Agresti/Franklin Statistics, 1 of 87 Chapter 5 Probability in Our Daily Lives Learn …. About probability – the way we quantify uncertainty How to measure the chances of the possible outcomes of random phenomena How to find and interpret probabilities

Agresti/Franklin Statistics, 2 of 87  Section 5.1 How Can Probability Quantify Randomness?

Agresti/Franklin Statistics, 3 of 87 Randomness Applies to the outcomes of a response variable Possible outcomes are known, but it is uncertain which will occur for any given observation

Agresti/Franklin Statistics, 4 of 87 Some Popular Randomizers Rolling dice Spinning a wheel Flipping a coin Drawing cards

Agresti/Franklin Statistics, 5 of 87 Random Phenomena Individual outcomes are unpredictable With a large number of observations, predictable patterns occur

Agresti/Franklin Statistics, 6 of 87 Random Phenomena With random phenomena, the proportion of times that something happens is highly random and variable in the short run but very predictable in the long run.

Agresti/Franklin Statistics, 7 of 87 Jacob Bernoulli: Law of Large Numbers As the number of trials of a random phenomenon increases, the proportion of occurrences of any given outcome approaches a particular number “in the long run”.

Agresti/Franklin Statistics, 8 of 87 Probability With a random phenomenon, the probability of a particular outcome is the proportion of times that the outcome would occur in a long run of observations.

Agresti/Franklin Statistics, 9 of 87 Roll a Die What is the probability of rolling a ‘6’? a..22 b..10 c..17

Agresti/Franklin Statistics, 10 of 87 Question about Random Phenomena If a family has four girls in a row and is expecting another child, does the next child have more than a ½ chance of being a boy?

Agresti/Franklin Statistics, 11 of 87 Independent Trials Different trials of a random phenomenon are independent if the outcome of any one trial is not affected by the outcome of any other trial.

Agresti/Franklin Statistics, 12 of 87  Section 5.2 How Can We Find Probabilities?

Agresti/Franklin Statistics, 13 of 87 Sample Space For a random phenomenon, the sample space is the set of all possible outcomes

Agresti/Franklin Statistics, 14 of 87 Example: Roll a Die Once The Sample Space consists of six possible outcomes: {1, 2, 3, 4, 5, 6}

Agresti/Franklin Statistics, 15 of 87 Example: Flip a Coin Twice The Sample Space consists of the four possible outcomes: {(H,H) (H,T) (T,H) (T,T)}

Agresti/Franklin Statistics, 16 of 87 Example: A 3-Question Multiple Choice Quiz Diagram of the Sample Space

Agresti/Franklin Statistics, 17 of 87 Tree Diagram An ideal way of visualizing sample spaces with a small number of outcomes As the number of trials or the number of possible outcomes on each trial increase, the tree diagram becomes impractical

Agresti/Franklin Statistics, 18 of 87 Event An event is a subset of the sample space

Agresti/Franklin Statistics, 19 of 87 Probabilities for a Sample Space The probability of each individual outcome is between 0 and 1 The total of all the individual probabilities equals 1

Agresti/Franklin Statistics, 20 of 87 Example: Assigning Subjects to Echinacea or Placebo for Treating Colds Experiment Multi-center randomized experiment to compare an herbal remedy to a placebo for treating the common cold Half of the volunteers are randomly chosen to receive the herbal remedy and the other half will receive the placebo Clinic in Madison, Wisconsin has four volunteers Two men: Jamal and Ken Two women: Linda and Mary

Agresti/Franklin Statistics, 21 of 87 Example: Assigning Subjects to Echinacea or Placebo for Treating Colds Sample Space to receive the herbal remedy: {(Jamal, Ken), (Jamal, Linda), (Jamal, Mary), (Ken, Linda), (Ken, Mary), (Linda, Mary)} These six possible outcomes are equally likely

Agresti/Franklin Statistics, 22 of 87 Example: Assigning Subjects to Echinacea or Placebo for Treating Colds What is the probability of the event that the sample chosen to receive the herbal remedy consists of one man and one woman?

Agresti/Franklin Statistics, 23 of 87 Probability of an Event The probability of an event A, denoted by P(A), is obtained by adding the probabilities of the individual outcomes in the event. When all the possible outcomes are equally likely:

Agresti/Franklin Statistics, 24 of 87 Example: What are the Chances of a Taxpayer being Audited? Each year, the Internal Revenue Service audits a sample of tax forms to verify their accuracy

Agresti/Franklin Statistics, 25 of 87 Example: What are the Chances of a Taxpayer being Audited?

Agresti/Franklin Statistics, 26 of 87 Example: What are the Chances of a Taxpayer being Audited? What is the sample space for selecting a taxpayer? {(under $25,000, Yes), (under $25,000, No), ($25,000 - $49,000, Yes) …}

Agresti/Franklin Statistics, 27 of 87 Example: What are the Chances of a Taxpayer being Audited? For a randomly selected taxpayer in 2002, what is the probability of an audit?

