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Презентация была опубликована 2 года назад пользователемЭдуард Водынский

1 © The McGraw-Hill Companies, Inc., Chapter 5 Probability

2 © The McGraw-Hill Companies, Inc., Outline 5-1 Introduction 5-2 Sample Spaces and Probability 5-3 The Addition Rules for Probability 5-4 The Multiplication Rules and Conditional Probability

3 © The McGraw-Hill Companies, Inc., Objectives Determine Sample Spaces and find the probability of an event using classical probability. Find the probability of an event using empirical probability. Find the probability of compound events using the addition rules.

4 © The McGraw-Hill Companies, Inc., Objectives Find the probability of compound events using the multiplication rules. Find the conditional probability of an event.

5 © The McGraw-Hill Companies, Inc., Sample Spaces and Probability probability experiment A probability experiment is a process that leads to well-defined results called outcomes. outcome An outcome is the result of a single trial of a probability experiment. NOTE: NOTE: A tree diagram can be used as a systematic way to find all possible outcomes of a probability experiment.

6 © The McGraw-Hill Companies, Inc., Tree Diagram for Tossing Two Coins First Toss H T H T H T Second Toss

7 © The McGraw-Hill Companies, Inc., Sample Spaces Sample Spaces - Examples

8 © The McGraw-Hill Companies, Inc., Formula for Classical Probability Classical probability assumes that all outcomes in the sample space are equally likely to occur. equally likely That is, equally likely events are events that have the same probability of occurring.

9 © The McGraw-Hill Companies, Inc., Formula for Classical Probability hisprobabilityiscalledclassicalprobability, anditusesthesamplespaceS..

10 © The McGraw-Hill Companies, Inc., Classical Probability Classical Probability - Examples For a card drawn from an ordinary deck, find the probability of getting (a) a queen (b) a 6 of clubs (c) a 3 or a diamond. Solution: P(queen) = 4/52 = 1/13 Solution: (a) Since there are 4 queens and 52 cards, P(queen) = 4/52 = 1/13. P(6 of clubs) = 1/52 (b) Since there is only one 6 of clubs, then P(6 of clubs) = 1/52.

11 © The McGraw-Hill Companies, Inc., Classical Probability Classical Probability - Examples P(3 or diamond) = 16/52 = 4/13 (c) There are four 3s and 13 diamonds, but the 3 of diamonds is counted twice in the listing. Hence there are only 16 possibilities of drawing a 3 or a diamond, thus P(3 or diamond) = 16/52 = 4/13.

12 © The McGraw-Hill Companies, Inc., Classical Probability Classical Probability - Examples When a single die is rolled, find the probability of getting a 9. Solution: P(9) = 0/6 = 0 Solution: Since the sample space is 1, 2, 3, 4, 5, and 6, it is impossible to get a 9. Hence, P(9) = 0/6 = 0. NOTE: NOTE: The sum of the probabilities of all outcomes in a sample space is one.

13 © The McGraw-Hill Companies, Inc., Complement of an Event E

14 © The McGraw-Hill Companies, Inc., Complement of an Event Complement of an Event - Example Find the complement of each event. Rolling a die and getting a 4. Solution: Solution: Getting a 1, 2, 3, 5, or 6. Selecting a letter of the alphabet and getting a vowel. Solution: Solution: Getting a consonant (assume y is a consonant).

15 © The McGraw-Hill Companies, Inc., Complement of an Event Complement of an Event - Example Selecting a day of the week and getting a weekday. Solution: Solution: Getting Saturday or Sunday. Selecting a one-child family and getting a boy. Solution: Solution: Getting a girl.

16 © The McGraw-Hill Companies, Inc., Rule for Complementary Event PEPE or PEPE PEPE ()() ()= () ()+()=.

17 © The McGraw-Hill Companies, Inc., Empirical Probability empirical probability The difference between classical and empirical probability is that classical probability assumes that certain outcomes are equally likely while empirical probability relies on actual experience to determine the probability of an outcome.

18 © The McGraw-Hill Companies, Inc., Formula for Empirical Probability Givenafrequencydistribution, theprobabilityofaneventbeing inagivenclassis PE frequencyfortheclass totalfrequenciesinthedistribution f n Thisprobabilityiscalledtheempirical probabilityandisbasedonobservation. ()=.

19 © The McGraw-Hill Companies, Inc., Empirical Probability Empirical Probability - Example In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had AB blood. Set up a frequency distribution.

