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

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

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.

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

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.

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.

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.

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.

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.

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.

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. So the probability that a randomly selected passenger on a U.S. carrier will temporarily lose luggage is 1/176 or about.006.

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.).

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

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

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.

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 15 7.3 Probability Definitions and Relationships Sample space: the collection of unique, nonoverlapping 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.

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}

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.

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.

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

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.

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. The definitions can apply either … to events within the same random circumstance or to events from two separate random circumstances.

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.

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.

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.

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

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

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

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

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 29 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).

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 30 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

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

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.

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

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?

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.

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

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 37 A sample is drawn with replacement if individuals are returned to the eligible pool for each selection. A sample is drawn without replacement if sampled individuals are not eligible for subsequent selection. Sampling with and without Replacement

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 38 7.5 Strategies for Finding Complicated Probabilities Example 7.2 Winning the Lottery Event A = winning number is 956. What is P(A)? Method 1: With physical assumption that all 1000 possibilities are equally likely, P(A) = 1/1000. Method 2: Define three events, B 1 = 1 st digit is 9, B 2 = 2 nd digit is 5, B 3 = 3 rd digit is 6 Event A occurs if and only if all 3 of these events occur. Note: P(B 1 ) = P(B 2 ) = P(B 3 ) = 1/10. Since these events are all independent, we have P(A) = (1/10) 3 = 1/1000. * Can be more than one way to find a probability.

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 39 P(A and B): define event in physical terms and see if know probability. Else try multiplication rule (Rule 3). Series of independent events all happen: multiply all individual probabilities (Extension of Rule 3b) One of a collection of mutually exclusive events happens: add all individual probabilities (Rule 2b extended). Check if probability of complement easier, then subtract it from 1 (applying Rule 1). Hints and Advice for Finding Probabilities

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 40 None of a collection of mutually exclusive events happens: find probability one happens, then subtract that from 1. Conditional probability: define event in physical terms and see if know probability. Else try Rule 4 or next bullet as well. Know P(B|A) but want P(A|B): Use Rule 3a to find P(B) = P(A and B) + P(A C and B), then use Rule 4. Hints and Advice for Finding Probabilities

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 41 Step 1: List each separate random circumstance involved in the problem. Step 2: List the possible outcomes for each random circumstance. Step 3: Assign whatever probabilities you can with the knowledge you have. Step 4: Specify the event for which you want to determine the probability. Step 5: Determine which of the probabilities from step 3 and which probability rules can be combined to find the probability of interest. Steps for Finding Probabilities

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 42 Example 7.17 Alicia Is Probably Healthy Steps 1 to 3: Random circumstances, outcomes, probabilities. Random circumstance 1: Alicia’s disease status Possible Outcomes: A = disease; A C = no disease Probabilities: P(A) = 1/1000 =.001; P(A C ) =.999 Random circumstance 2: Alicia’s test results Possible Outcomes: B = test is positive, B C = test is negative Probabilities: P(B|A) =.95 (positive test given disease) P(B C |A) =.05 (negative test given disease) P(B|A C ) =.05 (positive test given no disease) P(B C |A C ) =.95 (negative test given no disease) What is the probability that Alicia has the disease given that the test was positive?

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 43 Example 7.17 Alicia Healthy? (cont) Step 4: Specify event you want to determine the probability. P(disease | positive test) = P(A|B). Step 5: Determine which probabilities and probability rules can be combined to find the probability of interest. Note we have P(B|A) and we want P(A|B). Hints tell us to use P(B) = P(A and B) + P(A C and B). Note P(A and B) = P(B|A)P(A), similarly for P(A C and B). So … P(A C and B) = P(B|A C )P(A C ) = (.05)(.999) =.04995 P(A and B) = P(B|A)P(A) = (.95)(.001) =.00095 P(B) =.04995 +.00095 =.0509 There is less than a 2% chance that Alicia has the disease, even though her test was positive.