Agresti/Franklin Statistics, 28 of 87 Example: What are the Chances of a Taxpayer being Audited? For a randomly selected taxpayer in 2002, what is the probability of an income of $100,000 or more?

Agresti/Franklin Statistics, 29 of 87 Basic Rules for Finding Probabilities about a Pair of Events Complement of an Event Intersection of 2 Events Union of 2 Events

Agresti/Franklin Statistics, 30 of 87 Complement of an Event Complement of Event A: Consists of all outcomes in the sample space that are not in A Is denoted by A c The probabilities of A and A c add to 1 P(A c ) = 1 – P(A)

Agresti/Franklin Statistics, 31 of 87 Complement of an Event

Agresti/Franklin Statistics, 32 of 87 Disjoint Events Two events, A and B, are disjoint if they do not have any common outcomes

Agresti/Franklin Statistics, 33 of 87 Example: Disjoint Events Pop Quiz: 3 Multiple-Choice Questions Event A: Student answers exactly 1 question correctly Event B: Student answer exactly 2 questions correctly

Agresti/Franklin Statistics, 34 of 87 Example: Disjoint Events

Agresti/Franklin Statistics, 35 of 87 Intersection of Two Events The intersection of A and B: consists of outcomes that are in both A and B

Agresti/Franklin Statistics, 36 of 87 Union of Two Events The union of A and B: Consists of outcomes that are in A or B In probability, “A or B” denotes that A occurs or B occurs or both occur

Agresti/Franklin Statistics, 37 of 87 Intersection and Union of Two Events

Agresti/Franklin Statistics, 38 of 87 How Can We Find the Probability that A or B Occurs? Addition Rule: Probability of the Union of Two Events For the union of two events, P(A or B) = P(A) + P(B) – P(A and B) If the events are disjoint, P(A and B) = 0, so P(A or B) = P(A) + P(B)

Agresti/Franklin Statistics, 39 of 87 How Can We Find the Probability that A and B Occurs? Multiplication Rule: Probability of the Intersection of Independent Events For the intersection of two independent events, A and B: P(A and B) = P(A) x P(B)

Agresti/Franklin Statistics, 40 of 87 Example: Two Rolls of A Die P(6 on roll 1 and 6 on roll 2): 1/6 x 1/6 = 1/36

Agresti/Franklin Statistics, 41 of 87 Example: Guessing on a Pop Quiz Pop Quiz with 3 Multiple-choice questions Each question has 5 options A student is totally unprepared and randomly guesses the answer to each question

Agresti/Franklin Statistics, 42 of 87 Example: Guessing on a Pop Quiz The probability of selecting the correct answer by guessing = 0.20 Responses on each question are independent

Agresti/Franklin Statistics, 43 of 87 Tree Diagram for the Pop Quiz

Agresti/Franklin Statistics, 44 of 87 Example: Guessing on a Pop Quiz What is the probability that a student answers at least 2 questions correctly? P(CCC) + P(CCI) + P(CIC) + P(ICC) = (0.032) = 0.104

Agresti/Franklin Statistics, 45 of 87 Events Often Are Not Independent Example: A Pop Quiz with 2 Multiple Choice Questions Data giving the proportions for the actual responses of students in a class Outcome: II IC CI CC Probability:

Agresti/Franklin Statistics, 46 of 87 Events Often Are Not Independent Define the events A and B as follows: A: {first question is answered correctly} B: {second question is answered correctly}

Agresti/Franklin Statistics, 47 of 87 Events Often Are Not Independent P(A) = P{(CI), (CC)} = = 0.63 P(B) = P{(IC), (CC)} = = 0.69 P(A and B) = P{(CC)} = 0.58 If A and B were independent, P(A and B) = P(A) x P(B) = 0.63 x 0.69 = 0.43

Agresti/Franklin Statistics, 48 of 87 Question of Independence Don’t assume that events are independent unless you have given this assumption careful thought and it seems plausible

Agresti/Franklin Statistics, 49 of 87 Example: A family has two children If each child is equally likely to be a girl or boy, find the probability that the family has two girls. a. 1/2 b. 1/3 c. 1/4 d. 1/8

Agresti/Franklin Statistics, 50 of 87  Section 5.3 Conditional Probability: What’s the Probability of A, Given B?

Agresti/Franklin Statistics, 51 of 87 Conditional Probability For events A and B, the conditional probability of event A, given that event B has occurred is:

Agresti/Franklin Statistics, 52 of 87 Conditional Probability

Agresti/Franklin Statistics, 53 of 87 Example: What are the Chances of a Taxpayer being Audited?

