Introductory Statistics. Chapter 5

5.1 Models of Random Behavior Math 140 Introductory Statistics „ Outcome: Result or answer obtained from a chance process. „ Event: Collection of outcomes. „ Probability: Number between 0 and 1 (0% and 100%). It tells how likely it is for an outcome or event to happen. P = 0 The event cannot happen. „ P = 1 The event is certain to happen. „ 5.1 Models of Random Behavior Where do Probabilities come from? „ P(A) = the probability that event A happens „ Observed data (long-run relative frequencies). „ P(not A) = 1 − P(A) = the probability that event A doesn’t happen. „ The event not A is called the complement of event A. For example, observation of thousands of births has shown that about 51% of newborns are boys. You can use these data to say that the probability of the next newborn being a boy is about 0.51. „ Symmetry (equally likely outcomes). If you flip a fair coin, there is nothing about the two sides of the coin to suggest that one side is more likely than the other to land facing up. Relying on symmetry, it is reasonable to think that heads and tails are equally likely. So the probability of heads is 0.5. „ Subjective estimates. What’s the probability that you’ll get an A in this statistics class? That’s a reasonable, everyday kind of question, and the use of probability is meaningful, but you can’t gather data or list equally likely outcomes. However, you can make a subjective judgment. 1 Equally likely outcomes. Examples „ If we have a list of all possible outcomes and „ Flipping a coin. „ P (heads) = 1/2 „ P (tails) = ½ all of them are equally likely then P (specific outcome) = P (event) = 1 total number of equally likely outcomes number of outcomes in event total number of equally likely outcomes „ Rolling a fair die. „ P (2) = 1/6 „ P (5) = 1/6 „ P (odd) = P (1 or 2 or 3) = 3/6 „ P (even less than 5) = P (2 or 4) = 2/6 =1/3. Tap vs. Bottled Water: The problem Tap vs. Bottled Water: Who is right? „ Jack and Jill, just won a contract to determine Jack: There are three possible outcomes: Neither person chooses T, one chooses T, or both choose T. These three outcomes are equally likely, so each outcome has probability ⅓. In particular, the probability that both choose T is ⅓ . if people can tell tap water (T) from bottled water (B). „ They will give each person in their sample both kinds of water, in random order, and ask which is the tap water. „ Assuming that the tasters can’t identify tap water, what is the probability that two tasters will guess correctly and choose T? Jill: Jack, did you break your crown already? I say there are four equally likely outcomes: The first taster chooses T and the second also chooses T (TT); the first chooses T and the second chooses B (TB); the first chooses B and the second chooses T (BT); or both choose B (BB). Because these four outcomes are equally likely, each has probability ¼ . In particular, the probability that both choose T is ¼ , not ⅓. 2 Tap vs. Bottled Water: Simulation. Law of Large Numbers „ Jack and Jill use two flips of a coin to simulate the taste-test „ In a random sampling, the larger the sample, the experiment with two tasters who can’t identify tap water. „ Two tails represented neither person choosing the tap water, one heads and one tails represented one person choosing the tap water and the other choosing the bottled water, and two heads represented both people choosing the tap water. closer the proportion of successes in the sample tends to be the proportion in the population. „ Example, simulation of flipping a coin. Number of Flips 10 100 1000 10000 100000 Heads 2 45 525 4990 50246 Tails 8 55 475 5010 49754 „ So it seems that Jill is right. Sample Space Examples „ A Sample Space for a chance process is a complete „ Rolling a fair die. list of disjoint outcomes. „ Complete means that no possible outcomes are left off the list. „ Disjoint (or mutually exclusive) means that no two outcomes can occur at once. „ Often by symmetry we can assume that the outcomes on a sample space are equally likely. But to verify this we need to collect data and see if indeed each of the outcomes occurs the same number of times (approximately). Sample Space: {1,2,3,4,5,6} „ P (4)= 1/6 „ P (number is even)= 3/6 = 1/2 „ Selecting a card