1. Block 3 Discrete Systems Lesson 11 –Discrete Probability Models The world is an uncertain place

2. Random Process Random – happens by chance, uncertain, non-deterministic, stochastic Random Process – a process having an observable outcome which cannot be predicted with certainty one of several outcomes will occur at random

3. Uses of Probability Measures uncertainty Foundation of inferential statistics Basis for decision models under uncertainty A manager observing an uncertainty outcome

4. Two Approaches to modeling probability Sample Space and Random Events Uses sets and set theory Random Variables and Probability Distributions Discrete case – algebraic Continuous case - calculus

5. Sample Spaces and Random Events Let S = the set of all possible outcomes (events) from a random process. Then S is called the sample space. Let E = a subset of S. Then E is called a random event. Basic problem: Given a random process and the sample space S, what is the probability of the event E occurring - P(E).

6. Example Sample Space & Random Events Let S = the set of all outcomes from observing the number of demands on a given day for a particular product. S = {0, 1, 2,…n} Let E 1 = the random event, there is one demand. Then E 1 = {1} Let E 2 = the random event of no more than 3 demands. Then E 2 = {0, 1, 2, 3} Let E 3 = the random event, there are at least 4 demands. Then E 3 = {4, 5, …, n} or E 3 = E 2 c S x S = the set of outcomes from observing two days of demands = {(0,0), (1,0), (0,1), …}

7. What is a probability – P(E)? Let P(E) =Probability of the event E occurring, then 0 ≤ P(E) ≤ 1 If P(E) = 0, then event will not occur (impossible event) If P(E) = 1, then the event will occur, i.e. a certain event So the closer P(E) is to 1, then the more likely it is that the event E will occur?

8. The Sample Space The collection of all possible outcomes (events) relative to a random process is called the sample space, S where S = {E 1, E 2... E k } and P(S) = 1 sampling space

9. How are probabilities determined? Elementary or basic events 1. Empirical or relative frequency 2. A priori or equally-likely using counting methods 3. Subjectively –personal judgment or belief Compound events formed from unions, intersections, and complements of basic events Laws of probability

10. Example - Relative Frequency (empirical) A coin is tossed 2,000 times and heads appear 1,243 times. P(H) = 1243/2000 =.6215 The company’s Web site has been down 5 days out of the last month (30 days). P(site down) = 5/30 =.16667 2 out of every 20 units coming off the production line must be sent back for rework. P(rework) = 2/20 =.1

11. Will you die this year?

12. Example - A priori (equally-likely outcomes) A pair of fair dice are tossed. S x S = {(1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6), (2, 1),(2, 2),(2, 3),(2, 4),(2, 5),(2, 6),(3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6),(4, 1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6),(5, 1),(5, 2),(5, 3),(5, 4),(5, 5),(5, 6),(6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6, 6)} P(a seven) = 6/36 =.1667 A supply bin contains 144 bolts to be used by the manufacturing cell in the assembly of an automotive door panel. The supplier of the bolts has indicated that the shipment contains 7 defective bolts. Let E = the event, a defective bolt is selected P(E) = 7/144 =.04861.

13. A priori (equally-likely outcomes) versus Relative Frequency (empirical) A prior (knowable independently of experience): n(A) = the number of ways in which event A can occur n(S) = total number of outcomes from the random process Relative frequency: where n(E) = number of times event E occurs in n trials

14. Examples – Subjective Probability 8 out of 10 “leading” economists believe that the gross national product (GNP) will grow by at least 3% this year. P(GNP .03) = 8/10 =.8 Bigg Bosse, the CEO for a major corporation, consults his marketing staff. Together they make the assessment that there is a 50-50 chance that sales will increase next year. The House of Congress majority leader, after consulting with his staff, determines that there is only 25 percent chance that an important tax bill will be passed.

15. Computing Probabilities for Compound Events Finding the probability of the union, intersection, and complements of events

16. Mutually Exclusive Events P(A c ) = 1 - P(A) P( A  B) = P(A) + P(B) if A and B are mutually exclusive P(A  B) = P(  ) = 0 if A and B are mutually exclusive AB Note that A’ and B’ are not mutually exclusive.

17. More Mutually Exclusive Events Random process: draw a card at random from an ordinary deck of 52 playing cards Let A = the event, an ace is drawn Let B = the event, a king is drawn P(A) = 1/13 and P(B) = 1/13 Then P( A  B) = P(A) + P(B) = 1/13 + 1/13 = 2/13 P(A  B) = 0 P(A’) = 1 – 1/13 = 12/13; P(B’) = 12/13 P(A  B)’ = P( A’  B’) = ? the event is not an ace or not king It’s not me!

18. The Addition Formula P(A  B) = P(A) + P(B) - P(A  B) AB A B A  B

19. More of the Addition Formula Random process: draw a card at random from an ordinary deck of 52 playing cards Let A = the event, draw a spade P(A) = 13/52 =1/4 let B = the event draw an ace P(B) = 4/52 = 1/13 P(A  B) = 1/52 P(A  B) = P(A) + P(B) - P(A  B) = 1/4 + 1/13 – 1/52 = 13/52 + 4/52 – 1/52 = 16/52 =.3077 112 3 36

20. The Multiplication rule Independent Events Two events, A and B are independent if the P(A) is not affected by the event B having occurred (and vice-versa). If A and B are independent, then P(A  B) = P(A) P(B) An Independence event Note that A’ and B’ are independent if A and B are independent.

