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APPENDIX
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Probability Concepts and Applications PowerPoint presentation to accompany Balakrishnan/Render/Stair Managerial Decision Modeling with Spreadsheets, 3/e © 2013 Pearson Education, Inc. publishing as Prentice Hall
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Fundamental Concepts
• Basic statements of 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 0 ≤ P(event) ≤ 1 2. The sum of the simple probabilities for all possible outcomes of an activity must equal 1
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Two Laws of Probability
• Demand for paint at Diversey Paint QUANTITY DEMANDED (gallons) 0 1 2 3 4
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NUMBER OF DAYS 40 80 50 20 10 Total 200
PROBABILITY 0.20 (= 40/200) 0.40 (= 80/200) 0.25 (= 50/200) 0.10 (= 20/200) 0.05 (= 10/200) 1.00 (= 200/200)
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Types of Probability
• Objective Probability • Relative frequency
• Classical, or logical approach P(head) =
P(spade) = =
1
number of ways of getting a head
2
number of possible outcomes (head or tail) number of chances of drawing a spade
13 52 1 4
number of possible outcomes = 0.25 = 25%
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Types of Probability
• Subjective Probability
• Experience • Opinion polls • Judgment of individual
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Mutually Exclusive and Collectively Exhaustive
• Mutually exclusive – one and only one event
• Collectively exhaustive – all possible events
• Common
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Mutually Exclusive and Collectively Exhaustive
• Rolling a Die OUTCOME OF ROLL
PROBABILITY
1 2 3 4 5 6
1/6 1/6 1/6 1/6 1/6 1/6 Total 1
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Mutually Exclusive and Collectively Exhaustive
• Drawing a Card
DRAWS 1. 2. 3. 4. 5. 6.
MUTUALLY EXCLUSIVE?
COLLECTIVELY EXHAUSTIVE?
Yes Yes Yes Yes No No
No Yes No Yes No No
Draw a spade and a club Draw a face card and a number card Draw and ace and a 3 Draw a club and a nonclub Draw a 5 and a diamond Draw a red card and a diamond
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Adding Mutually Exclusive Events P(event A or event B) = P(event A) + P(event B) or P(A or B) = P(A) + P(B) P(spade or club) = P(spade) + P(club)
P(A)
P(B)
Figure A.1 © 2013 Pearson Education, Inc. publishing as Prentice Hall
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Law of Addition
• For events that are 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)
P(A)
P(five or diamond) = P(five) + P(diamond) – P(five and diamond)
P(B) Figure A.2
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Statistically Independent Events }
1.
(a) Your education (b) Your income level
2.
} (a) Chicago Cubs win the National League pennant } (b) Chicago Cubs win the World Series Independent (a) Snow in Santiago, Chile } events (b) Rain in Tel Aviv, Israel
3. 2.
Dependent events Can you explain why?
(a) Draw a jack of hearts from a full 52-card deck (b) Draw a jack of clubs from a full 52-card deck
Independent events Dependent events
• Marginal probability • Probability of an event occurring • Joint probability •
Product of marginal probabilities
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Statistically Independent Events P(AB) = P(A) x P(B) where P(AB) = joint probability of events A and B occurring together, or one after the other P(A) = marginal probability of event A P(B) = marginal probability of event B
Probability of tossing a 6 on the first roll and 2 on the second P(6 on first and 2 on second) = P(tossing a 6) x P(tossing a 2)
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Statistically Independent Events
• Conditional probability
• The probability of an event occurring
given that another event has taken place
P(B|A) P(A|B) = P(A) and P(B|A) = P(B)
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When Events are Independent 1. A black ball is drawn on the first draw: P(B) = 0.30 (marginal probability) 2. Two green balls are drawn: P(GG) = P(G) x P(G) = (0.7)(0.7) = 0.49 (joint probability – independent events)
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When Events are Independent 3. A black ball is drawn on the second draw if the first draw is green: P(B|G) = P(B) = 0.30 (conditional probability – independent events) 4. A green ball is drawn on the second draw if the forst draw was green: P(G|G) = P(G) = 0.70 (conditional probability – independent events) © 2013 Pearson Education, Inc. publishing as Prentice Hall
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Statistically Dependent Events
• The occurrence of one event effects the occurrence of some other event
• Marginal probability is unchanged • Conditional and joint change form Conditional probability (Bayes’ law)
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Dependent Events 4 balls White (W) and Lettered (L)
Urn contains 10 balls
2 balls White (W) and Numbered (N) 3 balls Yellow (Y) and Lettered (L)
Figure A.3
1 ball Yellow (Y) and Numbered (N)
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Dependent Events • If a yellow ball is drawn, what’s the probability it is lettered?
