1. Larson/Farber Ch. 3 Weather forecast Psychology Games Sports 3 Elementary Statistics Larson Farber Business Medicine Probability

2. Larson/Farber Ch. 3 Section 3.1 Basic Concepts of Probability

3. Larson/Farber Ch. 3 { 1 2 3 4 5 6 } { Die is even } = { 2 4 6 } {4} Roll a die Probability experiment: An action through which counts, measurements or responses are obtained Sample space: The set of all possible outcomes Event: A subset of the sample space. Outcome: Example The result of a single trial

4. Larson/Farber Ch. 3 Simple Event – an event that consists of a single outcome Example: “tossing heads” and “rolling a 3” is a simple event because there is 1 possible outcome. “tossing heads and “rolling an even number” is not a simple event because there are 3 possible outcomes Simple Event

5. Larson/Farber Ch. 3 Classical (theoretical) Based on educated guesses, intuition and estimates Empirical (actual, statistical, experimental) Subjective (intuition) Types of Probability

6. Larson/Farber Ch. 3 As you increase the number of times a probability experiment is repeated, the empirical probability of the event approaches the theoretical probability of the event. Law of Large Numbers

7. Larson/Farber Ch. 3 Two dice are rolled. Describe the sample space. 1st roll 36 outcomes 2nd roll Start 123456 1 2 3 4 5 6 Tree Diagrams

8. Larson/Farber Ch. 3 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 Find the probability the sum is 4. Find the probability the sum is 11. Find the probability the sum is 4 or 11. Two dice are rolled and the sum is noted. Sample Space and Probabilities

9. Larson/Farber Ch. 3 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 Find the probability the sum is 4. Find the probability the sum is 11. Find the probability the sum is 4 or 11. Two dice are rolled and the sum is noted. Sample Space and Probabilities 3/36 = 1/12 = 0.083 2/36 = 1/18 = 0.056 5/36 = 0.139

10. Larson/Farber Ch. 3 Range of Probabilities The probability of an event E is between 0 and 1, inclusive Probability cannot be negative Probability cannot be greater than 1 If probability is 1, the event is certain If probability is 0, the event is impossible

11. Larson/Farber Ch. 3 Complementary Events The complement of event E is event E´. E´ consists of all the events in the sample space that are not in event E. The day’s production consists of 12 cars, 5 of which are defective. If one car is selected at random, find the probability it is not defective. Solution: P(defective) = 5/12 P(not defective) = 1 - 5/12 = 7/12 = 0.583 P(E´) = 1 - P(E)

12. Larson/Farber Ch. 3 Section 3.2 Conditional Probability and the Multiplication Rule

13. Larson/Farber Ch. 3 The probability an event B will occur, given (on the condition) that another event A has occurred. Two cars are selected from a production line of 12 cars where 5 are defective. What is the probability the 2nd car is defective, given the first car was defective? We write this as P(B|A) and say “probability of B, given A.” Conditional Probability

14. Larson/Farber Ch. 3 Two cars are selected from a production line of 12 cars where 5 are defective. What is the probability the 2nd car is defective, given the first car was defective? Given a defective car has been selected, the conditional sample space has 4 defective out of 11. P(B|A) = 4/11 Conditional Probability

15. Larson/Farber Ch. 3 Two events A and B are independent if the probability of the occurrence of event B is not affected by the occurrence of event A. A = taking an aspirin each day B = having a heart attack A = being a female B = being under 64” tall Two events that are not independent are dependent. A = Being female B = Having type O blood A = 1st child is a boy B = 2nd child is a boy Independent / Dependent Events

16. Larson/Farber Ch. 3 Two dice are rolled. Find the probability the second die is a 4 given the first was a 4. Original sample space : {1, 2, 3, 4, 5, 6} Given the first die was a 4, the conditional sample space is: {1, 2, 3, 4, 5, 6} The conditional probability, P(B|A) = 1/6 Independent Events

17. Larson/Farber Ch. 3 If events A and B are independent, then P(B|A) = P(B) 12 cars are on a production line where 5 are defective and 2 cars are selected at random. A = first car is defective B = second car is defective. The probability of getting a defective car for the second car depends on whether the first was defective. The events are dependent. Dependent Events Conditional Probability Probability