21. Proof by Example Let E 1 = the event, a three or four is rolled on the toss of a single fair die, P(E 1 ) = 2/6 E 2 = the event, a head is tossed from a fair coin, P(E 2 ) = 1/2 then D x C = {(1,H), (1,T), (2,H), (2,T), (3,H), (3,T), (4,H), (4,T), (5,H), (5,T), (6,H), (6,T)}; P(E 1  E 2 ) = 2/12 Mult. rule: P(E 1  E 2 ) = P(E 1 ) P(E 2 ) = (2/6) (1/2) = 2/12

22. Another Example Let A = the event, prototype A fails heat stress test B = the event prototype B fails a vibration test Given P(A) =.1 ; P(B) =.3 Find P(A  B) = ? and P(A  B) = ? Assuming independence: P(A  B) = P(A) P(B) = (.1)(.3) =.03 P(A  B) = P(A) + P(B) - P(A  B) =.1 +.3 -.03 =.37 P(A  B)’ = P(A’  B’) = P(A’)P(B’) = (.9)(.7) =.63

23. Glee Laundry Detergent Each box of powered laundry detergent coming off of the final assembly line is subject to an automatic weighing to insure that the weight of the contents falls within specification. Each box is then visually inspected by a quality assurance technician to insure that is it properly sealed. Glee I have been rejected.

24. More Glee If three percent of the boxes fall outside the weight specifications and five percent are not properly sealed, what is the probability that a box will be rejected after final assembly? Let A = the event, a box does not meet the weight specification; P(A) =.03 Let B = the event, a box is not properly sealed; P(B) =.05 P(A  B) = P(A) + P(B) – P(A)P(B) =.03 +.05 -.0015 =.0785

25. A Reliability Problem An assembly is composed of 3 components as shown below. If A is the event, component A does not fail, B is the event, component B does not fail, and C is the event, component C does not fail, find the reliability of the assembly where P(A) =.8, P(B) =.9, and P(C) =.8. Assume independence among the components. AB C P(S) = P[ (A  B)  C] = P(A  B) + P(C) – P(A  B  C) = P(A) P(B) + P(C) – P(A)P(B)P(C) = (.8)(.9) +.8 – (.8)(.9)(.8) =.72 +.8 -.576 =.944

26. Next – random variables and their probability distributions! Variables that are random; what will they think of next?

27. Discrete Random Variables ¨ A random variable (RV) is a variable which takes on numerical values in accordance with some probability distribution. ¨ Random variables may be either continuous (taking on real numbers) or discrete (usually taking on non-negative integer values). ¨ The probability distribution which assigns probabilities to each value of a discrete random variable can be described in terms of a probability mass function (PMF), p(x) in the discrete case.

28. Random Variables - Examples « Y = a discrete random variable, the number of machines breaking down each shift « X = a discrete random variable, the monthly demand for a replacement part « Z = a discrete random variable, the number of hurricanes striking the Gulf Coast each year « X i = a discrete random variable, the number of products sold in month i

29. The PMF The Probability Mass Function (PMF), p(x), is defined as p(x) = Pr(X = x} and has two properties: By convention, capital letters represent the random variable while the corresponding small letters denote particular values the random variable may assume.

30. A Probability Distribution Let X = a RV, the number of customer complaints received each day The Probability Mass Function (PMF) is

31. The CDF The cumulative distribution function (CDF), F(x) is defined where Example:

32. Rolling the dice 2(1,1)1 3(1,2), (2,1)2 4(1,3), (2,2), (3,1)3 5(1,4), (2,3), (3,2), (4,1)4 6(1,5), (2,4), (3,3), (4,2), (5,1)5 7(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)6 8(2,6), (3,5), (4,4), (5,3), (6,2)5 9(3,6), (4,5), (5,4), (6,3)4 10(4,6), (5,5), (6,4)3 11(5,6), (6,5)2 12(6,6)1 Total 36 X = the outcome from rolling a pair of dice number of ways

33. An Example Let X = a random variable, the sum resulting from the toss of two fair dice; X = 2, 3, …, 12 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) = S

34. Probability Histogram for the Random Variable X

35. Expected Value or Mean The expected value of a random variable (or equivalently the mean of the probability distribution) is defined as Don’t you get it? The expected value is just a weighted average of the values that the random variable takes on where the probabilities are the weights.

36. Example – Expected Value E[X] = 0 (.1) + 1 (.3) + 2 (.5) + 3 (.1) = 1.6 I expected this to have a little value!

37. Example –Expected Value Dice example: = 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36) + 7(6/36) + 8(5/36) + 9(4/36) + 10(3/36) + 11(2/36) + 12(1/36) = (1/36) (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12) = (252/36) = 7

38. Yet Another Discrete Distribution Let X = a RV, the number of customers per day recall the arithmetic series: F(20) = (20)(21) / 420 = 1

39. More of yet another discrete distribution Let X = a RV, the number of customers per day Pr{X = 15} = p(15) = 15/210 =.0714 Pr{X  15} = F(15) = (15)(16)/420 =.5714 Pr{10 < X  15} = F(15) – F(10) =.5714 – (10)(11)/420 =.5714 -.2619 =.3095

40. So ends our discussion on Discrete Probabilities Coming soon to a classroom near you – discrete optimization models A most enjoyable lesson.