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 44 Two-Way Table: “Hypothetical Hundred Thousand” Example 7.8 Teens and Gambling (cont) Sample of 9 th grade teens: 49.1% boys, 50.9% girls. Results: 22.9% of boys and 4.5% of girls admitted they gambled at least once a week during previous year. Start with hypothetical 100,000 teens … (.491)(100,000) = 49,100 boys and thus 50,900 girls Of the 49,100 boys, (.229)(49,100) = 11,244 would be weekly gamblers. Of the 50,900 girls, (.045)(50,900) = 2,291 would be weekly gamblers.

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 45 Example 7.8 Teens and Gambling (cont) P(boy and gambler) = 11,244/100,000 =.1124 P(boy | gambler) = 11,244/13,535 =.8307 P(gambler) = 13,535/100,000 =.13535

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 46 Tree Diagrams Step 1: Determine first random circumstance in sequence, and create first set of branches for possible outcomes. Create one branch for each outcome, write probability on branch. Step 2: Determine next random circumstance and append branches for possible outcomes to each branch in step 1. Write associated conditional probabilities on branches. Step 3: Continue this process for as many steps as necessary. Step 4: To determine the probability of following any particular sequence of branches, multiply the probabilities on those branches. This is an application of Rule 3a. Step 5: To determine the probability of any collection of sequences of branches, add the individual probabilities for those sequences, as found in step 4. This is an application of Rule 2b.

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 47 Example 7.18 Alicia’s Possible Fates P(Alicia has D and has a positive test) =.00095. P(test is positive) =.00095 +.04995 =.0509. P(Alicia has D | positive test) =.00095/.0509 =.019

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 48 Example 7.8 Teens and Gambling (cont) P(boy and gambler) = (.491)(.229) =.1124 P(girl and not gambler) = (.509)(.955) =.4861 P(gambler) =.1124 +.0229 =.1353 P(boy | gambler) =.1124/.1353 =.8307

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 49 7.6 Using Simulation to Estimate Probabilities Some probabilities so difficult or time- consuming to calculate – easier to simulate. If you simulate the random circumstance n times and the outcome of interest occurs in x out of those n times, then the estimated probability for the outcome of interest is x/n.

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 50 Example 7.19 Getting All the Prizes Cereal boxes each contain one of four prizes. Any box is equally likely to contain each of the four prizes. If buy 6 boxes, what is the probability you get all 4 prizes? Shown above are 50 simulations of generating a set of 6 digits, each equally likely to be 1, 2, 3, or 4. There are 19 bold outcomes in which all 4 prizes were collected. The estimated probability is 19/50 =.38. (Actual probability is.3809.)

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 51 7.7 Coincidences & Intuitive Judgments about Probability Confusion of the Inverse Example: Diagnostic Testing Confuse the conditional probability “have the disease” given “a positive test result” -- P(Disease | Positive), with the conditional probability of “a positive test result” given “have the disease” -- P(Positive | Disease), also known as the sensitivity of the test. Often forget to incorporate the base rate for a disease.

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 52 Specific People versus Random Individuals In long run, about 50% of marriages end in divorce. At the beginning of a randomly selected marriage, the probability it will end in divorce is about.50. Does this statement apply to you personally? If you have had a terrific marriage for 30 years, your probability of ending in divorce is surely less than 50%. The chance that your marriage will end in divorce is 50%. Two correct ways to express the aggregate divorce statistics:

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 53 Coincidences Example 7.23 Winning the Lottery Twice A coincidence is a surprising concurrence of events, perceived as meaningfully related, with no apparent causal connection. In 1986, Ms. Adams won the NJ lottery twice in a short time period. NYT claimed odds of one person winning the top prize twice were about 1 in 17 trillion. Then in 1988, Mr. Humphries won the PA lottery twice. 1 in 17 trillion = probability that a specific individual who plays the lottery exactly twice will win both times. Millions of people play the lottery. It is not surprising that someone, somewhere, someday would win twice.

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 54 The Gambler’s Fallacy Primarily applies to independent events. Independent chance events have no memory. Example: Making ten bad gambles in a row doesn’t change the probability that the next gamble will also be bad. The gambler’s fallacy is the misperception of applying a long-run frequency in the short-run.

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