Probability

1. Probability

2. Assigning Probabilities to Events Random experiment a random experiment is a process or course of action, whose outcome is uncertain. Performing the same random experiment repeatedly, may result in different outcomes, therefore, the best we can do is talk about the probability of occurrence of a certain outcome. To determine the probabilities we need to define the possible outcomes first.

3. Determining the outcomes. Build an exhaustive list of all possible outcomes. Make sure the listed outcomes are mutually exclusive. A list of outcomes that meet the two conditions above, is called a sample space.

4. If the experimental outcomes are not equally likely, and no history of repetition exists, one needs to resort to subjective probability determination. This approach reflects the personal evaluation of the uncertainties involved.

5. Given a sample space S={E1,E2,�,En}, the following characteristics for the probability P(Ei) of the simple event Ei must hold: Probability of an event: The probability P(A) of event A is the sum of the probabilities assigned to the simple events contained in A. Assigning Probabilities

6. Experiment 1: Flip a coin Outcome 1: heads or tails Experiment 2: Roll a die Outcome 2: 1, 2, 3, 4, 5, 6 Experiment 3: Solicit a consumer's preference between product A and product B Outcome 3: Consumer prefers product A, B or indifferent With these experiments, the outcome cannot be determined in advance! If we repeat Experiment 3 tomorrow, the results will be different than those we experienced today. All we can talk about is the probability that an outcome will occur. Examples of Random Experiments

7. What is a repeatable experiment? Repeatable experiment � an experiment that, from the viewpoint of the experiments, may be repeated as many times as desired. � Say we want to determine, in advance of an experiment, the probabilities that various outcomes will occur. � What must we know? We must know what outcomes are possible. So, list all possible outcomes for an experiment. The list of possible outcomes must be exhaustive (each trial of the experiment must result in some outcome on the list) The list of outcomes must be mutually exclusive (no two outcomes on the list can both occur on any one trial of the experiment.) Repeatable Experiment

8. Sample Space

9. What are some examples of events that can be broken down into more outcomes? Event A: [An even number] can be broken down into [2, 4, or 6] To use mathematical tools to analyze a process we must make some assumptions, which simplify the process to be considered. The Assumptions form constraints on the possible unique outcomes. Sample Spaces

10. Discrete Continuous Discrete can be: Finite Countably infinite Example: Finite = toss of a coin, which can be heads or tails Example: Countably infinite = number of heads before first tail Continuous: Always uncountably infinite because it is defined on a continuum. Example: Number of time intervals before an event: time is a continuous phenomenon therefore the number of possible time intervals is infinite! Two Kinds of Sample Spaces

11. How do we assign probabilities? Three approaches: Classical Relative frequency Subjective Assigning Probabilities

12. Classical Approach

13. Relative Frequency

14. Subjective Approach

15. What are the requirements of probabilities?

16. Probability Trees

17. Probability Trees

18. Probability Trees

19. If A and B are two events, then P(A or B) = P(A occur or B occur or both) P(A and B) = P(A and B both occur) P(A|B) = P(A occurs given that B has occurred) Probability of Combinations of Events

20. Probability of Combinations of Events

21. (a) Define the sample space for this random experiment and assign probabilities to the simple events. Solution Sample space = S = {1, 2, 3, 4, 5, 6} Each simple event is equally likely to occur, thus, P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6

22. (b) Find P(A)

29. The probability of an event when partial knowledge about the outcome of an experiment is known, is called Conditional probability. We use the notation P(A|B) = The conditional probability that event A occurs, given that event B has occurred. Conditional Probability

30. Two events A and B are said to be independent if P(A|B) = P(A) or P(B|A) = P(B). Otherwise, the events are dependent. Note that, if the occurrence of one event does not change the likelihood of occurrence of the other event, the two events are independent Independent and Dependent Events

31. The personnel department of an insurance company has compiled data regarding promotion, classified by gender. Over the past 3 years, 8 out of 54 promotions went to women. The department claims that the decision to promote a manager (or not) is independent of the manager's gender. Would you agree?

32. What do we want to know? Let us check if P(A|M)=P(A). If this equality holds, there is no difference in probability of promotion between a male and a female manager.

34. Probability Rules and Trees Complement rule Each simple event must belong to either A or .Since the sum of the probabilities assigned to a simple event is one, we have for any event A The task of finding the probability that an event will not occur and then subtracting this probability from 1 is often easier than the task of directly computing the probability that it will occur.

35. For any two events A and B

36. For any two events A and B For the special case in which A and B are independent events, we have P(B/A) = P(B), so we can write:

37. A stock market analyst feels that the probability that a certain mutual fund will receive increased contributions from investors is 0.6. the probability of receiving increased contributions from investors becomes 0.9 if the stock market goes up. the probability of receiving increased contributions from investors drops below 0.6 if the stock market drops. there is a probability of 0.5 that the stock market rises. The events of interest are: A: The stock market rises; B: The company receives increased contribution.

