Quick Review of Probability

1. Quick Review of Probability Event state of the world that may or may not occur --- A, B, C, etc. Joint Event events occurring together --- “A and B”, AB = BA Conditional Event an event occurring given that another occurs --- “A given B”, A|B --- “B given A”, B|A Mut. Exclusive Events events that cannot occur together Coll. Exhaustive Events events that together comprise all possibilities Independent Events events that do not depend on one another

2. P(A) “single-event” probability P(AB) = P(BA) joint probability P(A|B), P(B|A) conditional probability IMPORTANT FACT If several events are mutually exclusive and collectively exhaustive, then their probabilities add to 1.0 “The MECE rule”

3. General Formulas Conditional Probability P(A|B) = P(AB) / P(B) P(B|A) = P(BA) / P(A) Multiplication Rule P(AB) = P(A|B) P(B) = P(B|A) P(A) Special Formulas for Independent Events If you know that A and B are independent… Conditional Probability P(A|B) = P(A) P(B|A) = P(B) Multiplication Rule P(AB) = P(A) P(B) Do not use special formulas unless you know A and B are independent!

4. mutually exclusive, collectively exhaustive D E A F G P(G|B) B H P(H|B) mutually exclusive, collectively exhaustive mutually exclusive, collectively exhaustive Probability Trees A probability tree is a graphical way to visualize the possible outcomes of a sequence of occurrences P(D|A) P(A) P(E|A) P(F|A) P(B) P(D|A) + P(E|A) + P(F|A) = 1.0 P(A) + P(B) = 1.0 P(G|B) + P(H|B) = 1.0

5. D P(D|A) P(A) E P(E|A) A F P(F|A) G P(G|B) B H P(H|B) P(B) Information on a Probability Tree • There are two types of probabilities on a probability tree: • “Single event” probabilities • Conditional probabilities Joint probabilities are not on the tree, but they are easy to calculate from the tree. For example: P(AD) = P(D|A) P(A)

6. Tips for Probability Problems Always be aware of: • the events (single, joint, or conditional?) • which probabilities are given • which probabilities are asked for • which special situations exist (independence?) (may be given, or you may have to deduce)

7. Bayes’ Rule and “Too Many Bugs” Public health scientists estimate that 1% of the general population regularly abuses illegal substances. As the president of a 2,000-employee company, you would like to use drug testing to eliminate this 1% from your work force. A new drug test has been devised that is relatively inexpensive and that the manufacturer claims has a 95% accuracy rate. Your idea is to test each employee, terminating or keeping the employee based on the results of the test. Is this a reasonable and fair plan?

8. 0.95 0.01 A D 0.05 E A 0.05 F E 0.95 0.99 D = event that a person uses Drugs F = event that a person is drug Free A = event that the test Accuses a person E = event that the test Exonerates a person What does “95% accuracy” mean? This is terminology for how often the test confirms the truth. P(A | D) = 0.95 and P(E | F) = 0.95

9. What are the probabilities of these events? • Person is accused correctly = AD • Person is accused falsely = AF • Person is exonerated correctly = EF • Person is exonerated falsely = ED • Using the multiplication rule (events not independent): • P(AD) = P(A | D) P(D) = 0.95 * 0.01 = 0.0095 • P(AF) = P(A | F) P(F) = 0.05 * 0.99 = 0.0495 • P(EF) = P(E | F) P(F) = 0.95 * 0.99 = 0.9405 • P(ED) = P(E | D) P(D) = 0.05 * 0.01 = 0.0005 Have 4.95% accused falsely, but only 1% actually use drugs!

10. What are P(A) and P(E)? Because mut. exclusive: • P(A) = P(AD) + P(AF) = 0.0095 + 0.0495 = 0.0590 • P(E) = P(EF) + P(ED) = 0.9405 + 0.0005 = 0.9410 • What are P(D | A) and P(F | E)? By the conditional probability rule (events not independent): • P(D | A) = P(AD) / P(A) = 0.0095 / 0.0590 = 0.1610 • P(F | E) = P(EF) / P(E) = 0.9405 / 0.9410 = 0.9995 What does this mean exactly?

