1. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. Probability Chapter 7

2. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 2 7.1 Random Circumstances Random circumstance is one in which the outcome is unpredictable. Case Study 1.1 Alicia Has a Bad Day Doctor Visit: Diagnostic test comes back positive for a disease (D). Test is 95% accurate. About 1 out of 1000 women actually have D. Statistics Class: Professor randomly selects 3 separate students at the beginning of each class to answer questions. Alicia is picked to answer the third question.

3. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 3 Random Circumstance 1: Disease status Alicia has D. Alicia does not have D. Random Circumstance 2: Test result Test is positive. Test is negative. Random Circumstances in Alicia’s Day

4. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 4 Random Circumstances in Alicia’s Day Random Circumstance 3: 1 st student’s name is drawn Alicia is selected. Alicia is not selected. Random Circumstance 4: 2 nd student’s name is drawn Alicia is selected. Alicia is not selected. Random Circumstance 5: 3 rd student’s name is drawn Alicia is selected. Alicia is not selected.

5. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 5 Assigning Probabilities A probability is a value between 0 and 1 and is written either as a fraction or as a decimal fraction. A probability simply is a number between 0 and 1 that is assigned to a possible outcome of a random circumstance. For the complete set of distinct possible outcomes of a random circumstance, the total of the assigned probabilities must equal 1.

6. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 6 7.2 Interpretations of Probability The Relative Frequency Interpretation of Probability In situations that we can imagine repeating many times, we define the probability of a specific outcome as the proportion of times it would occur over the long run -- called the relative frequency of that particular outcome.

7. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 7 Example 7.1 Probability of Male versus Female Births Long-run relative frequency of males born in the United States is about.512. Information Please Almanac (1991, p. 815). Table provides results of simulation: the proportion is far from.512 over the first few weeks but in the long run settles down around.512.

8. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 8 Determining the Relative Frequency Probability of an Outcome Method 1: Make an Assumption about the Physical World Example 7.2 A Simple Lottery Choose a three-digit number between 000 and 999. Player wins if his or her three-digit number is chosen. Suppose the 1000 possible 3-digit numbers (000, 001, 002,..., 999) are equally likely. In long run, a player should win about 1 out of 1000 times. This does not mean a player will win exactly once in every thousand plays.

9. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 9 Determining the Relative Frequency Probability of an Outcome Method 1: Make an Assumption about the Physical World Example 7.3 Probability Alicia has to Answer a Question There are 50 student names in a bag. If names mixed well, can assume each student is equally likely to be selected. Probability Alicia will be selected to answer the first question is 1/50.

10. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 10 Determining the Relative Frequency Probability of an Outcome Method 2: Observe the Relative Frequency Example 7.4 The Probability of Lost Luggage “1 in 176 passengers on U.S. airline carriers will temporarily lose their luggage.” This number is based on data collected over the long run. The probability that a randomly selected passenger on a U.S. carrier will temporarily lose luggage is 1/176 or about.006.

11. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 11 Proportions and Percentages as Probabilities Ways to express the relative frequency of lost luggage: The proportion of passengers who lose their luggage is 1/176 or about.006. About 0.6% of passengers lose their luggage. The probability that a randomly selected passenger will lose his/her luggage is about.006. The probability that you will lose your luggage is about.006. Last statement is not exactly correct – your probability depends on other factors (how late you arrive at the airport, etc.).

12. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 12 Estimating Probabilities from Observed Categorical Data Assuming data are representative, the probability of a particular outcome is estimated to be the relative frequency (proportion) with which that outcome was observed. Approximate margin of error for the estimated probability is

13. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 13 Example 7.5 Nightlights and Myopia Revisited Assuming these data are representative of a larger population, what is the approximate probability that someone from that population who sleeps with a nightlight in early childhood will develop some degree of myopia? Note: 72 + 7 = 79 of the 232 nightlight users developed some degree of myopia. So we estimate the probability to be 79/232 =.34. This estimate is based on a sample of 232 people with a margin of error of about.066

14. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 14 The Personal Probability Interpretation Personal probability of an event = the degree to which a given individual believes the event will happen. Sometimes subjective probability used because the degree of belief may be different for each individual. Restrictions on personal probabilities: Must fall between 0 and 1 (or between 0 and 100%). Must be coherent (consistent).

15. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 15 7.3 Probability Definitions and Relationships Sample space: the collection of unique, non- overlapping possible outcomes of a random circumstance. Simple event: one outcome in the sample space; a possible outcome of a random circumstance. Event: a collection of one or more simple events in the sample space; often written as A, B, C, and so on.

16. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 16 Example 7.6 Days per Week of Drinking Random sample of college students. Q: How many days do you drink alcohol in a typical week? Simple Events in the Sample Space are: 0 days, 1 day, 2 days, …, 7 days Event “4 or more” is comprised of the simple events {4 days, 5 days, 6 days, 7 days}

17. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 17 Assigning Probabilities to Simple Events P(A) = probability of the event A Conditions for Valid Probabilities 1.Each probability is between 0 and 1. 2.The sum of the probabilities over all possible simple events is 1. Equally Likely Simple Events If there are k simple events in the sample space and they are all equally likely, then the probability of the occurrence of each one is 1/k.

18. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 18 Example 7.2 A Simple Lottery (cont) Random Circumstance: A three-digit winning lottery number is selected. Sample Space: {000,001,002,003,...,997,998,999}. There are 1000 simple events. Probabilities for Simple Event: Probability any specific three-digit number is a winner is 1/1000. Assume all three-digit numbers are equally likely. Event A = last digit is a 9 = {009,019,...,999}. Since one out of ten numbers in set, P(A) = 1/10. Event B = three digits are all the same = {000, 111, 222, 333, 444, 555, 666, 777, 888, 999}. Since event B contains 10 events, P(B) = 10/1000 = 1/100.

19. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 19 Complementary Events Note: P(A) + P( A C ) = 1 One event is the complement of another event if the two events do not contain any of the same simple events and together they cover the entire sample space. Notation: A C represents the complement of A. Example 7.2 A Simple Lottery (cont) A = player buying single ticket wins A C = player does not win P(A) = 1/1000 so P(A C ) = 999/1000

20. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 20 Mutually Exclusive Events Two events are mutually exclusive, or equivalently disjoint, if they do not contain any of the same simple events (outcomes). Example 7.2 A Simple Lottery (cont) A = all three digits are the same. B = the first and last digits are different The events A and B are mutually exclusive (disjoint), but they are not complementary.

21. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 21 Independent and Dependent Events Two events are independent of each other if knowing that one will occur (or has occurred) does not change the probability that the other occurs. Two events are dependent if knowing that one will occur (or has occurred) changes the probability that the other occurs.

22. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 22 Example 7.7 Winning a Free Lunch Customers put business card in restaurant glass bowl. Drawing held once a week for free lunch. You and Vanessa put a card in two consecutive weeks. Event A = You win in week 1. Event B = Vanessa wins in week 1. Event C = Vanessa wins in week 2. Events A and B refer to the same random circumstance and are not independent. Events A and C refer to to different random circumstances and are independent.

23. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 23 Example 7.3 Alicia Answering (cont) Event A = Alicia is selected to answer Question 1. Event B = Alicia is selected to answer Question 2. P(A) = 1/50. If event A occurs, her name is no longer in the bag, so P(B) = 0. If event A does not occur, there are 49 names in the bag (including Alicia’s name), so P(B) = 1/49. Events A and B refer to different random circumstances, but are A and B independent events? Knowing whether A occurred changes P(B). Thus, the events A and B are not independent.

24. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 24 Conditional Probabilities Conditional probability of the event B, given that the event A occurs, is the long-run relative frequency with which event B occurs when circumstances are such that A also occurs; written as P(B|A). P(B) = unconditional probability event B occurs. P(B|A) = “probability of B given A” = conditional probability event B occurs given that we know A has occurred or will occur.

25. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 25 Example 7.8 Probability That a Teenager Gambles Depends upon Gender Notice dependence between “weekly gambling habit” and “gender.” Knowledge of a 9 th grader’s gender changes probability that he/she is a weekly gambler. Survey: 78,564 students (9 th and 12 th graders) The proportions of males and females admitting they gambled at least once a week during the previous year were reported. Results for 9 th grade: P(student is weekly gambler | teen is boy) =.20 P(student is weekly gambler | teen is girl) =.05

26. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 26 7.4 Basic Rules for Finding Probabilities Rule 1 (for “not the event”): P(A C ) = 1 – P(A) Example 7.9 Probability a Stranger Does Not Share Your Birth Date P(next stranger you meet will share your birthday) = 1/365. P(next stranger you meet will not share your birthday) = 1 – 1/365 = 364/365 =.9973. Probability an Event Does Not Occur

27. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 27 Rule 2 (addition rule for “either/or”): Rule 2a (general): P(A or B) = P(A) + P(B) – P(A and B) Rule 2b (for mutually exclusive events): If A and B are mutually exclusive events, P(A or B) = P(A) + P(B) Probability That Either of Two Events Happen

28. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 28 Example 7.10 Roommate Compatibility Brett is off to college. There are 1000 male students. Brett hopes his roommate will not like to party and not snore. A = likes to partyP(A) = 250/1000 =.25 B = snoresP(B) = 350/1000 =.35 Probability Brett will be assigned a roommate who either likes to party or snores, or both is: P(A or B) = P(A) + P(B) – P(A and B) =.25 +.35 –.15 =.45 So the probability his roommate is acceptable is 1 –.45 =.55

29. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 29 Rule 3 (multiplication rule for “and”): Rule 3a (general): P(A and B) = P(A)P(B|A) Rule 3b (for independent events): If A and B are independent events, P(A and B) = P(A)P(B) Extension of Rule 3b (for > 2 indep events): For several independent events, P(A 1 and A 2 and … and A n ) = P(A 1 )P(A 2 )…P(A n ) Probability That Two or More Events Occur Together

30. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 30 Example 7.11 Probability of Two Boys or Two Girls in Two Births What is the probability that a woman who has two children has either two girls or two boys? Recall that the probability of a boy is.512 and probability of a girl is.488. Then we have (using Rule 3b): Event A = two girls P(A) = (.488)(.488) =.2381 Event B = two boys P(B) = (.512)(.512) =.2621 Probability woman has either two boys or two girls is: P(A or B) = P(A) + P(B) =.2381 +.2621 =.5002 Note: Events A and B are mutually exclusive (disjoint).

31. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 31 Example 7.8 Probability of Male and Gambler (cont) About 11% of all 9 th graders are males and weekly gamblers. For 9 th graders, 22.9% of the boys and 4.5% of the girls admitted they gambled at least once a week during the previous year. The population consisted of 50.9% girls and 49.1% boys. P(male and gambler) = P(A and B) = P(A)P(B|A) = (.491)(.229) =.1124 Event A = male Event B = weekly gambler P(A) =.491P(B|A) =.229

32. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 32 Example 7.12 Probability Two Strangers Both Share Your Birth Month Note: The probability that 4 unrelated strangers all share your birth month would be (1/12) 4. Assume all 12 birth months are equally likely. What is the probability that the next two unrelated strangers you meet both share your birth month? P(both strangers share your birth month) = P(A and B) = P(A)P(B) = (1/12)(1/12) =.007 Event A = 1 st stranger shares your birth month P(A) = 1/12 Event B = 2 nd stranger shares your birth month P(B) = 1/12 Note: Events A and B are independent.

33. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 33 Rule 4 (conditional probability): P(B|A) = P(A and B)/P(A) P(A|B) = P(A and B)/P(B) Determining a Conditional Probability

34. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 34 Example 7.13 Alicia Answering A = Alicia selected to answer Question 1, P(A) = 1/50 B = Alicia is selected to answer any one of the questions, P(B) = 3/50 Since A is a subset of B, P(A and B) = 1/50 P(A|B) = P(A and B)/P(B) = (1/50)/(3/50) = 1/3 If we know Alicia is picked to answer one of the questions, what is the probability it was the first question?

35. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 35 In Summary … When two events are mutually exclusive and one happens, it turns the probability of the other one to 0. When two events are independent and one happens, it leaves the probability of the other one alone. Students sometimes confuse the definitions of independent and mutually exclusive events.

36. Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 36 In Summary …