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Theoretical Probability Models
Dr. Yan LiuDepartment of Biomedical, Industrial & Human Factors Engineering
Wright State University
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Introduction
Use Theoretical Probability Models When They Describe the Physical Model “Adequately” The results of intelligent tests ~ Normal Distribution The length of a telephone call ~ Exponential Distribution The number of people arriving at a bank within an hour ~ Poisson Distribution The number of defects in a bottle production line ~ Binomial Distribution
Discrete Distributions Binomial and Poisson distributions
Continuous Distributions Exponential, Normal, and Beta distributions
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Binomial Distribution Characteristics
Totally n trials Dichotomous outcomes
Each trial results in one of two possible outcomes (e.g. yes/no, true/false) Constant Probability
Each trial has the same probability of success, p Independence
Different trials are independent
Probability Mass Function (PMF)
),,1,0( nx
X = “# of successes in a sequence of n independent trials, and the probability of success in each trial is p” (Success means the occurrence of an event)
xnx ppx
npnxXpnBX
)(1),|Pr(),(~
: number of ways you can choose x successes from n trials
(See Appendix A)
x
n)!(!
!xnx
n
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Binomial Distribution (Cont.)
Expected Value
Variance
npXE )(
)1()var()var( pnpYX
Y = “# of failures in a sequence of n independent trials”, then Y = n – X
yny ppy
npnyYpnBY
)()1())1(,|Pr())1(,(~
Cumulative Distribution Function
x
i
ini ppi
n
pnxXpnxFpnBX
0
)(1
),|Pr(),|(),(~
(See Appendix B)
)1()( pnYE
5
Pr(Hit| X=5)=?
Let X = number of people out of 20 tasters who preferred the new pretzel
You are planning to sell a new pretzel, and you want to know whether it will be a success or not. If your pretzel is a “Hit”, you expect to gain 30% of the market. If it is a “Flop”, on the other hand, the market share is only 10%. Initially, you judged these outcomes to be equally likely. You decided to test the market first and found out that 5 out of 20 people preferred your pretzel to the competing product. Given the new data, what do you think of the chance of your pretzel being a Hit?
(Bayes Theorem)
)Flop5Pr()Hit5Pr()5Pr( XXX
Pr(Flop)Flop)|5Pr(Pr(Hit)Hit)|5Pr( XX
0.5 0.5? ?
Pretzel Example
)5Pr()HitPr()Hit|5Pr(
XX
Pr(Hit | X = 5) =
6
)5|HitPr( X
In conclusion, the new data suggest that the new pretzel is very likely to be a hit
179.07.03.05
20)Hit|5Pr()3.0,20(~Hit| 155
XpnBX
032.09.01.05
20)Flop|5Pr()1.0,20(~Flop| 155
XpnBX
1055.05.0032.05.0179.0
)FlopPr()Flop|5Pr()HitPr()Hit|5Pr()5Pr(
XXX
848.0== 1055.05.0•179.0
)5=Pr()HitPr()Hit|5=Pr(
XX
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Poisson Distribution Represent occurrences of events over a unit of measure (time or
space) e.g. number of customers arriving, number of breakdowns occurring
Assumptions Events can happen at any point along a continuum At any particular point, the probability of an event is small (i.e. events do not
happen frequently) Events happen independently of one another The average number of events is constant over a unit of measure
Probability Mass Function
=)|=Pr(⇔)(Poisson~ mxXmXX = “# of events in a unit of measure”
m is the average number of events in a unit of measure
(See Appendix C)
mx
m ex !
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Poisson Distribution (Cont.)
Expected Value
Variance
mXE )(
mX )var(
=)•|=Pr(⇔)•(Poisson~ tmyYtmY
Y = “# of events in t units of measure”tm
ytm e
y !)(
tmYE )( tmY )var(
Cumulative Distribution Function
mx
ii
m emxXmxFmXi -
0=!∑=)|≤Pr(=)|(⇔)(Poisson~
(See Appendix D)
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Based on your previous market research, you decide to invest in a pretzel stand. Now you need to select a good location. You consider a location to be “good”, “bad”, or “dismal” if you sell 20, 10, or 6 pretzels per hour, respectively. You have found a new stand and your initial judgment is that the probabilities of the location being good, bad, and dismal are 0.7, 0.2, and 0.1, respectively. After having the stand for a week, you decided to run a test. Within 30 minutes, you sold 7 pretzels. Now, what are your probabilities regarding the quality of the stand?
