9/26/12 Learning Objectives Probability in our Daily Lives 1. Random Phenomena 2. Law of Large Numbers 3. Probability 4. Independent Trials 5. Finding probabilities Section 5.1: How can Probability Quantify Randomness? 6. Types of Probabilities: Relative Frequency and Subjective Learning Objective 1: Random Phenomena Learning Objective 2: Law of Large Numbers n For random phenomena, the outcome is n As the number of trials increase, the proportion of uncertain n In the short-run, the proportion of times that something happens is highly random n In the long-run, the proportion of times that something happens becomes very predictable occurrences of any given outcome approaches a particular number “in the long run” n For example, as one tosses a die, in the long run 1/6 of the observations will be a 6. Probability quantifies long-run randomness Learning Objective 3: Probability Learning Objective 4: Independent Trials n With random phenomena, the probability of a n Different trials of a random phenomenon are particular outcome is the proportion of times that the outcome would occur in a long run of observations n Example: independent if the outcome of any one trial is not affected by the outcome of any other trial. n Example: n When rolling a die, the outcome of “6” has probability = 1/6. In other words, the proportion of times that a 6 would occur in a long run of observations is 1/6. n If you have 20 flips of a coin in a row that are “heads”, you are not “due” a “tail” - the probability of a tail on your next flip is still 1/2. The trial of flipping a coin is independent of previous flips. 1 9/26/12 Learning Objective 5: How can we find Probabilities? Learning Objective 6: Types of Probability: Relative Frequency vs. Subjective n Calculate theoretical probabilities based on n The relative frequency definition of probability is the assumptions about the random phenomena. For example, it is often reasonable to assume that outcomes are equally likely such as when flipping a coin, or a rolling a die. n Observe many trials of the random phenomenon and use the sample proportion of the number of times the outcome occurs as its probability. This is merely an estimate of the actual probability. long run proportion of times that the outcome occurs in a very large number of trials - not always helpful/ possible. n When a long run of trials is not feasible, you must rely on subjective information. In this case, the subjective definition of the probability of an outcome is your degree of belief that the outcome will occur based on the information available. n Bayesian statistics is a branch of statistics that uses subjective probability as its foundation Learning Objectives Chapter 5: Probability in Our Daily Lives 1. Sample Space 2. Event 3. Probabilities for a sample space 4. Probability of an event 5. Basic rules for finding probabilities about a pair of events Section 5.2: How Can We Find Probabilities? Learning Objectives Learning Objective 1: Sample Space 6. Probability of the union of two events n For a random phenomenon, the sample 7. Probability of the intersection of two events space is the set of all possible outcomes. 2 9/26/12 Learning Objective 2: Event Learning Objective 3: Probabilities for a sample space n An event is a subset of the sample space Each outcome in a sample space has a probability n The probability of each individual outcome is between 0 and 1 n The total of all the individual probabilities equals 1. n An event corresponds to a particular outcome or a group of possible outcomes. n For example; n Event A = student answers all 3 questions correctly = (CCC) n Event B = student passes (at least 2 correct) = (CCI, CIC, ICC, CCC) Learning Objective 4: Probability of an Event Learning Objective 4: Example: What are the Chances of being Audited? n The probability of an event A, denoted by P(A), is obtained by adding the probabilities of the individual outcomes in the event. n When all the possible outcomes are equally likely: n What is the sample space for selecting a P ( A) = number of outcomes in event A number of outcomes in the sample space Learning Objective 4: Example: What are the Chances of being Audited? For a randomly selected taxpayer in 2002, n What is the probability of an audit? n 310/80200=0.004 n What is the probability of an income of $100,000 or more? n 10700/80200=0.133 n What income level has the greatest probability of being audited? n $100,000 taxpayer? {(under $25,000, Yes), (under $25,000, No), ($25,000 - $49,000, Yes) …} Learning Objective 5: Basic rules for finding probabilities about a pair of events n Some events are expressed as the outcomes that n Are not in some other event (complement of the event) n Are in one event and in another event (intersection of two events) n Are in one event or in another event (union of two events) or more = 80/10700= 0.007 3 9/26/12 Learning Objective 5: Complement of an event Learning Objective 5: Disjoint events n The