38. The probability that both A and B will occur is P(A and B). [Sharp increase in earnings]. The probability that either A or B will occur isP(A or B). [At least moderate increase in earning]. Solution P(A) = 0.5; P(B) = 0.6; P(B|A) = 0.9 P(A and B) = P(A)P(B|A) = (.5)(.9) = 0.45 P(A or B) = P(A) + P(B) - P(A and B) = .5 + .6 - .45 =0.65

39. A computer software supplier has developed a new record keeping package for use by hospitals. The company feels that the probability that the new package will show a profit in its first year is .6 unless a competitor introduces a product of comparable quality this year, in which case the probability of a first year profit drops to .3. The suppliers suggests that there is a 50% chance that a comparable product will be introduced this year. � What are the events? A: A competitor introduces a comparable product in its first year B: the record keeping package is profitable in its first year

40. What is the probability that both A and B will occur? P(A) = .5 P(B) = .6 P(B/A) = .3 � P(A and B) = P(A) * P(B/A) = (.5)(.3) = .15 � What is the probability that either A or B will occur? P(A or B) = P(A) + P(B) - P(A and B) = .5 + .6 - .15 = .95

41. Suppose we are interested in the condition of a machine that produces a particular item. Information From experience it is known that the machine is in good conditions 90% of the time. When in good conditions, the machine produces a defective item 1% of the time. When in bad conditions, the machine produces a defective 10% of the time. An item selected at random from the current production run was found defective.

42. Let us define the two events of interest:A: The machine is in good conditionsB: The item is defective The prior probability that the machine is in good conditions is P(A) = 0.9. With the new information, (the selected item is defective, or, event B has occurred) we can reevaluate this probability by calculating P(A|B).

45. Random Variables and Probability Distributions A random experiment is a function that assigns a numerical value to each simple event in a sample space. A random variable reflects the aspect of a random experiment that is of interest to us. There are two types of random variables Discrete random variable Continuous random variable.

46. A random variable is discrete if it can assume only a countable number of values. A random variable is continuous if it can assume an uncountable number of values. Discrete and Continuous Random Variables

47. Random Variables What is a discrete random variable? A Discrete Random variable has a countable number of possible values. We can identify the first value, the second, and so on. In most practical situations, a discrete random variable counts the number of times a particular attribute is observed. Examples of Discrete random variables: Number of defective items in a production batch The number of telephone calls received in a given hour The number of shoppers who prefer a particular product

48. Random Variables Say X represents the number of respondents in a survey of 400 shoppers, who state a preference for a particular product, then X can take any one of the values x = 0, 1, 2, �, 400. � Just because discrete random variables are countable, does not mean that there is a finite number of values it can assume.

49. Continuous Random Variables What is a Continuous Random variable? A Continuous random variable has an uncountably infinite number of possible values. It can take on any value in one or more intervals of values. � What are the typical applications of random variables? Recording the value of a measurement such as time, weight, or length. � Examples of Continuous random Variables: The amount of time workers on an assembly line take to complete a particular task

50. Continuous Random Variables How do we assign probabilities to a discrete random variable? With a discrete probability distribution � What is a discrete probability distribution? A table, formula, or graph that lists all possible values a discrete random variable can assume, together with their associated probabilities.

51. A table, formula, or graph that lists all possible values a discrete random variable can assume, together with associated probabilities, is called a discrete probability distribution.. To calculate P(X = x), the probability that the random variable X assumes the value x, add the probabilities of all the simple events for which X is equal to x. Discrete Probability Distribution

52. Find the probability distribution of the random variable describing the number of heads that turn-up when a coin is flipped twice. Solution

53. Requirements of discrete probability distribution If a random variable can take values xi, then the following must be true:

54. In practice, often probabilities are estimated from relative frequencies Example The number of cars a dealer is selling daily were recorded in the last 100 days. This data was summarized as follows:

55. From the table of frequencies we can calculate the relative frequencies, which becomes our estimated probability distribution

56. Expected Value and Variance The expected value Given a discrete random variable X with values xi, that occur with probabilities p(xi), the expected value of X is

57. E(c) = c E(cX) = cE(X) E(X + Y) = E(X) + E(Y)E(X - Y) = E(X) - E(Y) E(XY) = E(X)E(Y) if X and Y are independent random variables. Laws of Expected Value

58. Expected Value Example

59. Expected Value Example

60. Expected Value Example

61. Let X be a discrete random variable with possible  values xi that occur with probabilities p(xi), and let E(xi) = m. The variance of X is defined to be Variance

62. What about our chip example? � ?2 = (20 - 35)2 (1/2) + (40 - 35)2 (1/4) + (60 - 35)2 (1/4) = 275 (dollars)2 � When will this information be useful? The variance will be used to compare the variabilities of other distributions (such as a measure of risk). � A shortcut formula for the variance: ?2 = (X2) - ?2 Variance

63. The standard deviation of a random variable X, denoted s, is the positive square root of the variance of X. Example 6.5 The total number of cars to be sold next week is described by the following probability distribution Determine the expected value and standard deviation of X, the number of cars sold.

65. With the probability distribution of cars sold per week (example 6.5), assume a salesman earns a fixed weekly wages of $150 plus $200 commission for each car sold. What is his expected wages and the variance of the wages for the week? Solution The weekly wages is Y = 200X + 150 E(Y) = E(200X+150) = 200E(X)+150= 200(2.4)+150=630 $. V(Y) = V(200X+150) = 2002V(X) = 2002(1.24) = 49,600 $2