(Bayes Theorem)
Let X=number of pretzels sold within 30 minutes or 0.5 hour
Pr(Good | X = 7) = ?
)DismalPr()Dismal|7Pr()BadPr()Bad|7Pr()GoodPr()Good|7Pr(
)Dismal7Pr()Bad7Pr()Good7Pr()7Pr(
XXX
XXXX
=)7=|GoodPr( X)7=Pr(
)GoodPr()Good|7=Pr(X
X
0.7? ? ?
Pretzel Example (Cont.)
0.2 0.1
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In light of the new data, you feel that the chance of the current stand being a good location has slightly increased and thus you should stay.
09.0)Good|7Pr()10205.0(~Good| 10!7
107 eXmPoissonX
104.0)Bad|7Pr()5105.0(~Bad| 5!7
57 eXmPoissonX
022.0)Dismal|7Pr()365.0(~Dismal| 3!7
37 eXmPoissonX
086.01.0022.02.0104.07.009.0)DismalPr()Dismal|7Pr(
)BadPr()Bad|7Pr()GoodPr()Good|7Pr()7Pr(
X
XXX
)7|GoodPr( X 733.0== 086.07.0•09.0
)7=Pr()GoodPr()Good|7=Pr(
XX
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Exponential Distribution If the number of events occurring within a unit of measure follows a
Poisson distribution, then the time or space between the occurrence of two events follows an exponential distribution
Exponential distribution has the same assumptions as Poisson distribution
Probability Density Function
Let T =“Time (space) between two consecutive events”
mtemtTmtF 1)|Pr()|(
)0≥(=)|(⇔)(Exp~ - tmemtfmT mt
m is the same average rate used in Poisson distribution
Cumulative Distribution Function
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Exponential Distribution (Cont.)
Expected Value
Variance
Other Important Probabilities
mTE /1)(
2/1)var( mT
btat
mtmt
eemaTmbTmbTaeemtTmtT
--
--
-=)|≤Pr(-)|≤Pr(=)|≤<Pr(=)-1(-1=)|<Pr(-1=)|>Pr(
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You wonder if you can provide fast service to your customers. It takes 3.5 minutes to cook a pretzel, so what is the probability that the next customer arrives before the pretzel is finished?As in the previous example, you assume customers arrive according to a Poisson process, and you consider your location being good, bad or dismal if you sell 20, 10, 6 pretzels per hour, respectively. Your prior belief is that Pr(Good)=0.7, Pr(Bad)=0.2, and Pr(Dismal)=0.1.
Let T=the time between two consecutive customers
)DismalPr()Dismal|5.3Pr()BadPr()Bad|5.3Pr()GoodPr()Good|5.3Pr(
)Dismal5.3Pr()Bad5.3Pr()Good5.3Pr()5.3Pr(
TTT
TTTT
Pr(T<3.5) = ?
0.7 0.2 0.1? ? ?
Pretzel Example (Cont.)
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295.0=-1=)Dismal|5.3<Pr(⇔min)/=min/=(~Dismal|
442.0=-1=)Bad|5.3<Pr(⇔min)/=min/=(~Bad|
689.0=-1=)Good|5.3<Pr(⇔min)/=min/=(~Good|
5.3•-101
606
5.3•-61
6010
5.3•-31
6020
101
61
31
eTmExpT
eTmExpT
eTmExpT
6.01.0295.02.0442.07.0689.0
)DismalPr()Dismal|5.3Pr(
)BadPr()Bad|5.3Pr()GoodPr()Good|5.3Pr()5.3Pr(
T
TTT
In other words, about 60% of your customers will have to wait until the pretzel is ready. Therefore, the fast service does not seem very appealing.