18. Larson/Farber Ch. 3 The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is: 1. P(Yes) 2. P(Seattle) 3. P(Miami) 4. P(No, given Miami) OmahaSeattleMiamiTotal Yes100150 400 No12513095 350 Undecided 75170 5 250 Total3004502501000 One of the responses is selected at random. Find: Contingency Table

19. Larson/Farber Ch. 3 1. P(Yes) 2. P(Seattle) 3. P(Miami) 4. P(No, given Miami) 100150 12513095 350 75170 5 250 OmahaSeattleMiamiTotal Yes No Undecided Total300450250 400 1000 = 95 / 250 = 0.38 = 250 / 1000 = 0.25 Answers: 1) 0.4 2) 0.45 3) 0.25 4) 0.38 = 450 / 1000 = 0.45 Solutions = 400 / 1000 = 0.4

20. Larson/Farber Ch. 3 To find the probability that two events, A and B will occur in sequence, multiply the probability A occurs by the conditional probability B occurs, given A has occurred. P(A and B) = P(A) x P(B|A) If events A and B are independent then the rule can be simplified to Multiplication Rule P(A and B) = P(A) x P(B)

21. Larson/Farber Ch. 3 Two cars are selected from a production line of 12 where 5 are defective. Find the probability both cars are defective. A = first car is defective B = second car is defective. P(A) = 5/12 P(B|A) = 4/11 P(A and B) = 5/12 x 4/11 = 5/33 = 0.152 Multiplication Rule

22. Larson/Farber Ch. 3 Two dice are rolled. Find the probability both are 4’s. Multiplication Rule

23. Larson/Farber Ch. 3 Two dice are rolled. Find the probability both are 4’s. P(A and B) = 1/6 x 1/6 = 1/36 = 0.028 When two events A and B are independent, then P (A and B) = P(A) x P(B) Multiplication Rule

24. Larson/Farber Ch. 3 Section 3.3 The Addition Rule

25. Larson/Farber Ch. 3 Mutually Exclusive Events Two events, A and B, are mutually exclusive if they cannot occur in the same trial. A = A person is under 21 years old B = A person is running for the U.S. Senate A = A person was born in Philadelphia B = A person was born in Houston A B Mutually exclusive P(A and B) = 0 When event A occurs it excludes event B in the same trial.

26. Larson/Farber Ch. 3 Non-Mutually Exclusive Events If two events can occur in the same trial, they are non-mutually exclusive. A = A person is under 25 years old B = A person is a lawyer A = A person was born in Philadelphia B = A person watches Revenge on TV A B Non-mutually exclusive P(A and B) ≠ 0 A and B

27. Larson/Farber Ch. 3 Compare “A and B” to “A or B” The compound event “A and B” means that A and B both occur in the same trial. Use the multiplication rule to find P(A and B). The compound event “A or B” means either A can occur without B, B can occur without A or both A and B can occur. Use the addition rule to find P(A or B). A B A or B A and B A B

28. Larson/Farber Ch. 3 The Addition Rule The probability that one or the other of two events will occur is: P(A or B) = P(A) + P(B) – P(A and B)

29. Larson/Farber Ch. 3 The Addition Rule A card is drawn from a standard deck. Find the probability it is a king or it is red. A = the card is a king B = the card is red. P(A or B) = P(A) + P(B) – P(A and B) P(A) = 4/52, P(B) = 26/52, P(A and B) = 2/52 P(A or B) = 4/52 + 26/52 – 2/52 = 28/52 = 0.538

30. Larson/Farber Ch. 3 The Addition Rule A card is drawn from a standard deck. Find the probability the card is a king or a 10.

