1. Lecture 7

2. 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)

3. 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

4. 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).

5. 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: 

6. 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.

7. 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.

8. 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 

9. 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

10. 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

11. R code that generated table!
N < 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

12. 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.

13. 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 .

14. 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

15. 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).

16. 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.

17. 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).

18. 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?

19. 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.