from a poker deck. „ „ Sample Space: {A♥,2♥,3♥,…,Q♥,K♥, A♦,2♦,3♦,…,Q♦,K♦, A♣,2♣,3♣,…,Q♣,K♣, A♠,2♠,3♠,…,Q♠,K♠} 3 Examples A random process is repeated several times „ Selecting a card from a poker deck. „ To list the total list of outcomes when a random process is made Sample Space: {A♥,2♥,3♥,…,Q♥,K♥, A♦,2♦,3♦,…,Q♦,K♦, A♣,2♣,3♣,…,Q♣,K♣, A♠,2♠,3♠,…,Q♠,K♠} „ P (3♥) = 1/52 „ up of many repetitions of another random process we can make a tree diagram. „ Example. Jack and Jill give samples of tap water (T) or bottled water (B) at random to three persons so that they taste it and see if they recognize tap water or not. Person 1 Person 2 B P (Ace) = 4/52 = 1/13 „ P (♦) = 13/52 = 1/4 „ B T B T T Person 3 B T B T B T B T Outcome BBB BBT BTB BTT TBB TBT TTB TTT Fundamental Counting Principle Discussion D8 (p. 296) „ For a two-stage process, with „ Suppose you flip a fair coin five times. n1 possible outcomes for stage 1 and n2 possible outcomes for stage 2, the number of possible outcomes for the two stages together is n1n2 „ More generally, if there are k stages, with ni possible outcomes for stage i, then the number of possible outcomes for all k stages taken together is n1n2n3 . . . nk . a. How many possible outcomes are there? „ b. What is the probability you get five heads? „ c. What is the probability you get four heads and one tail? „ 4 5.3 Addition Rule and Disjoint Events „ “OR” in mathematics means one, the other, or both. „ Two events A and B are called disjoint (mutually exclusive) if they have no outcomes in common. „ If A and B are disjoint then P (A or B) = P (A) + P (B) „ Similarly if A, B, and C are mutually Discussion A or B Type of Fishing All freshwater fishing Saltwater fishing Total „ Are the categories in the Number table of Display 6.8 (Thousands) complete? Are they disjoint? 31,041 „ What is the probability that a randomly selected person who fishes does their fishing in freshwater or in 8,885 saltwater? How many people fish in both 35,578 freshwater and saltwater? exclusive then P (A or B) = P (A) + P (B) + P (C) Discussion A or B Type of Fishing All freshwater fishing Saltwater fishing Total „ Are the categories in the Number table of Display 6.8 (Thousands) complete? Are they disjoint? 31,041 „ Complete: YES, any person that fishes does so in either fresh water or salt water (maybe both) 8,885 „ Disjoint: NO, the events “Saltwater” and “Freshwater” 35,578 have outcomes in common. Discussion A or B Type of Fishing Number (Thousands) All freshwater 31,041 fishing Saltwater fishing 8,885 Total 35,578 „ What is the probability that a randomly selected person who fishes does their fishing in freshwater or in saltwater? How many people fish in both freshwater and saltwater? „ P (“Fresh” or “Salt”) = 1 However, 31041 P (“Fresh”) = 35578 P 8885 (“Salt”) = 35578 and then, 39926 P (“Fresh”) + P (“Salt”) = 35578 >1 5 Discussion A or B Type of Fishing Number (Thousands) All freshwater fishing 31,041 Saltwater fishing 8,885 Total 35,578 Fresh Water Salt Water A Property of Disjoint Events „ What is the probability that a randomly selected person who fishes does their fishing in freshwater or in saltwater? How many people fish in both freshwater and saltwater? Similarly # (“Only Fresh”) = 35578 – 8885 # (“Only Fresh”) = 26693. # (“Only Salt” or “Only Fresh”) = 31230 # (“Only Salt”) + #(“Fresh”) = 35578 # (“Only Salt”) + 31041 = 35578 # (“Only Salt”) = 35578 – 31041 = 4537 Then, #(“Fresh” and “Salt”) = 35578 – 31230 #(“Fresh” and “Salt”) = 4348 „ If A and B are disjoint then P (A and B) = 0 „ Example. Suppose two dice are rolled. the event of getting a sum of 12, B is the event of getting two odd numbers. What is P (A and B) ? „ The only way to get a sum of 12 is (6,6) and both numbers are even. So A and B are disjoint and then P (A and B) = 0. Discussion D13 (p. 318) General Addition Rule „ Suppose you select a person at random from „ For any two events your school. Which of these pairs of events must be disjoint? a. the person has ridden a roller coaster; the person has ridden a Ferris wheel „ b. owns a classical music CD; owns a jazz CD „ c. is a senior; is a junior „ d. has brown hair; has brown eyes „ e. is left-handed; is right-handed „ f. has shoulder-length hair; is a male A is A and B, P (A or B) = P (A) + P (B) – P (A and B) „ „ In particular if A and B are disjoint then P (A and B) = 0 and then, P (A or B) = P (A) + P (B) 6 Example: The Addition Rule for Events that are not Disjoint (p. 320) Example: The Addition Rule for Events that are not Disjoint (p. 320) „ Use the Addition Rule to find the probability that if „ Use the Addition Rule to find the probability First Solution. that if you roll two dice, you get doubles or a sum of 8. „ Second Solution. you roll two dice, you get doubles or a sum of 8. „ A = getting doubles „ B = getting a sum of 8 Doubles? „ P (A) = P ({1,1}, {2,2}, {3,3}, {4,4}, {5,5}, {6,6}) = 6/36 = 1/6 „ P (B) = P ( {2,6}, {3,5}, {4,4}, {5,3}, {6,2}) = 5/36 „ P (A and B) = P ( {4,4}) = 1/36 „ P (A or B) = P (A) + P (B) – P (A and B) = 6/36 + 5/36 - 1/36 = 10/36 Sum of 8? Yes No Total Yes 1 4 5 No 5 26 31 Total 6 30 36 P = 1 + 5 + 4 10 = 36 36 Example: Computing P (A and B) (p. 320) 5.4 Conditional Probability „ In a local school, 80% of the students carry a „ The Titanic sank in 1912 without enough lifeboats for backpack, B, or a wallet, W. Also, 40% carry a backpack and 50% carry a wallet. If a student is selected at random, find the probability that the student carries both a backpack and a wallet. B No B Total 10% 40% 50% No W 30% 20% 50% Total 40% 60% 100% W the passengers and crew. Almost 1500 people died, most of them men. Was that because a man was less likely than a woman to survive? Or did more men die simply because men outnumbered women by more than 3 to 1 on the Titanic? Gender Survived? Male Female Total Yes 367 344 711 No 1364 126 1490 Total 1731 470 2201 7 Definition of Conditional Probability Example: The Titanic and Conditional Probability (p. 326) „ Suppose you pick a person at „ Conditional Probability refers to the probability of a particular event where additional information is known. „ We write it the following way P (S |T ) = “probability of S given that T is known to have happened” „ For short we refer to P (S |T ) as the probability of S given T. random from the list of people aboard the Titanic. Let S be the event that this person survived, and let F be the event that the person was female. Find P (S |F ) Gender Survived? Discussion P23. (p. 334) „ P (F) = Probability of a female passenger event a person survived and F 470 be the event a person was P (F ) = ≈ 0.213 female. Find and interpret these 2201 probabilities. „ a. P (F) and P (F |S ) „ P (F |S ) = Probability that a survivor „ b. P (not F), P (not F |S), and is female. P(S |not F) P (F | S ) = Yes Survived? Male Female Total 367 344 711 1364 126 1490 Total 1731 470 2201 No 344 ≈ 0.483 711 „ P (not F) = Probability of a male passenger P (not F ) = Female Total Yes 367 344 711 No 1364 126 1490 Total 1731 470 2201 sample space to only the 470 females (the outcomes for which we know the condition F is true). „ Then among these the favorable outcomes are 344. Thus P (S | F ) = 344 ≈ 0.732 472 „ So 73.2 % of the females survived even though only 32.3 % of the people survived. Discussion P23. (p. 334) „ For the Titanic data, let S be the Gender Male „ To find P (S |F ) restrict your 1731 ≈ 0.786 2201 „ For the Titanic data, let S be the „ P (not F |S ) = Probability that a survivor is male. event a person survived and F be the event a person was 367 female. Find and interpret these P (not F | S ) = ≈ 0.516 711 probabilities. „ a. P (F) and P (F |S ) „ b. P (not F), P (not F |S), and „ P (S | not F ) = Probability of P(S |not F) surviving given that the person selected is male. Gender Survived? Male Female Total Yes 367 344 711 No 1364 126 1490 Total 1731 470 2201 P (S | not F ) = 367 ≈ 0.212 1731 8 The Multiplication Rule for P (A and B) (p. 327) The Multiplication Rule „ The probability that event A and event B both happen is given by P (A and B) = P (A).P (B | A) or alternatively P (A and B) = P (B).P (A | B) We can interpret the top branch as: P (F ) ⋅ P (S | F ) = 470 344 344 ⋅ = = P (F and S ) 2201 470 2201 Definition of Conditional Probability „ For any two events Important Uses of Conditional Probability A and B, P (B)>0, P (A and B ) P (B ) „ Example: Suppose you roll two dice. Use the definition of conditional probability to find the probability that you get a sum of 8 given that you rolled doubles. P (A | B ) = P (sum 8 and doubles ) P (sum 8 | doubles ) = P (doubles) 1 / 36 1 = = 6 / 36 6 „ To compare sampling with or without replacement. „ To study effectiveness of medical tests „ To study effectiveness of statistical testing 9 Conditional Probability and Medical Tests Conditional Probability and Medical Tests Test Result „ In medicine, screening tests give a quick indication of whether or not a person is likely to have a particular disease. Because screening tests are intended to be relatively quick and noninvasive, they are often not as accurate as other tests that take longer or are more invasive. „ A two-way table is often used to show the four possible outcomes of a screening test. Test Result Disease Positive Negative Total Present a b a+b Absent c d c+d Total a+c b+d a+b+c+d Test Result Present Negative Total Present a b a+b Absent c d c+d Total a+c b+d a+b+c+d P (no disease | test positive) = „ False Negative Rate = c a+c P (disease present | test negative) = „ Sensitivity = P (test positive | disease present) = „ Specificity = P (test negative | no disease) = a a+b b b +d d c +d Example P34 (p. 335) „ A laboratory technician is being tested on her ability to detect Positive Negative Total 9 1 10 Absent 50 9,940 9,990 Total 59 9,941 10,000 50 ≈ 0.8474 59 1 ≈ 0.0001 „ False Neg. Rate = P (disease present | test negative) = 9941 9 = 0.90 „ Sensitivity = P (test positive | disease present) = 10 9940 „ Specificity = P (test negative | no disease) = ≈ .9949 9990 „ False Pos. Rate = Disease „ False Positive Rate = Example of a Rare Disease on 10,000 patients Disease Positive contaminated blood samples. Among 100 samples given to her, 20 are contaminated, each with about the same degree of contamination. Suppose the technician makes the correct decision 90% of the time (regardless of contamination or not). Make a table showing what you would expect to happen. What is her false positive rate? What is her false negative rate? How would these rates change if she were given 100 samples with 50 contaminated? Detection of Contamination (test) P (no disease | test positive) = Yes Contaminated? Positive Negative Total 18 2 20 No 8 72 80 Total 26 74 100 10 Conditional Probability and Statistical Inference Example P34 (p. 335) Detection of Contamination (test) Positive Contaminated? Negative „ Statistician: Suppose you draw Total Yes 18 2 20 No 8 72 80 Total 26 74 100 „ „ „ False Pos. Rate = 8/26 = 0.3076 „ „ False Neg. Rate = 2/74 = 0.027 „ „ „ Sensitivity = 18/20 = 0.90 „ 3 workers at random from the set of 10 hourly workers. This establishes random sampling as the model for the study. Lawyer: Okay. Statistician: It turns out that there are 120, possible samples of size 3, and only 6 of them give an average age of 58 or more. Lawyer: So the probability is 6/120 , or .05. Statistician: Right. Lawyer: There’s only a 5% chance the company didn’t discriminate and a 95% chance that they did. Statistician: No, that’s not true. „ Specificity = 72/80 = 0.9 5.5 Independent Events A and B are independent if and only if P (A | B)= P (A). Equivalently, A and B are independent if and only if P (B | A)= P (B). P (av. age ≥ 58 | random draws ) = .05 Example: Water, Gender, and Independence (p.340) Identified Tap Water ? Drinks Bottled Water ? Identified Tap Water ? Yes No Total Yes No Total Yes No 24 36 Total 60 B happened does not affect the probability of A happening, and conversely knowing that A happened does not affect the probability of B. age-neutral model of random draws is correct, then there’s only a 5% chance of getting an average age of 58 or more. „ Lawyer: So the chance the company is guilty must be 95%. „ Statistician: Slow down. If you start by assuming the model is true, you can compute the chances of various results. But you’re trying to start from the results and compute the chance that the model is right or wrong. You can’t do that. P (no discr. | av. age = 58) ?? „ Events „ In other words, knowing that „ Lawyer: But you said . . . „ Statistician: I said that if the 6 30 34 70 40 100 21 14 35 Female 39 26 65 Male Gender Total 60 40 100 „ Show that the events is a male and correctly identifies tap water are independent and that the events drinks bottled water and correctly identifies tap water are not independent. 11