20 © The McGraw-Hill Companies, Inc., Empirical Probability Empirical Probability - Example TypeFrequency A B AB O = n

21 © The McGraw-Hill Companies, Inc., Empirical Probability Empirical Probability - Example Find the following probabilities for the previous example. A person has type O blood. Solution: Solution: P(O) = f /n = 21/50. A person has type A or type B blood. Solution: Solution: P(A or B) = 22/50+ 5/50 = 27/50.

22 © The McGraw-Hill Companies, Inc., The Addition Rules for Probability mutually exclusive Two events are mutually exclusive if they cannot occur at the same time (i.e. they have no outcomes in common).

23 © The McGraw-Hill Companies, Inc., The Addition Rules for Probability AB A and B are mutually exclusive

24 © The McGraw-Hill Companies, Inc., Addition Rule 1 When two events A and B are mutually exclusive, the probability that A or B will occur is PAorBPAPB ()()()

25 © The McGraw-Hill Companies, Inc., Addition Rule Addition Rule 1- Example At a political rally, there are 20 Republicans (R), 13 Democrats (D), and 6 Independents (I). If a person is selected, find the probability that he or she is either a Democrat or an Independent. Solution: Solution: P(D or I) = P(D) + P(I) = 13/39 + 6/39 = 19/39.

26 © The McGraw-Hill Companies, Inc., Addition Rule Addition Rule 1- Example A day of the week is selected at random. Find the probability that it is a weekend. Solution: Solution: P(Saturday or Sunday) = P(Saturday) + P(Sunday) = 1/7 + 1/7 = 2/7.

27 © The McGraw-Hill Companies, Inc., Addition Rule 2 When two events A and B are not mutually exclusive, the probabilityythatAorBwill occuris PAorBPAPBPAandB ()()()()

28 © The McGraw-Hill Companies, Inc., Addition Rule 2 A B A and B (common portion)

29 © The McGraw-Hill Companies, Inc., Addition Rule Addition Rule 2- Example In a hospital unit there are eight nurses and five physicians. Seven nurses and three physicians are females. If a staff person is selected, find the probability that the subject is a nurse or a male. The next slide has the data.

30 © The McGraw-Hill Companies, Inc., Addition Rule Addition Rule 2 - Example

31 © The McGraw-Hill Companies, Inc., Addition Rule Addition Rule 2 - Example Solution: Solution: P(nurse or male) = P(nurse) + P(male) – P(male nurse) = 8/13 + 3/13 – 1/13 = 10/13.

32 © The McGraw-Hill Companies, Inc., Addition Rule Addition Rule 2 - Example On New Years Eve, the probability that a person driving while intoxicated is 0.32, the probability of a person having a driving accident is 0.09, and the probability of a person having a driving accident while intoxicated is What is the probability of a person driving while intoxicated or having a driving accident?

33 © The McGraw-Hill Companies, Inc., Addition Rule Addition Rule 2 - Example Solution: Solution: P(intoxicated or accident) = P(intoxicated) + P(accident) – P(intoxicated and accident) = – 0.06 = 0.35.

34 © The McGraw-Hill Companies, Inc., independent Two events A and B are independent if the fact that A occurs does not affect the probability of B occurring. Example: Example: Rolling a die and getting a 6, and then rolling another die and getting a 3 are independent events. 5-4 The Multiplication Rules and Conditional Probability

35 © The McGraw-Hill Companies, Inc., Multiplication Rule 1 When two events A and B are independent, the probability of both occurring is P(A and B) = P(A) · P(B)

36 © The McGraw-Hill Companies, Inc., Multiplication Rule Multiplication Rule 1 - Example A card is drawn from a deck and replaced; then a second card is drawn. Find the probability of getting a queen and then an ace. Solution: Solution: Because these two events are independent (why?), P(queen and ace) = (4/52) (4/52) = 16/2704 = 1/169.

37 © The McGraw-Hill Companies, Inc., Multiplication Rule Multiplication Rule 1 - Example A Harris pole found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer stress at least once a week. Solution: Solution: Let S denote stress. Then P(S and S and S) = (0.46) 3 =

38 © The McGraw-Hill Companies, Inc., Multiplication Rule Multiplication Rule 1 - Example The probability that a specific medical test will show positive is If four people are tested, find the probability that all four will show positive. Solution: Solution: Let T denote a positive test result. Then P(T and T and T and T) = (0.32) 4 =

39 © The McGraw-Hill Companies, Inc., When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed, the events are said to be dependent. Example: Example: Having high grades and getting a scholarship are dependent events. 5-4 The Multiplication Rules and Conditional Probability

