Lecture 7

Paper 1 Select an area of human endeavor (e.g., a particular sport, marketing, politics, car manufacturing …) and write a 5 page paper on how statistics influenced it. Due this Thursday 9/29. Make sure you clearly reference your sources (including websites)

Probability models Probability models Rules for probabilities sample space (set of possible outomes) events probabilities Rules for probabilities Assigning probabilities when the sample space is finite Assigning probabilities when all outcomes are equally likely Independence and the multiplication rule

Probability models Definition: A probability model for a random phenomenon consists of The set of all possible outcomes of the phenomenon. This set is called the sample space and denoted S. An assignment of probabilities to events, which are subsets of S. The probability of event A is denoted P(A).

Probability models: Assignment of probabilities to events Assignment of probabilities to events can be done in any of several ways: If S is a finite set, assign probabilities to the individual outcomes; then probability of an event is the sum of the probabilities of the outcomes in it. If S is an infinite set of real numbers, such as the set of all real numbers or all positive numbers, assign probabilities to intervals in some way. In any case, the probabilities must be assigned in a way that follows four basic rules. These rules are necessary because proportions follow them. Ex. 1: 3 tosses of a coin. Ex. 2: D,R,I. Ex. 3: random member of a normal population. The rules: 

Probability models: Rules for assignment of probabilities to events Rule 1: Any probability must be a number between 0 and 1. This is because the proportion of times an event occurs is always between 0 and 1. Rule 2: The probability of the set of all possible outcomes together must total 1. This is because the proportions of times that all outcomes occur must add to 1 (= 100%). Rule 3: If two events have no outcomes in common, then the probability that one or the other occurs must be the sum of their probabilities. They can never occur together, so the proportion of times that one or the other occurs must be the sum of their individual proportions. Rule 4: The probability that an event does not occur must be 1 minus the probability that it does occur. The proportion of times it doesn’t occur, plus the proportion of times it does occur, must total 1.

Probability models: Rules for assignment of probabilities to events The four rules, restated briefly: P(A) denotes the probability of event A in sample space S. Rule 1: For any event A, 0 ≤ P(A) ≤ 1. Rule 2: P(S) = 1. Rule 3: If events A and B are disjoint (have no outcomes in common, i.e. cannot occur together), then P(A or B) = P(A) + P(B). Rule 4: If Ac denotes the complement of A (the event that A does not occur), then P(Ac) = 1 – P(A). Note: Any assignment satisfying these is a prob model. But not all prob models are meaningful models of real-life situations.

If the set of possible outcomes is finite: Probability models: Assigning probabilities when the sample space is finite If the set of possible outcomes is finite: Assign a probability between 0 and 1 to each individual outcome, making sure that the assigned probabilities add to 1. Then the probability of any event is the sum of the probabilities assigned to its outcomes. The four rules are then automatically satisfied, and we can use Rules 3 and 4 to help find probabilities. Examples 

Example Assigning probabilities when the sample space is finite For numbers in tables of financial, demographic, and other data, the first digits are not equally likely! Their proportions typically are close to Benford’s Law: digit 1 2 3 4 5 6 7 8 9 prob .301 .176 .125 .097 .079 .067 .058 .051 .046

Benford’s Law is approximately valid for populations of NC counties in 2010 County Population Mecklenburg 919625 Henderson 106740 Halifax 54691 Currituck 23547 Wake 900993 Craven 103505 Pender 52203 Martin 24505 Guilford 488406 Cleveland 98078 Watauga 51079 Caswell 23719 Forsyth 350670 Nash 95836 Hoke 46952 Northampton 22098 Cumberland 319431 Rockingham 93640 Beaufort 47770 Greene 21362 Durham 269974 Moore 88247 Stokes 47401 Madison 20764 Buncombe 238319 Burke 90914 Richmond 46639 Bertie 21293 New Hanover 202681 Caldwell 83029 Vance 45419 Warren 20975 Gaston 206086 Wilson 81234 McDowell 44996 Polk 20510 Union 201292 Lincoln 78265 Davie 41222 Yancey 17818 Onslow 177772 Surry 73673 Jackson 40271 Avery 17797 Cabarrus 178014 Wilkes 69340 Pasquotank 40661 Mitchell 15579 Johnston 168878 Carteret 66469 Person 39464 Chowan 14793 Pitt 168148 Rutherford 67809 Yadkin 38406 Swain 13981 Davidson 162878 Chatham 63493 Alexander 37193 Perquimans 13453 Iredell 159442 Sampson 63431 Scotland 36157 Pamlico 13144 Catawba 154356 Franklin 60619 Bladen 35190 Washington 13221 Alamance 151087 Stanly 60585 Dare 33920 Gates 12186 Randolph 141752 Duplin 58505 Macon 33922 Alleghany 11155 Orange 133857 Lenoir 59495 Transylvania 33090 Clay 10587 Rowan 138446 Haywood 59036 Montgomery 27798 Jones 10153 Robeson 134168 Lee 57866 Cherokee 27444 Camden 9980 Wayne 122623 Columbus 58098 Ashe 27281 Graham 8861 Harnett 114678 Granville 57532 Anson 26948 Hyde 5810 Brunswick 107431 Edgecombe 56552 Hertford 24658 Tyrrell 4407 1st digit 1 2 3 4 5 6 7 8 9 prop. .29 .19 .10 .11 .07 .02 .04 B. Law .301 .176 .125 .097 .079 .067 .058 .051 .046