31. Larson/Farber Ch. 3 The Addition Rule A card is drawn from a standard deck. Find the probability the card is a king or a 10. A = the card is a king B = the card is a 10. P(A) = 4/52 and P(B) = 4/52 and P(A and B) = 0/52 P(A or B) = 4/52 + 4/52 – 0/52 = 8/52 = 0.154 When events are mutually exclusive, P(A or B) = P(A) + P(B)

32. Larson/Farber Ch. 3 The results of responses when a sample of adults in 3 cities was asked if they liked a new juice is: Contingency Table 3. P(Miami or Yes) 4. P(Miami or Seattle) OmahaSeattleMiamiTotal Yes100150 400 No12513095 350 Undecided 75170 5 250 Total3004502501000 One of the responses is selected at random. Find : 1. P(Miami and Yes) 2. P(Miami and Seattle)

33. Larson/Farber Ch. 3 Contingency Table 1.P(Miami and Yes) = 250/1000 150/250 = 0.15 2.P(Miami and Seattle) = 0 3.P(Miami or Yes) = 250/1000 + 400/1000 – 150/1000 = 0.5 4. P(Miami or Seattle) = 250/1000 + 450/1000 – 0/1000 = 0.7 OmahaSeattleMiamiTotal Yes100150 400 No12513095 350 Undecided 75170 5 250 Total3004502501000 One of the responses is selected at random. Find:

34. Larson/Farber Ch. 3 See page 144 in the text book for a summary of probability Summary

35. Larson/Farber Ch. 3 Section 3.4 Counting Principles

36. Larson/Farber Ch. 3 If one event can occur m ways and a second event can occur n ways, then number of ways the two events can occur in sequence is m n. This rule can be extended for any number of events occurring in a sequence. Fundamental Counting Principle

37. Larson/Farber Ch. 3 If a meal consists of 2 choices of soup, 3 main dishes and 2 desserts, how many different meals can be selected? = 12 meals Start 2 Soup 3 Main 2 Dessert Fundamental Counting Principle

38. Larson/Farber Ch. 3 How many license plates can you make if a license plate consists of a. 6 letters which can be repeated b. 6 letters which cannot be repeated Fundamental Counting Principle

39. Larson/Farber Ch. 3 Factorials Suppose you want to arrange n objects in order. There are n choices for 1 st place, n – 1 choices for second, then n – 2 choices for third place and so on until there is one choice for last place. Using the Fundamental Counting Principle, the number of ways of arranging n objects is: This is called n factorial and written as n! n(n – 1)(n – 2)…1

40. Larson/Farber Ch. 3 Factorials The starting lineup for a baseball team consists of nine players. How many different batting orders are possible using the starting lineup?

41. Larson/Farber Ch. 3 A permutation is an ordered arrangement. The number of permutations for n objects is n! n! = n (n – 1) (n – 2)…..3 2 1 The number of permutations of n objects taken r at a time is: Permutations

42. Larson/Farber Ch. 3 You are required to read 5 books from a list of 8. In how many different orders can you do so? There are 6720 permutations of 8 books reading 5. Permutations

43. Larson/Farber Ch. 3 where n 1 +n 2 +n 3 +…….+n k = n A contractor wants to plant 6 oak trees, 9 maple trees and 5 poplar trees along the street. In how many distinguishable ways can they be planted? Distinguishable Permutations

44. Larson/Farber Ch. 3 A combination is a selection of r objects from a group of n objects. The number of combinations of n objects taken r at a time is: Combinations

45. Larson/Farber Ch. 3 You are required to read 5 books from a list of 8. In how many different ways can you choose the books if order does not matter. There are 56 combinations of 8 objects taking 5. Combinations

46. Larson/Farber Ch. 3 2 3 Combinations of 4 objects choosing 2 1 4 Each of the 6 groups represents a combination. 1 2 1 3 1 4 3 4 2 4 2 3

47. Larson/Farber Ch. 3 2 3 Permutations of 4 objects choosing 2 1 4 Each of the 12 groups represents a permutation. 1 2 1 2 1 3 1 3 2 3 2 3 1 4 3 4 2 4 1 4 3 4 2 4

48. Larson/Farber Ch. 3 A word consists of one L, two Es, two Ts, and one R. If the letters are randomly arranged in order, what is the probability that the arrangement spells the word letter? Applications

49. Larson/Farber Ch. 3 Find the probability of being dealt five diamonds from a standard deck of playing cards. Applications