Statistics for Managers Using Microsoft Excel (5th Edition)

1. Statistics for Managers Using Microsoft Excel (5th Edition) Chapter 4 Basic Probability and Discrete Probability Distributions

2. Chapter Topics • Basic probability concepts • Sample spaces and events, simple probability, joint probability • Conditional probability • Statistical independence, marginal probability • Bayes’s Theorem

3. Sample Spaces • Collection of all possible outcomes • e.g.: All six faces of a die: • e.g.: All 52 cards in a deck:

4. Events • Simple event • Outcome from a sample space with one characteristic • e.g.: A red card from a deck of cards • Joint event • Involves two outcomes simultaneously • e.g.: An ace that is also red from a deck of cards

5. Visualizing Events • Contingency Tables • Tree Diagrams Ace Not Ace Total Black 2 24 26 Red 2 24 26 Total 4 48 52 Ace Red Cards Not an Ace Full Deck of Cards Ace Black Cards Not an Ace

6.   Special Events Null Event • Impossible event e.g.: Club & diamond on one card draw • Complement of event • For event A, all events not in A • Denoted as A’ • e.g.: A: queen of diamonds A’: all cards in a deck that are not queen of diamonds

7. Special Events (continued) • Mutually exclusive events • Two events cannot occur together • e.g.: -- A: queen of diamonds; B: queen of clubs • Events A and B are mutually exclusive • Collectively exhaustive events • One of the events must occur • The set of events covers the whole sample space • e.g.: -- A: all the aces; B: all the black cards; C: all the diamonds; D: all the hearts • Events A, B, C and D are collectively exhaustive • Events B, C and D are also collectively exhaustive

8. Probability • Probability is the numerical measure of the likelihood that an event will occur • Value is between 0 and 1 • Sum of the probabilities of all mutually exclusive and collective exhaustive events is 1 1 Certain .5 0 Impossible

9. Computing Probabilities • The probability of an event E: • Each of the outcomes in the sample space is equally likely to occur For example, the probability of drawing an ace from a standard deck of cards is: P(Ace) = 4/52

10. Computing Joint Probability • The probability of a joint event, A and B: If we draw a card from a standard deck or cards, what is the probability it will be a red ace (red and ace)?

11. Joint Probability Using Contingency Table Event Total B1 B2 Event A1 P(A1 and B1) P(A1 and B2) P(A1) P(A2 and B2) A2 P(A2) P(A2 and B1) 1 Total P(B1) P(B2) Marginal (Simple) Probability Joint Probability

12. Marginal and Joint probabilities with a Contingency Table A Deck of 52 Cards Joint Probability of a Red Ace is 2/52 or 1/26 Not an Ace Total Ace Red 2 24 26 Black 2 24 26 Total 4 48 52 Marginal (Simple) Probability of an Ace is 4/52 Sample Space

13. Computing Compound Probability • Probability of a compound event, A or B: P(A or B) = P(A) + P(B) - P(A and B) For Mutually Exclusive Events: P(A or B) = P(A) + P(B)

14. Compound Probability with a Contingency Table P(A or B) = P(A) + P(B) - P(A and B)

15. Computing Conditional Probability • The probability of event A given that event B has occurred: E.g. Suppose we draw a red card – what is the probability it is an Ace? P(Ace given Red Card) = P (Ace and Red Card) / P (Red Card)

16. Conditional Probability Using Contingency Table Color Type Total Red Black 2 2 4 Ace 24 24 48 Non-Ace 26 26 52 Total Revised Sample Space

17. Conditional Probability and Statistical Independence • Conditional probability: • Multiplication rule:

18. Conditional Probability and Statistical Independence (continued) • Events A and B are independent if • Events A and B are independent when the probability of one event, A, is not affected by another event, B

19. Bayes’ Theorem Adding up the parts of A in all the B’s Same Event

20. Bayes’ Theorem Using Contingency Table A study of borrowers that had taken an educational loan found that only fifty percent had repaid their loans. Out of those who repaid, 40% had a college degree. Ten percent of those who defaulted had a college degree. What is the probability that a borrower who has a college degree will repay the loan? Solution: A contingency table helps solve these types of problems. Let’s use the information provided above to set up a contingency table. Since fifty percent of borrowers repaid their loans, we know that the remaining fifty percent of borrowers defaulted on their loans. Of those who repaid, 40% had a college degree – thus, percentage of borrowers who repaid and have a college degree in the sample is .5*.4=.2, or 20%. Ten percent of those who defaulted had a college degree – thus, percentage of borrowers who defaulted and have a college degree in the sample is .5*.1=.05. This information is organized using a contingency table on the next slide

21. Bayes’ Theorem Using Contingency Table (continued) Loan Repaid Not Repaid Total Note: Whether or not the borrower repaid the loan forms one dimension of the contingency table. The other dimension is whether or not the borrower has a college degree. College Degree .2 .05 .25 No College Degree .3 .45 .75 Total .5 .5 1.0 What is the probability that a borrower who has a college degree will repay the loan? Using the contingency table, the probability that a borrower with a college degree will repay the loan is .2/.25, or 80%.

22. Medical Diagnosis problem (Example 4.10 on p. 167/168) The probability a person has a certain disease is 3%. Medical diagnostics tests are available to determine whether the person actually has the disease. If the disease is present, the probability the diagnostic will give a positive result (correctly identify the disease) is 90%. If the disease is absent, the probability of a positive test result is 2%. Suppose a person is administered the test and the result is positive. What is the probability this person has the disease? Solution: Since probability of a person having the disease is 3%, we know that the chance this person does not have the disease is 97%. For a person with disease, the chance of a positive test result is 90% -- thus, the probability that a person has the disease and tests positive is .03X.9=.027 If the disease is absent, the chance of a positive test result is 2% -- thus, the probability that a person has the disease and tests positive is .97X.02=.0194 This information is organized using a contingency table (next slide)

23. Using Contingency Table for the Medical Diagnosis problem on p. 167/168 Has Disease No Disease Total Note: Whether or not the patient has the disease forms one dimension of the contingency table. The other dimension is whether the test is positive or negative. Test Positive .027 .0194 .0464 Test Negative .003 .9506 .9536 Total .03 .97 1.0 What is the probability that a person who tests positive has the disease? Using the contingency table, the probability that a person who tests positive has the disease is .027/.0464, or 58.2%. An alternate way of solving this problem using formulas is shown on pages 167/168

24. More examples similar to the last two problems • See problems 4.30, 4.31, 4.32, 4.33, 4.36 and 4.37 (problems are on pages 170 / 171) • Note: I recommend students use contingency tables because it is a better and a more visual approach to solving these problems – of course, if you are more comfortable using formulas, please feel free – either way, the student is responsible in arriving at the correct response

25. Chapter Summary • Discussed basic probability concepts • Sample spaces and events, simple (marginal) probability, and joint probability • Defined conditional probability • Statistical independence • Discussed Bayes’ theorem