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Normal Distribution Bell-Shaped Curve Particularly good for modeling situations in which the uncertain
quantity is subject to many different sources of errors many measured biological phenomena (e.g. height, weight, length)
Probability Density Function
Expected Value: Variance: Some Handy Empirical Rules
),|(),(~ xfNX
)(XE2=)var( σX
99.0)33Pr(
95.0)22Pr(
68.0)Pr(
X
X
X
22
2)(
21
x
e
μ=2μ-σ μ+σ
μ-2σ
μ+2σ
μ-3σ
μ+3σ
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Normal Distribution (Cont.) Standard Normal Distribution Convert to Standard Normal Distribution
)1,0|Pr(),|Pr( aZaX
,
)1,0(~),(~ NZNX X
(See Appendix E for Cumulative Probability)
z P(Z≤z)
X ~ N(μ=10, σ2=400), then the probability X is less than or equal to 35 is (Appendix E)8944.0)25.1ZPr()ZPr()35XPr(
4001035
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Normal Distribution (Cont.) Other Important Probabilities
)Pr()Pr(
)Pr()Pr()Pr(
ab
babXa
ZZ
ZbXa
Because standard normal distribution is symmetric around zero,
7888.01056.08944.0)25.1Pr()25.1Pr(
)25.125.1Pr()Pr()3515Pr(400
1035400
1015
ZZ
ZZX
)<Pr(-1=)≥Pr(=)-≤Pr( zZzZzZ
1056.0=8944.0-1=)25.1≤Pr(-1=)25.1≥Pr(=)25.1-≤Pr(=)≤Pr(=)15-≤Pr( 400
1015 ZZZZX
X ~ N(μ=10, σ2=400), then
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Standard Normal Distribution
-z
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Your plant manufactures disk drivers for personal computers. One of your machines produces a part that is used in the final assembly. The width of the part is important to the proper functioning of the disk driver. If the width falls below 3.995 or above 4.005 mm, the disk driver will not work properly and must be repaired at a cost of $10.40. The machine can be set to produce parts with width of 4mm, but it is not perfectly accurate. In fact, the width is normally distributed with mean 4 and the variance depends on the speed of the machine. The standard deviation of the width is 0.0019 at the lower speed and 0.0026 at the higher speed. Higher speed means lower overall cost of the disk driver. The cost of the driver is $20.45 at the higher speed and $20.75 at the lower speed. Should you run the machine at the higher or lower speed?
Quality Control Example
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Let X = width of a disk driver
(P1=?) $20.75+$10.40=$31.15
$20.75
$20.45+$10.40=$30.85
$20.45
Cost/Driver
Low Speed
High Speed
X≤3.995 or X≥4.005 (Defective)
3.995 ≤ X≤4.005 (Not Defective)
X≤3.995 or X≥4.005 (Defective)
3.995 ≤ X≤4.005 (Not Defective)(P2=?)
P1=Pr(Defective | Low Speed) P2=Pr(Defective | High Speed)
0086.09914.01
)SpeedLow|DefectiveNotPr(1)SpeedLow|DefectivePr(
)0019.0,4|005.4995.3Pr()SpeedLow|DefectiveNotPr(
)0019.0,4(~SpeedLow|
X
NX
9914.00043.09957.0
)63.2Pr()63.2Pr()63.263.2Pr()Pr( 0019.04005.4
0019.04995.3
ZZZZ
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E(Cost|Low Speed)=0.0086∙31.15+0.9914∙20.75=$20.84
E(Cost|High Speed)=0.0548∙30.85+0.9452∙20.45=$21.02
Conclusion: Because E(Cost|Low Speed)<E(Cost|High Speed), you should run the machine at the lower speed
9452.00274.09726.0
)92.1Pr()92.1Pr()92.192.1Pr()Pr( 0026.04005.4
0026.04995.3
ZZZZ
)0026.0,4|005.4995.3Pr()SpeedHigh|DefectiveNotPr(
)0026.0,4(~SpeedHigh|
X
NX
0548.09452.01
)SpeedHigh|DefectiveNotPr(1)SpeedHigh|DefectivePr(
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Beta Distribution Useful in modeling an uncertain ratio or proportion (ranging from 0 to 1)
e.g the proportion of voters who will vote for the Republican candidate
Probability Density Function
Let Q=“the proportion of interest”
n, r are parameters that determine the shape of f(q|n,r). n determines the “tightness” of the distribution; the larger n is, the tighter the distribution is. r determines the “skewness” of the distribution. In particular, When r = n/2, the distribution is symmetric around 0.5. Otherwise, the distribution is skewed to the right and left when r < n/2 and r > n/2, respectively.
(See Appendix F for Cumulative Probability)
,3,2,1,)!1()(
)1(),|(),(~ 11)()(
)(
nnn
qqrnqfrnbetaQ rnrrnr
n
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Beta Distribution
q
f(q)
Some Symmetric Beta Distributions
Some Asymmetric Beta Distributions
q
f(q)
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Beta Distribution (Cont.)