40 © The McGraw-Hill Companies, Inc., conditional probability The conditional probability of an event B in relationship to an event A is the probability that an event B occurs after event A has already occurred. The notation for the conditional probability of B given A is P(B|A). NOTE: NOTE: This does not mean B A. 5-4 The Multiplication Rules and Conditional Probability

41 © The McGraw-Hill Companies, Inc., Multiplication Rule 2 WhentwoeventsAandB aredependentthe probabilityofboth occurringis PAandBPAPBA, ()()(|). ·

42 © The McGraw-Hill Companies, Inc., In a shipment of 25 microwave ovens, two are defective. If two ovens are randomly selected and tested, find the probability that both are defective if the first one is not replaced after it has been tested. Solution: Solution: See next slide. 5-4 The Multiplication Rules and Conditional Probability The Multiplication Rules and Conditional Probability - Example

43 © The McGraw-Hill Companies, Inc., Solution: Solution: Since the events are dependent, P(D 1 and D 2 ) = P(D 1 ) P(D 2 | D 1 ) = (2/25)(1/24) = 2/600 = 1/ The Multiplication Rules and Conditional Probability The Multiplication Rules and Conditional Probability - Example

44 © The McGraw-Hill Companies, Inc., The WW Insurance Company found that 53% of the residents of a city had homeowners insurance with its company. Of these clients, 27% also had automobile insurance with the company. If a resident is selected at random, find the probability that the resident has both homeowners and automobile insurance. 5-4 The Multiplication Rules and Conditional Probability The Multiplication Rules and Conditional Probability - Example

45 © The McGraw-Hill Companies, Inc., Solution: Solution: Since the events are dependent, P(H and A) = P(H) P(A|H) = (0.53)(0.27) = The Multiplication Rules and Conditional Probability The Multiplication Rules and Conditional Probability - Example

46 © The McGraw-Hill Companies, Inc., Box 1 contains two red balls and one blue ball. Box 2 contains three blue balls and one red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the probability of selecting a red ball. 5-4 The Multiplication Rules and Conditional Probability The Multiplication Rules and Conditional Probability - Example

47 © The McGraw-Hill Companies, Inc., Tree Diagram for 5-4 Tree Diagram for Example P(B 1 ) 1/2 Red Blue Box 1 P(B 2 ) 1/2 Box 2 P(R|B 1 ) 2/3 P(B|B 1 ) 1/3 P(R|B 2 ) 1/4 P(B|B 2 ) 3/4 (1/2)(2/3) (1/2)(1/3) (1/2)(1/4) (1/2)(3/4)

48 © The McGraw-Hill Companies, Inc., Solution: Solution: P(red) = (1/2)(2/3) + (1/2)(1/4) = 2/6 + 1/8 = 8/24 + 3/24 = 11/ The Multiplication Rules and Conditional Probability The Multiplication Rules and Conditional Probability - Example

49 © The McGraw-Hill Companies, Inc., Conditional Probability - Formula s occurredbytheprobabilitythatthefirsteventhas occurredTheformulais PBA PAandB PA second.

50 © The McGraw-Hill Companies, Inc., The probability that Sam parks in a no- parking zone and gets a parking ticket is 0.06, and the probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.2. On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a ticket. 5-4 Conditional Probability Conditional Probability - Example

51 © The McGraw-Hill Companies, Inc., Solution: Solution: Let N = parking in a no-parking zone and T = getting a ticket. Then P(T |N) = [P(N and T) ]/P(N) = 0.06/0.2 = Conditional Probability Conditional Probability - Example

52 © The McGraw-Hill Companies, Inc., A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results are shown in the table on the next slide. 5-4 Conditional Probability Conditional Probability - Example

53 © The McGraw-Hill Companies, Inc., Conditional Probability Conditional Probability - Example

54 © The McGraw-Hill Companies, Inc., Find the probability that the respondent answered yes given that the respondent was a female. Solution: Solution: Let M = respondent was a male; F = respondent was a female; Y = respondent answered yes; N = respondent answered no. 5-4 Conditional Probability Conditional Probability - Example

55 © The McGraw-Hill Companies, Inc., P(Y|F) = [P( F and Y) ]/P(F) = [8/100]/[50/100] = 4/25. Find the probability that the respondent was a male, given that the respondent answered no. Solution: P(M|N) = [P(N and M)]/P(N) = [18/100]/[60/100] = 3/ Conditional Probability Conditional Probability - Example

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