1. Formal study of uncertainty The engine that drives statistics
Probability
Formal study of uncertainty
The engine that drives statistics

2. Introduction Nothing in life is certain
We gauge the chances of successful outcomes in business, medicine, weather, and other everyday situations such as the lottery (recall the birthday problem)

3. History
For most of human history, probability, the formal study of the laws of chance, has been used for only one thing: gambling

4. History (cont.)
Nobody knows exactly when gambling began; goes back at least as far as ancient Egypt where 4-sided “astragali” (made from animal heelbones) were used

5. History (cont.)
The Roman emperor Claudius (10BC-54AD) wrote the first known treatise on gambling.
The book “How to Win at Gambling” was lost.
Rule 1: Let Caesar win IV
out of V times

6. Approaches to Probability
Relative frequency
event probability = x/n,
where x=# of occurrences of event of interest, n=total # of observations
Coin, die tossing; nuclear power plants?
Limitations
repeated observations not practical

7. Approaches to Probability (cont.)
Subjective probability
individual assigns prob. based on personal experience, anecdotal evidence, etc.
Classical approach
every possible outcome has equal probability (more later)

8. Basic Definitions
Experiment: act or process that leads to a single outcome that cannot be predicted with certainty
Examples:
1. Toss a coin
2. Draw 1 card from a standard deck of cards
3. Arrival time of flight from Atlanta to RDU

9. Basic Definitions (cont.)
Sample space: all possible outcomes of an experiment. Denoted by S
Event: any subset of the sample space S;
typically denoted A, B, C, etc.
Simple event: event with only 1 outcome
Null event: the empty set F
Certain event: S

10. Examples 1. Toss a coin once
S = {H, T}; A = {H}, B = {T} simple events
2. Toss a die once; count dots on upper face
S = {1, 2, 3, 4, 5, 6}
A=even # of dots on upper face={2, 4, 6}
B=3 or fewer dots on upper face={1, 2, 3}

11. Laws of Probability

12. Laws of Probability (cont.)
3. P(A’ ) = 1 - P(A)
For an event A, A’ is the complement of A; A’ is everything in S that is not in A. S A' A

13. Birthday Problem
What is the smallest number of people you need in a group so that the probability of 2 or more people having the same birthday is greater than 1/2?
Answer: 23
No. of people
Probability

14. Example: Birthday Problem
A={at least 2 people in the group have a common birthday}
A’ = {no one has common birthday}

15. Unions and Intersections
AÇB A B
AÈB

16. Mutually Exclusive Events
Mutually exclusive events-no outcomes from S in common
A Ç B = Æ S A B

17. Laws of Probability (cont.)
Addition Rule for Disjoint Events:
4. If A and B are disjoint events, then
P(A  B) = P(A) + P(B)

18. 5. For two independent events A and B
P(A  B) = P(A) × P(B)

19. Laws of Probability (cont.)
General Addition Rule
6. For any two events A and B
P(A  B) = P(A) + P(B) – P(A  B)

20. P(AÈB)=P(A) + P(B) - P(A Ç B)S
AÇB A B

21. Example: toss a fair die once
A = even # appears = {2, 4, 6}
B = 3 or fewer = {1, 2, 3}
P(A È B) = P(A) + P(B) - P(A Ç B)
=P({2, 4, 6}) + P({1, 2, 3}) - P({2})
= 3/ / /6 = 5/6

22. Laws of Probability: Summary
1. 0  P(A)  1 for any event A
2. P() = 0, P(S) = 1
3. P(A’) = 1 – P(A)
4. If A and B are disjoint events, then
P(A  B) = P(A) + P(B)
5. If A and B are independent events, then
P(A  B) = P(A) × P(B)
6. For any two events A and B,
P(A  B) = P(A) + P(B) – P(A  B)

23. The Equally Likely Approach (also called the Classical Approach)
Probability Models
The Equally Likely Approach
(also called the Classical Approach)

24. Assigning Probabilities
If an experiment has N outcomes, then each outcome has probability 1/N of occurring
If an event A1 has n1 outcomes, then
P(A1) = n1/N

25. We Need Efficient Methods for Counting Outcomes

26. Product Rule for Ordered Pairs
A student wishes to commute to a junior college for 2 years and then commute to a state college for 2 years. Within commuting distance there are 4 junior colleges and 3 state colleges. How many junior college-state college pairs are available to her?

27. Product Rule for Ordered Pairs
junior colleges: 1, 2, 3, 4
state colleges a, b, c
possible pairs:
(1, a) (1, b) (1, c)
(2, a) (2, b) (2, c)
(3, a) (3, b) (3, c)
(4, a) (4, b) (4, c)

28. Product Rule for Ordered Pairs
junior colleges: 1, 2, 3, 4
state colleges a, b, c
possible pairs:
(1, a) (1, b) (1, c)
(2, a) (2, b) (2, c)
(3, a) (3, b) (3, c)
(4, a) (4, b) (4, c)
4 junior colleges
3 state colleges
total number of possible
pairs = 4 x 3 = 12

29. Product Rule for Ordered Pairs
junior colleges: 1, 2, 3, 4
state colleges a, b, c
possible pairs:
(1, a) (1, b) (1, c)
(2, a) (2, b) (2, c)
(3, a) (3, b) (3, c)
(4, a) (4, b) (4, c)
In general, if there are n1 ways
to choose the first element of
the pair, and n2 ways to choose
the second element, then the
number of possible pairs is
n1n2. Here n1 = 4, n2 = 3.