complement of an event A consists of all n Two events, A and B, are disjoint if they do outcomes in the sample space that are not in A. n The probabilities of A and of Ac add to 1 n P(Ac) = 1 – P(A) not have any common outcomes Learning Objective 5: Intersection of two events Learning Objective 5: Union of two events n The intersection of A and B consists of n The union of A and B consists of outcomes outcomes that are in both A and B that are in A or B or in both A and B. Learning Objective 6: Probability of the Union of Two Events Learning Objective 6: Example Addition Rule: For the union of two events, P(A or B) = P(A) + P(B) – P(A and B) n 80.2 million tax payers (80,200 thousand) If the events are disjoint, P(A and B) = 0, so P(A or B) = P(A) + P(B) n Event A = being audited n Event B = income greater than $100,000 n P(A and B) = 80/80200=.001 4 9/26/12 Learning Objective 7: Probability of the Intersection of Two Events Learning Objective 7: Example Multiplication Rule: n What is the probability of getting 3 questions For the intersection of two independent events, A and B, P(A and B) = P(A) x P(B) correct by guessing? n Probability of guessing correctly is .2 What is the probability that a student answers at least 2 questions correctly? P(CCC) + P(CCI) + P(CIC) + P(ICC) = 0.008 + 3(0.032) = 0.104 Learning Objective 7: Assuming independence Learning Objective 7: Events Often Are Not Independent n Don’t assume that events are independent n Example: A Pop Quiz with 2 Multiple Choice unless you have given this assumption careful thought and it seems plausible. Questions n Data giving the proportions for the actual responses of students in a class Outcome: II IC Probability: 0.26 0.11 Learning Objective 7: Events Often Are Not Independent n Define the events A and B as follows: A: {first question is answered correctly} n B: {second question is answered correctly} n P(A) = P{(CI), (CC)} = 0.05 + 0.58 = 0.63 n P(B) = P{(IC), (CC)} = 0.11 + 0.58 = 0.69 n P(A and B) = P{(CC)} = 0.58 n n If A and B were independent, P(A and B) = P(A) x P(B) = 0.63 x 0.69 = 0.43 Thus, in this case, A and B are not independent! CI CC 0.05 0.58 Chapter 5 Probability in Our Daily Lives Section 5.3: Conditional Probability: What’s the Probability of A, Given B? 5 9/26/12 Learning Objectives Learning Objective 1: Conditional Probability 1. Conditional probability n For events A and B, the conditional probability of 2. Multiplication rule for finding P(A and B) event A, given that event B has occurred, is: 3. Independent events defined using P( A | B ) = conditional probability P ( A and B ) P( B) n P(A|B) is read as “the probability of event A, given event B.” The vertical slash represents the word “given”. Of the times that B occurs, P(A|B) is the proportion of times that A also occurs Learning Objective 1: Conditional Probability Learning Objective 1: Example 1 Learning Objective 1: Example 1 Learning Objective 1: Example 1 n What was the probability of being audited, given that the income was ≥ $100,000? n Event n Event A: Taxpayer is audited B: Taxpayer’s income ≥ $100,000 P(A | B) = P(A and B) 0.0010 = = 0.007 P(B) 0.1334 6 9/26/12 Learning Objective 1: Example 1 Learning Objective 1: Example 2 n What is the probability of being audited given that the n A study of 5282 women aged 35 or over income level is < $25,000 n Let A =Event being Audited n Let B = Income < $25,000 P( A | B ) = analyzed the Triple Blood Test to test its accuracy P ( A and B ) P( B) n P(A and B) = .0011 n P(B)=.1758 n .0011/.1758=.0063 Learning Objective 1: Example 2 n A positive test result states that the condition is present n A negative test result states that the condition is not present Learning Objective 1: Example 2 n Assuming the sample is representative of the population, find the estimated probability of a positive test for a randomly chosen pregnant woman 35 years or older n P(POS) = 1355/5282 = 0.257 n False Positive: Test states the condition is present, but it is actually absent n False Negative: Test states the condition is absent, but it is actually present Learning Objective 1: Example 2 Learning Objective 2: Multiplication Rule for Finding P(A and B) n Given that the diagnostic test result is positive, find the estimated probability that Down syndrome truly is present P(D and POS) 48 / 5282 P(D | POS) = = = P(POS) 1355 / 5282 0.009 = 0.035 0.257 n For events A and B, the probability that A and B both occur equals: n P(A and B) = P(A|B) x P(B) also n P(A and B) = P(B|A) x P(A) n Summary: Of the women who tested positive, fewer than 4% actually had fetuses with Down syndrome 7 9/26/12 Learning Objective 2: Example Learning Objective 2: Example n Roger Federer – 2006 men’s champion in n Assuming these are typical of his serving the Wimbledon tennis tournament n He made 56% of his first serves n He faulted on the first serve 44% of the time n Given that he made a fault with his first serve, he made a fault on his second serve