R code that generated table! N <- 1000000 maxT <- 100 x <- maxT*runif(N) hist(x) firstdigitx <- floor(x/10^floor(log(x,10))) table(firstdigitx)/N y <- 1.2^x hist(y) firstdigity <- floor(y/10^floor(log(y,10))) table(firstdigity)/N

Homework (due on Thursday 10/6) Select your favorite data set. (Must contain 100+ items) Does it follow Benford’s law? Explain (Make a table computed from your data and compare to the Benford’s law table.) Use real data set. Do NOT generate the data on the computer.

Finite sample spaces with equally likely outcomes If we can assume that all possible outcomes are equally likely, then the assignment of probabilities to individual outcomes is easy: If the sample space S has k outcomes in all, then each must have probability 1/k. So in the case of equally likely outcomes, if A is any event, then P(A) = (number of outcomes in A) / k .

Probability models: Example Finite sample spaces with equally likely outcomes SRS of size 2 from a set of 5 people Label the people 1,2,3,4,5. S has 10 equally likely members: 12, 13, 14, 15, 23, 24, 25, 34, 35, 45 (Why can we assume they are equally likely?) So each is assigned a probability of _____. Event A = “2 is chosen but 3 is not.” P(A) = _____ Event B = “either 2 or 3 or both is chosen.” P(B) = _____ 3/10 7/10

Independence and the multiplication rule Events A and B are independent if knowing that one occurs does not change the probability that the other occurs. Rule 5: If events A and B are independent, then P(A and B) = P(A)P(B).

Independence and the multiplication rule: Example 1 Toss a coin twice and let A = “heads on the first toss,” B = “heads on the second toss.” These are independent, because we assume the outcome of one toss can’t affect the outcome of another toss. (If this assumption needs verification, we could test it with many repetitions of the two-toss experiment.) So, since we know P(A) = 1/2 and P(B) = 1/2, Rule 5 says that P(A and B) = (1/2)(1/2) = 1/4. Exercise: In the sample space for this experiment, identify the three events (1) A; (2) B; (3) A and B; and check that their probabilities are as given above.

Independence and the multiplication rule: Example 2 Take a SRS of size 2 from a group of 5 people. Number the people 1,2,3,4,5. Then the sample space S contains 10 outcomes: 12, 13, , , , , , , , . Let A = “Person #1 is chosen,” B = “Person #2 is chosen.” These are not independent. Knowing that #1 is chosen decreases the chance that #2 is chosen. Exercise: In the sample space, identify the three events (1) A, (2) B, (3) A and B. Find their probabilities and verify that P(A and B) does not equal P(A)P(B).

Independence and the multiplication rule: Example 3 Draw 2 cards at random from a standard deck, but replace the first card and reshuffle before drawing the second. What’s the probability that they’re both hearts? Standard deck has 52 cards; 13 are hearts. Let A = “1st card is a heart” and B = “2nd card is a heart.” We want P(A and B). P(A) = _______ and P(B) = _______. So P(A and B) = ________. How would the answer change if we did not return the first card back?

4.2 Probability models: ndependence and the multiplication rule Independence and the multiplication rule: Example 4 4.2 Probability models: ndependence and the multiplication rule From a very large population in which 40% identify themselves as Dog Lovers, take a SRS of size 3. What’s the probability that they’re all Dog Lovers? Let A = “First one chosen is a Dog Lover,” B = “second one is a Dog Lover,” and C = “third one is a Dog Lover”. They’re not really independent. For example, if we know A occurred, then B and C are very slightly less likely than if A had not occurred. But because the population is so large, it is nearly true (and we can safely assume) that They’re independent, and The probability of each is .40. So the probability that they’re all Dog Lovers is ______ P(A and B and C) = (.4)3 = .064.