Agresti/Franklin Statistics, 54 of 87 Example: Probabilities of a Taxpayer Being Audited

Agresti/Franklin Statistics, 55 of 87 Example: Probabilities of a Taxpayer Being Audited What was the probability of being audited, given that the income was ≥ $100,000? Event A: Taxpayer is audited Event B: Taxpayer’s income ≥ $100,000

Agresti/Franklin Statistics, 56 of 87 Example: Probabilities of a Taxpayer Being Audited

Agresti/Franklin Statistics, 57 of 87 Example: The Triple Blood Test for Down Syndrome A positive test result states that the condition is present A negative test result states that the condition is not present

Agresti/Franklin Statistics, 58 of 87 Example: The Triple Blood Test for Down Syndrome False Positive: Test states the condition is present, but it is actually absent False Negative: Test states the condition is absent, but it is actually present

Agresti/Franklin Statistics, 59 of 87 Example: The Triple Blood Test for Down Syndrome A study of 5282 women aged 35 or over analyzed the Triple Blood Test to test its accuracy

Agresti/Franklin Statistics, 60 of 87 Example: The Triple Blood Test for Down Syndrome

Agresti/Franklin Statistics, 61 of 87 Example: The Triple Blood Test for Down Syndrome 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

Agresti/Franklin Statistics, 62 of 87 Example: The Triple Blood Test for Down Syndrome P(POS) = 1355/5282 = 0.257

Agresti/Franklin Statistics, 63 of 87 Given that the diagnostic test result is positive, find the estimated probability that Down syndrome truly is present Example: The Triple Blood Test for Down Syndrome

Agresti/Franklin Statistics, 64 of 87 Example: The Triple Blood Test for Down Syndrome

Agresti/Franklin Statistics, 65 of 87 Summary: Of the women who tested positive, fewer than 4% actually had fetuses with Down syndrome Example: The Triple Blood Test for Down Syndrome

Agresti/Franklin Statistics, 66 of 87 Multiplication Rule for Finding P(A and B) For events A and B, the probability that A and B both occur equals: P(A and B) = P(A|B) x P(B) also P(A and B) = P(B|A) x P(A)

Agresti/Franklin Statistics, 67 of 87 Example: How Likely is a Double Fault in Tennis? Roger Federer – 2004 men’s champion in the Wimbledon tennis tournament He made 64% of his first serves He faulted on the first serve 36% of the time Given that he made a fault with his first serve, he made a fault on his second serve only 6% of the time

Agresti/Franklin Statistics, 68 of 87 Example: How Likely is a Double Fault in Tennis? Assuming these are typical of his serving performance, when he serves, what is the probability that he makes a double fault?

Agresti/Franklin Statistics, 69 of 87 Example: How Likely is a Double Fault in Tennis? P(F1) = 0.36 P(F2|F1) = 0.06 P(F1 and F2) = P(F2|F1) x P(F1) = 0.06 x 0.36 = 0.02

Agresti/Franklin Statistics, 70 of 87 Sampling Without Replacement Once subjects are selected from a population, they are not eligible to be selected again

Agresti/Franklin Statistics, 71 of 87 Example: How Likely Are You to Win the Lotto? In Georgia’s Lotto, 6 numbers are randomly sampled without replacement from the integers 1 to 49 You buy a Lotto ticket. What is the probability that it is the winning ticket?

Agresti/Franklin Statistics, 72 of 87 Example: How Likely Are You to Win the Lotto? P(have all 6 numbers) = P(have 1 st and 2 nd and 3 rd and 4 th and 5 th and 6 th ) = P(have 1 st )xP(have 2 nd |have 1 st )xP(have 3 rd | have 1 st and 2 nd ) …P(have 6 th |have 1st, 2 nd, 3 rd, 4 th, 5 th )

Agresti/Franklin Statistics, 73 of 87 Example: How Likely Are You to Win the Lotto? 6/49 x 5/48 x 4/47 x 3/46 x 2/45 x 1/44 =

Agresti/Franklin Statistics, 74 of 87 Independent Events Defined Using Conditional Probabilities Two events A and B are independent if the probability that one occurs is not affected by whether or not the other event occurs

Agresti/Franklin Statistics, 75 of 87 Independent Events Defined Using Conditional Probabilities Events A and B are independent if: P(A|B) = P(A) If this holds, then also P(B|A) = P(B) Also, P(A and B) = P(A) x P(B)

Agresti/Franklin Statistics, 76 of 87 Checking for Independence Here are three ways to check whether events A and B are independent: Is P(A|B) = P(A)? Is P(B|A) = P(B)? Is P(A and B) = P(A) x P(B)? If any of these is true, the others are also true and the events A and B are independent

Agresti/Franklin Statistics, 77 of 87 Example: How to Check Whether Two Events are Independent The diagnostic blood test for Down syndrome: POS = positive result NEG = negative result D = Down Syndrome D C = Unaffected

Agresti/Franklin Statistics, 78 of 87 Example: How to Check Whether Two Events are Independent Blood Test: StatusPOSNEGTotal D D c Total

Agresti/Franklin Statistics, 79 of 87 Are the events POS and D independent or dependent? Is P(POS|D) = P(POS)? Example: How to Check Whether Two Events are Independent

Agresti/Franklin Statistics, 80 of 87 Example: How to Check Whether Two Events are Independent Is P(POS|D) = P(POS)? P(POS|D) =P(POS and D)/P(D) = 0.009/0.010 = 0.90 P(POS) = The events POS and D are dependent