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Chapter 5

Chapter 5. Survey of Probability Concepts SOLVED PROBLEMS. Problem 5-1. Outcome 1 2 1 A A 2 A F 3 F A 4 F F. Problem 5-3, 4. 3. Accounting 10 Finance 5 Economics 3

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Chapter 5

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  1. Chapter 5 Survey of Probability Concepts SOLVED PROBLEMS

  2. Problem 5-1 Outcome 1 2 1 A A 2 A F 3 F A 4 F F

  3. Problem 5-3, 4 3. Accounting 10 Finance 5 Economics 3 Management 6 Marketing 10 Total 34 a. P(Mgt) = b. Empirical 4. a P(minority) = b. classical

  4. Problem 5-8 • a. Number of violations • b. at least one violation • c. P(x=2) = • d. empirical

  5. Problem 5-11

  6. Problem 5-12

  7. Problem 5-14 • P(P) = .50 • P(BE) = .30 • P(L) = .20 • a. P( No loss) = P(P) + P(BE) = .50+.30=.80 • b. P(No loss) = 1-P(L)= 1-.20=.80

  8. Problem 5-17

  9. Problem 5-18

  10. Problem 5-26

  11. Problem 5-27

  12. Problem 5-27 Continued

  13. Problem 5-28

  14. Problem 5-29(modified) • Potential for Advancement Remember probability of an event is the numberof ways the event can occur divided by the total possible outcomes. You can find the probabilities of an event by the counts contained in a contingency table.

  15. Find the following probabilities:1. P(AA and E) = 135/500=.272. P(F or A)= 154/500+150/500-45/500=259/500=.518  - events are not  mutually exclusive - general rule of addition.3. P(F or E)=154/500+202/500=356/500=.712   - events are mutually exclusive - special rule of addition.4. P(A and G)=60/500=.12  -  the person must have both attributes of being A and G and there are only 60 of them. 5.P(E/A)=45/150=.30  -   remember the / indicates given or on the condition that we have an average person on sales ability. So our denominator becomes 150. This is a conditional probability. 6. P(BA/G)=12/144=.08333 7. P(A and E )=45/500=.098. P(A or E) = 150/500 + 202/500 - 45/500=307/500=.6149. P (BA/E)=22/202=.108910. P( E and BA)= 22/500=.044 Problem 5-29(cont’d)

  16. Problem 5-41

  17. Problem 5-42 • Exchange No. Each digit can range is fixed from 0-9. 5370000 9999

  18. Problem 5-43 nPr nCr

  19. Problem 5-44

  20. Problem 5-45 15P10

  21. Problem 5-45 Continued • 15C10

  22. Problem 5-66(Previous Edition)

  23. Problem 5-66 (Previous Edition) An example of the special rule of addition of probabilities

  24. Problem 5-66 Continued (Previous Edition) An example of the general rule of the addition of probabilities

  25. Problem 5-66 Continued (Previous Edition) AN EXAMPLE OF THE GENERAL RULE OF MULTIPLICATION

  26. Problem 5-66 ContinuedExtra questions

  27. Problem 5-66Joint Probability Table

  28. Problem 5-67(Previous Edition)

  29. Problem 5-67Continued(Previous Edition)

  30. Problem 5-67 Continued(Previous Edition)

  31. Problem 5-78 Given: Smoke Do not smoke Male 75 250 Female 150 100 300 400 Completed table: Smoke Do not smoke Male 75 175 250 Female 25 125 150 100 300 400

  32. Problem 5-78 Contd.. Joint probability Table: Smokers Non-Smokers Male .1875 .4375 .625 Female .0425 .3125 .375 .25 .75 1.00 • P(M) = .625 • P(S) = .25 • P(M and S) = .1875 • P(M or S) = P(M) + P(S) – P (M and S) = .625 + .25 - .1875 =.6875 • Extra problem : given that you selected a smoker, What is the probability that it is a male? P(M/S) =P(M and S) / P(S) = .1875 / .25 = .75

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