only 2% of the time Learning Objective 3: Independent Events Defined Using Conditional Probabilities performance, when he serves, what is the probability that he makes a double fault? n P(F1) = 0.44 n P(F2|F1) = 0.02 n P(F1 and F2) = P(F2|F1) x P(F1) = 0.02 x 0.44 = 0.009 Learning Objective 3: Checking for Independence n To determine whether events A and B are n Two events A and B are independent if the probability that one occurs is not affected by whether or not the other event occurs n Events A and B are independent if: P(A|B) = P(A), or equivalently, P(B|A) = P(B) independent: n Is P(A|B) = P(A)? P(B|A) = P(B)? n Is P(A and B) = P(A) x P(B)? n Is n If any of these is true, the others are also true and the events A and B are independent n If events A and B are independent, P(A and B) = P(A) x P(B) Learning Objective 3: Example Learning Objective 3: Example: Checking for Independent n The diagnostic blood test for Down syndrome: Status n Are the events POS and D independent or dependent? Is P(POS|D) = P(POS)? POS = positive result NEG = negative result D = Down Syndrome DC = Unaffected n P(POS|D) =P(POS and D)/P(D) POS NEG Total D 0.009 0.001 0.010 Dc 0.247 0.742 0.990 Total 0.257 0.743 1.000 = 0.009/0.010 = 0.90 n P(POS) = 0.257 n The events POS and D are dependent 8 9/26/12 Learning Objectives Chapter 5: Probability in Our Daily Lives 1. Is a “Coincidence” Truly an Unusual Event? 2. Probability Model 3. Probabilities and Diagnostic Testing 4. Simulation Section 5.4: Applying the Probability Rules Learning Objective 1: Is a “Coincidence” Truly an Unusual Event? n The law of very large numbers states that Learning Objective 1: Example: Is a Matching Birthday Surprising? n What is the probability that at least two if something has a very large number of opportunities to happen, occasionally it will happen, even if it seems highly unusual students in a group of 25 students have the same birthday? Learning Objective 1: Example: Is a Matching Birthday Surprising? Learning Objective 1: Example: Is a Matching Birthday Surprising? n P(at least one match) = 1 – P(no matches) n P(no matches) = P(students 1 and 2 and 3 …and 25 have different birthdays) 9 9/26/12 Learning Objective 1: Example: Is a Matching Birthday Surprising? Learning Objective 1: Example: Is a Matching Birthday Surprising? n P(no matches) = (365/365) x (364/365) x (363/365) x … x (341/365) n P(no matches) = 0.43 n P(at least one match) = 1 – P(no matches) = 1 – 0.43 = 0.57 Not so surprising when you consider that there are 300 pairs of students who can share the same birthday! Learning Objective 2: Probability Model Learning Objective 2: Probability Model n We’ve dealt with finding probabilities in many n A probability model specifies the possible idealized situations n In practice, it’s difficult to tell when outcomes are equally likely or events are independent n In most cases, we must specify a probability model that approximates reality outcomes for a sample space and provides assumptions on which the probability calculations for events composed of these outcomes are based n Probability models merely approximate reality Learning Objective 2: Example: Probability Model Learning Objective 2: Example: Probability Model n Out of the first 113 space shuttle missions n This answer relies on the assumptions of there were two failures n What is the probability of at least one failure in a total of 100 missions? n P(at least 1 failure)=1-P(0 failures) =1-P(S1 and S2 and S3 … and S100) =1-P(S1)xP(S2)x…xP(S100) =1-[P(S)]100=1-[0.971]100=0.947 n n Same success probability on each flight Independence These assumptions are suspect since other variables (temperature at launch, crew experience, age of craft, etc.) could affect the probability 10 9/26/12 Learning Objective 3: Probabilities and Diagnostic Testing Learning Objective 3: Example: Probabilities and Diagnostic Testing Random Drug Testing of Air Traffic Controllers n Sensitivity of test = 0.96 n Specificity of test = 0.93 n Probability of drug use at a given time ≈ 0.007 (prevalence of the drug) n Sensitivity = P(POS|S) n Specificity = P(NEG|SC) Learning Objective 3: Example: Probabilities and Diagnostic Testing Learning Objective 4: Simulation What is the probability of a positive test result? Some probabilities are very difficult to find with ordinary reasoning. In such cases, we can approximate an answer by simulation. P(POS)=P(S and POS)+P(SC and POS) n P(S and POS)=P(S)P(POS|S) = 0.007x0.96=0.0067 n P(SC and POS)=P(SC)P(POS|SC) n P(POS)=.0067+.0695=0.0762 = 0.993x0.07=0.0695 Even though the prevalence is < 1%, there is an almost 8% chance of the test suggesting drug use! Learning Objective 4: Simulation Carrying out a Simulation: n Identify the random phenomenon to be simulated n Describe how to simulate observations n Carry out the simulation many times (at least 1000 times) n Summarize results and state the conclusion 11
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