 0.0022

5 0.0000 0.0000 0.0001

Table 2.8 (partial)

We find the three probabilities in the tablefor n = 5, p = 0.15, and r = 3, 4, and 5 andadd them together.




Table 2.8 (partial)

We find the three probabilities in the tablefor n = 5, p = 0.15, and r = 3, 4, and 5 andadd them together


B inom ial Tables

P

n r 0.05 0.10 0.15

5 0 0.7738 0.5905 0.4437

1 0.2036 0.3281 0.3915

2 0.0214 0.0729 0.13823 0.0011 0.0081 0.0244

4 0.0000 0.0005 0.0022

5 0.0000 0.0000 0.0001

)()()()( 543defectsmoreor 3 P P P P

02670000100022002440 ....




The Normal Distr ibu t ion

The no rmal dis t ribut ion is the one of themost popular and useful continuousprobability distributions.

The formula for the probability density

function is rather complex:

2

2

2

2

1

)(

)(

x

e X f

The normal distribution is specifiedcompletely when we know the mean, µ ,and the standard deviation, .





The normal distribution is symmetrical,with the midpoint representing the mean.

Shifting the mean does not change the

shape of the distribution. Values on the X axis are measured in the

number of standard deviations away fromthe mean.

As the standard deviation becomes larger,the curve flattens.

As the standard deviation becomessmaller, the curve becomes steeper.




µ


Figure 2.9

Same µ , smaller

Same µ , larger




Haynes Cons truct ion Company

Haynes builds three- and four-unit apartmentbuildings (called triplexes and quadraplexes,respectively).

Total construction time follows a normal

distribution. For triplexes, µ = 100 days and = 20 days.

Contract calls for completion in 125 days,and late completion will incur a severe

penalty fee. What is the probability of completing in 125

days?





From Appendix A, for Z = 1.25the area is 0.89435.

The probability is about

0.89 that Haynes will notviolate the contract.

20

100125

X Z

251

20

25.

µ = 100 days X = 125 days

= 20 daysFigure 2.11





Suppose that completion of a triplex in 75days or less will earn a bonus of $5,000.

What is the probability that Haynes will getthe bonus?





But Appendix A has onlypositive Z values, and the

probability we are looking for is in the negative tail.

20

10075

X Z

251

20

25.

Figure 2.12

µ = 100 daysX = 75 days

P(X < 75 days)Area of

Interest





Because the curve issymmetrical, we can look atthe probability in the positivetail for the same distance

away from the mean.

20

10075

X Z

251

20

25.


P(X > 125 days)Area of

Interest






We know the probabilitycompleting in125 days is 0.89435.

So the probability

completing in morethan 125 days is1 – 0.89435 = 0.10565.





µ = 100 daysX = 75 days

The probability of completing in less than 75days is 0.10565.

Going back to theleft tail of thedistribution:

The probabilitycompleting in more than125 days is1 – 0.89435 = 0.10565.





What is the probability of completing a triplexwithin 110 and 125 days?

We know the probability of completing in 125days, P (X < 125) = 0.89435.

We have to complete the probability of completing in 110 days and find the area

between those two events.





From Appendix A, for Z = 0.5 thearea is 0.69146.

P (110 < X < 125) = 0.89435 – 0.69146 = 0.20289.

20

100110

X Z

50

20

10.

Figure 2.13

µ = 100days

125days

= 20 days

110days




The Empir ical Rule

For a normally distributed random variable withmean µ and standard deviation , then

1. About 68% of values will be within±1 of the

mean.

2. About 95.4% of values will be within±2 of themean.

3. About 99.7% of values will be within±3 of themean.




The Empir ical Rule

Figure 2.14

68%16% 16%

–1 +1

a µ b

95.4%2.3% 2.3%

–2 +2 a µ b

99.7%0.15% 0.15%

–3 +3 a µ b




The Exponen t ial Distr ibu t ion

The exponent ial dist r ibu t ion (also calledthe negat ive exponent ial d ist r ibut ion ) is acontinuous distribution often used inqueuing models to describe the timerequired to service a customer. Itsprobability function is given by:

x e X f

)(

where

X = random variable (service times)

µ = average number of units the service facility canhandle in a specific period of time

e = 2.718 (the base of natural logarithms)




The Exponen t ial Distr ibu t ion

timeserviceAverage1

valueExpected

2

1Variance

f (X )

X

Figure 2.17




Arnold’s Muffler Shop

Arnold’s Muffler Shop installs newmufflers on automobiles and small trucks.

The mechanic can install 3 new mufflers

per hour. Service time is exponentially distributed.

What is the probability that the time to install a new

muffler would be ½ hour or less?





Here:

X = Exponentially distributed service time

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

t = ½ hour = 0.5hour

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





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

Note also that if:

Then it must be the case that:

P(X>0.5) = 1 - 0.7769 = 0.2231




The Poisson Distr ibut ion

The Poisson dis t r ibut ion is a discrete distribution that is often used in queuingmodels to describe arrival rates over time.Its probability function is given by:

!)(

X

e X P

x

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 = specific value (0, 1, 2, 3, …) of the random variable




The Poisson Distr ibut ion

The mean and variance of the distribution are both.

Expected value =

Variance =




Poisson Dist r ibu t ion

We can use Appendix C to find Poisson probabilities.Suppose that λ = 2. Some probability calculations are:

2706.02

)1353.0(4

!2

2)2(

2706.01

)1353.0(2

!1

2

)1(

1353.01

)1353.0(1

!0

2)0(

!)(

22

21

20

e P

e

P

e P

X

e X P

x




Poisson Dist r ibu t ion

Figure 2.19

Sample Poisson Distributions with λ = 2 and λ = 4




Exponent ial and Poisson

Together

If the number of occurrences per timeperiod follows a Poisson distribution, thenthe time between occurrences follows anexponential distribution: Suppose the number of phone calls at a

service center followed a Poisson distributionwith a mean of 10 calls per hour.

Then the time between each phone call would

be exponentially distributed with a mean timebetween calls of 6 minutes (1/10 hour).



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photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the UnitedStates of America.

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