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Dependent Events • If a yellow ball is drawn, what’s the probability it is lettered? Conditional probability
Joint probability
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Revising Probabilities with Bayes’
• Revised or Posterior Probabilities Prior Probabilities Bayes’ Process
Posterior Probabilities
New Information
Figure A.4
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Revising Probabilities with Bayes’
• Revised or Posterior Probabilities
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General Form of Bayes’ Theorem
we can show that
where
A = the complement of the event A; for example, if A is the event “fair die,” then A is “unfair” or “loaded die”
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General Form of Bayes’ Theorem Solving the “fair die” problem
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Further Probability Revisions
• Second Probability Revision
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Random Variables
• Assigns a real number to every • •
possible outcome or event in an experiment Discrete random variables can assume only a finite or limited set of values Continuous random variables have an infinite or an unlimited set of values
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Random Variables EXPERIMENT
OUTCOME
Stock 50 Christmas trees Inspect 600 items
Number of Christmas trees sold Number of acceptable items Number of people responding to the letters Percent of building competed after 4 months Length of time the bulb lasts up to 80,000 minutes
Send out 5,000 sales letters Build and apartment building Test the lifetime of a lightbulb
RANGE OF RANDOM VARIABLES X = number of Christmas tress sold Y = number of acceptable items Z = number of people responding to letters R = percent of building competed after 4 months S = time the bulb burns
VARIABLES 0, 1, 2,…, 50 0, 1, 2,…, 600 0, 1, 2,…, 5,000 0 ≤ R ≤ 100 0 ≤ S ≤ 80,000
Table A.1
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Random Numbers
• For variables where outcomes are not numbers
EXPERIMENT
OUTCOME
Students respond to a questionnaire
Strongly agree (SA) Agree (A) Neutral (N) Disagree (D) Strongly disagree (SD) Defective Not defective Good Average Poor
One machine is inspected Consumers respond to how much they like a product
RANGE OF RANDOM VARIABLES
X=
Y= Z=
5 if SA 4 if A 3 if N 2 if D 1 if SD 0 if defective 1 if not defective 3 if good 2 if average 1 if poor
VARIABLES 1, 2, 3, 4, 5
0, 1 1, 2, 3
Table A.2 © 2013 Pearson Education, Inc. publishing as Prentice Hall
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Probability Distributions
• Discrete Random Variable
• Mutually exclusive events • Probability values between 0 and 1 • Probabilities sum to 1
OUTCOME
RANDOM VARIABLE (X)
NUMBER RESPONDING
PROBABILITY P(X)
5 4 3 2 1
10 20 30 30 10
0.1 = 10/100 0.2 = 20/100 0.3 = 30/100 0.3 = 30/100 0.1 = 10/100
Total 100
1.0 = 100/100
Strongly agree Agree Neutral Disagree Strongly disagree
Table A.3 © 2013 Pearson Education, Inc. publishing as Prentice Hall
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Probability Distributions 0.4
Probability
0.3
0.2
0.1
0.0 1
2
3
4
5
Possible Values of the Random Variable X Figure A.5 © 2013 Pearson Education, Inc. publishing as Prentice Hall
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Expected Value
• Discrete probability distribution
where = random variable’s possible values = probability of each of the random variable’s possible values = summation sign indicating we are adding all n possible values = expected value of the random variable © 2013 Pearson Education, Inc. publishing as Prentice Hall
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Expected Value
• For the textbook question
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Variance
where = random variable’s possible values = expected value of random variable = difference between each value of variable and expected value = probability of each possible value of the random variable
Standard Deviation
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Variance
• For textbook problem
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Probability Distributions
• Continuous Random Variable
• Mutually exclusive events • Probability values of each variable = 0 • Probabilities sum to 1 • Probability density function f(X)
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Probability
Density Function
|
|
|
|
|
|
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5.06
5.10
5.14
5.18
5.22
5.26
5.30
Weight (grams)
Figure A.6
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Normal Distribution
• Specified by mean, μ, and standard deviation, σ
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Normal Distribution |
|
|
μ = 50
40
60
Smaller μ, same σ |
μ = 40
|
|
50
60
Larger μ, same σ |
|
40
50
|
μ = 60 Same μ, smaller σ
Figure A.7
Same μ, larger σ
μ
Figure A.8 © 2013 Pearson Education, Inc. publishing as Prentice Hall
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Area Under the Normal Curve 68.26%
15.87%
–1σ a
μ
15.87%
+1σ b
95.44%
2.28%
–2σ a
μ
2.28%
+2σ b
99.74%
0.13%
Figure A.9
–3σ a
μ
0.13%
+3σ b
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Standard Normal Table Step 1 – Standard Normal Distribution
X = μ = σ = Z =
value of the random variable we want to measure mean of the distribution standard deviation of the distribution number of standard deviations from X to the mean, μ
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Standard Normal Table
μ = 100 σ = 15
P(X < 130)
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55
70
85
100
115
130
145
μ |
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–3
–2
–1
0
1
2
3 Figure A.10
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Standard Normal Table For μ = 100, σ = 15
Step 2 – Find probability in Appendix C
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Haynes Construction Figure A.11
μ = 100 days, σ = 20 Probability of completing in 125 days
μ = 100 days
X = 125 days
σ = 20 days
Area under the curve = 0.8944 or 89.44%
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Haynes Construction μ = 100 days, σ = 20 Probability of completing in 75 days or less
Figure A.12
P(X < 75 days) Area of Interest
X = 75 days
μ = 100 days
P(X > 125) = 1.0 – P(X < 125) = 1.0 – 0.8944 = 0.1056 or 10.56%
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Haynes Construction Figure A.13
Probability of completing between 110 and 125 days
σ = 20 days
μ= 100 days
110 days
125 days
P(110 < X < 125) = 0.8944 – 0.6915 = 0.2029
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Exponential Distribution
• Negative exponential distribution • Queuing models 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
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Exponential Distribution Figure A.14
f(X)
(X)
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Poisson Distribution
where P(X) = probability of exactly X arrivals or occurrences λ = average number of arrivals per unit of time (the mean arrival rate), pronounced “lambda” e = 2.718, the base of the natural logarithms X = specific value of the random variable
Expected value = λ
Variance = λ
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Poisson Distribution P(X)
Figure A.15
0.25 0.20 0.15 0.10 0.05 0 0
1
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