11. Consider the following situation. Suppose John Smith has been accused by the drug test. You have no way of knowing for sure if John uses drugs. Do you fire John based on the test alone? P( John uses | John has been accused ) = P( D | A) = 0.1610 There is a 16% chance John abuses drugs even though the test has accused him Now suppose John Smith has been exonerated by the test. Do you keep John as an employee? P( John is free | John has been exonerated ) = P( F | E) = 0.9995 It’s 99.95% sure that John does not use drugs based on the result of his test

12. The moral of the “Bugs” article: Even if a person is accused by a test, the conditional probabilities show that there is a good chance that the person has been falsely accused. On the other hand, if a person is exonerated by a test, then it is nearly certain that the person deserves to be exonerated. The process we went through is Bayes’ Rule, and it can be used in several additional ways (not just in “drug testing” scenarios)

13. Test Marketing a New Product New products introduced in the marketplace have high sales 8% of the time and low sales 92% of the time. A marketing test has the following accuracies: if sales are high, then consumer test reaction is positive 70%, neutral 25%, and negative 5%; if sales are low, then consumer test reaction is positive 15%, neutral 35%, and negative 50%. Your company is developing a new product and will be test marketing to better gauge the sales of the new product. Based on positive, neutral, or negative reactions, what are the probabilities of high and low sales?

14. Decision Analysis We will now be using probability trees to help us make decisions in the face of uncertainty • Ingredients for a quantitative decision • Possible decisions • Uncertain events • Payoffs or costs to decisions and events • Probabilities of the events

15. One day each weekend, you rent an indoor booth to sell your homemade crafts at the J&J Flea Market. From your experience, you know the following: This Saturday, there is a 70% chance of rain, and this Sunday, there is a 30% chance. Which day do you sell at the flea market? Actions: sell on Saturday, sell on Sunday Events: rain or not on Saturday, rain or not on Sunday Payoffs: dollar amounts Probs: 70%, 30%, etc.

16. How to Evaluate a Decision • There are two stages to evaluating a decision • FORWARD stage • Make a decision tree that lays out all possible sequences of decisions and events, along with the payoffs and probs • BACKWARD stage • Evaluate EMVs or EMCs from the end to the beginning What are EMVs and EMCs? Will explain shortly…

17. 0.7 Rain $500 0.3 Sat No Rain $1000 0.3 Rain $350 Sun 0.7 No Rain $700 Forward Stage for J&J $0 $0

18. EMVs and EMCs EMV = Expected Monetary Value = the average payoff of a series of future decisions and events, assuming that, at any time during the decision-making process, the decision-maker makes the wisest decisions given the pending uncertainties EMC = Expected Monetary Cost = the average cost of a series of future decisions and events, assuming that, at any time during the decision-making process, the decision-maker makes the wisest decisions given the pending uncertainties Goal: maximize EMV and minimize EMC!

19. Backward Stage for J&J 0.7 $500 Rain $500 $650 0.3 Sat No Rain $1000 $0 $1000 $650 0.3 $350 Rain $350 $595 Sun 0.7 $0 $700 No Rain $700

20. Rules for Evaluating EMVs • The EMVs at the ends of the paths in the tree are the total cumulative payoffs gotten over each path • The EMV of an event (circle) is the average of the EMVs of the event’s branches, weighted by the probs • The EMV of a decision (square) is the maximum of the EMVs of the decision’s branches

21. Rules for Evaluating EMCs • The EMCs at the ends of the paths in the tree are the total cumulative costs gotten over each path • The EMC of an event (circle) is the average of the EMCs of the event’s branches, weighted by the probs • The EMC of a decision (square) is the minimum of the EMCs of the decision’s branches EMVs and EMCs are similar except “payoffs vs costs” and “maximize vs minimize”

22. Example for EMV of Event P(A) EMVA A $X EMV P(B) EMVB B $Y P(C) EMVC C $Z EMV = P(A)*EMVA + P(B)*EMVB + P(C)*EMVC Note: payoffs $X, $Y, and $Z are not in the formula

23. Fruit Flies Discussion(preparation for Homework 2) • In 1981, Mediterranean fruit flies infested crops in Santa Clara County, CA, threatening California business • Spraying for the flies was an option, but spraying was environmentally questionable • Governor Brown and others dismissed spraying • Instead, a control program of “fruit stripping” was begun, depositing 750 tons of fruit in California landfills • Then, the USDA came up with the “sterile male” solution, attempting to capitalize on a biological trait of the flies

24. Fruit Flies Discussion(preparation for Homework 2) • Female fruit flies only mate once • Radiation was used to sterilize a large quantity of male fruit flies • The males were then released to halt fly population growth • Problem! The irradiation was done incorrectly, and the males weren’t sterile • More flies, more flies, and more flies • Embargoes against California fruit • Governor Brown decides to spray • But it’s too late, his career is over