30. Counting in “Either-Or” Situations
NCAA Basketball Tournament: how many ways can the “bracket” be filled out?
How many games?
2 choices for each game
Number of ways to fill out the bracket:
263 = 9.2 × 1018
Earth pop. about 6 billion; everyone fills out 1 million different brackets
Chances of getting all games correct is about 1 in 1,000

31. Counting Example
Pollsters minimize lead-in effect by rearranging the order of the questions on a survey
If Gallup has a 5-question survey, how many different versions of the survey are required if all possible arrangements of the questions are included?

32. Solution
There are 5 possible choices for the first question, 4 remaining questions for the second question, 3 choices for the third question, 2 choices for the fourth question, and 1 choice for the fifth question.
The number of possible arrangements is therefore
5  4  3  2  1 = 120

33. Efficient Methods for Counting Outcomes
Factorial Notation:
n!=12 … n
Examples
1!=1; 2!=12=2; 3!= 123=6; 4!=24;
5!=120;
Special definition: 0!=1

34. Factorials with calculators and Excel
non-graphing: x ! (second function)
graphing: bottom p. 9 T I Calculator Commands
(math button)
Excel:
Paste: math, fact

35. Factorial Examples 20! = 2.43 x 1018 1,000,000 seconds?
About 11.5 days
1,000,000,000 seconds?
About 31 years
31 years = 109 seconds
1018 = 109 x 109
31 x 109 years = 109 x 109 = 1018 seconds
20! is roughly the age of the universe in seconds

36. Permutations
A B C D E
How many ways can we choose 2 letters from the above 5, without replacement, when the order in which we choose the letters is important?
5  4 = 20

37. Permutations (cont.)

38. Permutations with calculator and Excel
non-graphing: nPr
Graphing
p. 9 of T I Calculator Commands
(math button)
Excel
Paste: Statistical, Permut

39. Combinations
A B C D E
How many ways can we choose 2 letters from the above 5, without replacement, when the order in which we choose the letters is not important?
5  4 = 20 when order important
Divide by 2: (5  4)/2 = 10 ways

40. Combinations (cont.)

41. ST 101 Powerball Lottery From the numbers 1 through 20,
choose 6 different numbers.
Write them on a piece of paper.

42. Chances of Winning?

43. North Carolina Powerball Lottery
Prior to Jan. 1, 2009
After Jan. 1, 2009

44. Visualize Your Lottery Chances
How large is 195,249,054?
$1 bill and $100 bill both 6” in length
10,560 bills = 1 mile
Let’s start with 195,249,053 $1 bills and one $100 bill …
… and take a long walk, putting down bills end-to-end as we go

45. Raleigh to Ft. Lauderdale…
… still plenty of bills remaining, so continue from …

46. … Ft. Lauderdale to San Diego
… still plenty of bills remaining, so continue from…

47. … San Diego to Seattle
… still plenty of bills remaining, so continue from …

48. … Seattle to New York
… still plenty of bills remaining, so continue from …

49. … New York back to Raleigh
… still plenty of bills remaining, so …

50. Go around again! Lay a second path of bills
Still have ~ 5,000 bills left!!

51. Chances of Winning NC Powerball Lottery?
Remember: one of the bills you put down is a $100 bill; all others are $1 bills
Your chance of winning the lottery is the same as bending over and picking up the $100 bill while walking the route blindfolded.

52. Example: Illinois State Lottery

53. Virginia State Lottery

54. A Graphical Method for Complicated Probability Problems
Probability Trees
A Graphical Method for Complicated Probability Problems

55. Example: AIDS Testing V={person has HIV}; CDC: P(V)=.006
+: test outcome is positive (test indicates HIV present)
-: test outcome is negative
clinical reliabilities for a new HIV test:
If a person has the virus, the test result will be positive with probability .999
If a person does not have the virus, the test result will be negative with probability .990

56. Question 1
What is the probability that a randomly selected person will test positive?

57. Probability Tree Approach
A probability tree is a useful way to visualize this problem and to find the desired probability.

58. Probability Tree
clinical reliability
clinical reliability

59. Probability Tree Multiply clinical reliability branch probs

60. Question 1 Answer
What is the probability that a randomly selected person will test positive?
P(+) = =

61. Question 2
If your test comes back positive, what is the probability that you have HIV?
(Remember: we know that if a person has the virus, the test result will be positive with probability .999; if a person does not have the virus, the test result will be negative with probability .990).
Looks very reliable

62. Question 2 Answer Answer
two sequences of branches lead to positive test; only 1 sequence represented people who have HIV.
P(person has HIV given that test is positive) =.00599/( ) = .376

63. Summary Question 1: P(+) = .00599 + .00994 = .01593
Question 2: two sequences of branches lead to positive test; only 1 sequence represented people who have HIV.
P(person has HIV given that test is positive) =.00599/( ) = .376

64. Recap We have a test with very high clinical reliabilities:
If a person has the virus, the test result will be positive with probability .999
If a person does not have the virus, the test result will be negative with probability .990
But we have extremely poor performance when the test is positive:
P(person has HIV given that test is positive) =.376
In other words, 62.4% of the positives are false positives! Why?
When the characteristic the test is looking for is rare, most positives will be false.

65. examples
1. P(A)=.3, P(B)=.4; if A and B are mutually exclusive events, then P(AB)=?
A  B = , P(A  B) = 0
2. 15 entries in pie baking contest at state fair. Judge must determine 1st, 2nd, 3rd place winners. How many ways can judge make the awards?
15P3 = 2730