Expected Value
VariancenrQE )(
)1(
)(2)var(
nn
rnrQ
Loosely speaking, r and n can interpreted as r successes in n trials
Suppose your guess for the preference of the Republican candidate is that 40% people would vote for the Republican candidate.
What if you set n=100, r=40?
You can set n=10, r=4. This coincides with the expected proportion of 40%.
This still coincides with the expected proportion of 40%. However, the variances of the two cases are different.
When n=10, r=4, 022.0)Qvar()110(10
)410(42
When n=100, r=40, 0024.0)Qvar()1100(100
)40100(402
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You want to re-evaluate your decision to invest in a pretzel stand. At this point, you estimate that you are 50% sure that your market share is less than 20% and 75% sure that your market share is less than 38%.
Using the table in Appendix F, you find that
76.0)1,4|38.0Pr(,49.0)1,4|20.0Pr( rnQrnQ
Let Q= market share, you decide to model the uncertainty in Q as a Beta distribution Pr(Q≤0.20)=0.5, Pr(Q≤0.38)=0.75
You think the beta distribution is close enough and thus should proceed with the analysis
)1=,4=(Beta~ rnQ
The expected value of Q, E(Q)=0.25
Pretzel Example (Cont.)
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However, as a careful person, suppose you also want to evaluate your chances of losing money.
Net Profit <0 => 40,000Q-8,000<0 => Q≤0.2
49.0)1,4|20.0Pr( rnQ (Appendix F)
Therefore, there is about 50% chance of losing money. Are you willing to continue to take this risk?
You estimate that the total market is 100,000 pretzels. You sell a pretzel at $0.50. It costs you $0.10 to produce a pretzel, in addition to $8,000 fixed cost for marketing, financing, and overhead.
Net Profit =Revenue – Cost =100,000*Q*0.5 – (100,000*Q*0.1+8,000) = 40,000Q – 8,000
E(Net Profit) =40,000*0.25 – 8,000 =$2,000 > 0
So it seems to be a good idea to start a pretzel career.
Pretzel Example (Cont.)
27
Exercises
Bottle ProductionIn bottle production, bubbles that appear in the glass are considered defects. Any bottle that has more than two bubbles is classified as “nonconforming” and is sent to recycling. Suppose that a particular production line produces bottles with bubbles at a rate of 1.1 bubbles per bottle. Bubbles occur independently of one another.
a.What is probability that a randomly chosen bottle is nonconforming?b.Bottles are packed in cases of 12. An inspector chooses one bottle from each case and examines it for defects. If it is nonconforming, she inspects the entire case, replacing nonconforming bottles with good ones. If the chosen one conforms, then she passes the case. In total, 20 cases are produced. What is the probability that at least 18 of them pass?
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a.X=# of bubbles in a bottleX~ Possion(m=1.1)Pr(X > 2 |m = 1.1) = 1 - Pr(X ≤ 2 |m = 1.1) = 1.00 - 0.90 = 0.1
b. Y=# of cases out of 20 cases that do not pass Y~ Binomial (n=20, p=0.1)Pr(Y≤2|n=20,p=0.1) = 0.677
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Exercises
Greeting CardA greeting card shop makes cards that are supposed to fit into 6 in. envelopes. The paper cutter, however, is not perfect. The length of a cut card is normally distributed with mean 5.9 in. and standard deviation 0.0365 in. If a card is longer than 5.975 in., it will not fit into a 6 in. envelope.
a.Find the probability that a card will not fit into a 6 in. envelopeb.The cards are sold in boxes of 20. what is the probability that in one box there will be two or more cards that do not fit in 6 in. envelopes?
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a. L= the uncertain length of an envelope.L~ N(µ = 5.9, σ = 0.0365)Pr (L > 5.975 | µ = 5.9, σ = 0.0365) = Pr(Z >(5.975-5.9)/0.0365) = Pr(Z > 2.055) =1-Pr(Z≤2.055)=1-0.98=0.02
b.X=# of cards in one box that do not fit in the envelopesX~ Binomial(n=20, p=0.02)Pr(X≥2|n=20,p=0.02) = 1-Pr(X≤1|n=20,p=0.02) =1-0.94=0.06
