1. Probability and counting
Chapters 13 and 14
Probability and counting

2. 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
We will solve this problem a few slides later using the laws of probability

3. Probability Formal study of uncertainty
The engine that drives Statistics
Primary objectives:
use the rules of probability to calculate appropriate measures of uncertainty.
Learn the probability basics so that we can do Statistical Inference

4. 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 or the birthday problem

5. Randomness and probability
Randomness ≠ chaos
A phenomenon is random if individual outcomes are uncertain, but there is nonetheless a regular distribution of outcomes in a large number of repetitions.

6. Coin toss
The result of any single coin toss is random. But the result over many tosses is predictable, as long as the trials are independent (i.e., the outcome of a new coin flip is not influenced by the result of the previous flip).
The probability of heads is 0.5 = the proportion of times you get heads in many repeated trials.
First series of tosses
Second series

7. 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

8. 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)

9. 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

10. 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.
Null event: the empty set F
Certain event: S

11. Examples 1. Toss a coin once S = {H, T}; A = {H}, B = {T}
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}
Select 1 card from a
deck of 52 cards.
S = {all 52 cards}

12. Laws of Probability

13. Probability rules (cont’d)
Coin Toss Example:
S = {Head, Tail}
Probability of heads = 0.5
Probability of tails = 0.5
Probability rules (cont’d)
3) The complement of any event A is the event that A does not occur, written as A.
The complement rule states that the probability of an event not occurring is 1 minus the probability that is does occur.
P(not A) = P(A) = 1 − P(A)
Tail  = not Tail = Head
P(Tail ) = 1 − P(Head) = 0.5 
Venn diagram:
Sample space made up of an event A and its complementary A , i.e., everything that is not A.

14. 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

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

16. Unions: , or Intersections: , and
AÇB
AÈB

17. Mutually Exclusive (Disjoint) Events
Venn Diagrams
Mutually Exclusive (Disjoint) Events
A and B disjoint: A  B=
Mutually exclusive or
disjoint events-no outcomes
from S in common
AÇB
AÈB
A and B not disjoint

18. Addition Rule for Disjoint Events
4. If A and B are disjoint events, then P(A or B) = P(A) + P(B)

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

20. General Addition Rule _ For any two events A and B +
P(A or B) = P(A) + P(B) - P(A and B)
P(A) =6/13
A or B + A
P(B) =5/13 _ B
P(A and B) =3/13
P(A or B) = 8/13

21. Laws of Probability - 5 Multiplication Rule
6. For two independent events A and B
P(A and B) = P(A) × P(B)
Note: assuming events are independent doesn’t make it true.

22. Multiplication Rule
The probability that you encounter a green light at the corner of Dan Allen and Hillsborough is 0.35, a yellow light 0.04, and a red light What is the probability that you encounter a red light on both Monday and Tuesday?
It’s reasonable to assume that the color of the light you encounter on Monday is independent of the color on Tuesday. So
P(red on Monday and red on Tuesday) =
P(red on Monday) × P(red on Tuesday) = 0.61 × 0.61 =

23. 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 or B) = P(A) + P(B)
5. For any two events A and B,
P(A or B) = P(A) + P(B) – P(A and B)
6. For two independent events A and B
P(A and B) = P(A) × P(B)

24. M&M candies
If you draw an M&M candy at random from a bag, the candy will have one of six colors. The probability of drawing each color depends on the proportions manufactured, as described here:
What is the probability that an M&M chosen at random is blue?
Color
Brown
Red
Yellow
Green
Orange
Blue
Probability
0.3
0.2
0.1 ?
S = {brown, red, yellow, green, orange, blue}
P(S) = P(brown) + P(red) + P(yellow) + P(green) + P(orange) + P(blue) = 1
P(blue) = 1 – [P(brown) + P(red) + P(yellow) + P(green) + P(orange)]
= 1 – [ ] = 0.1
What is the probability that a random M&M is any of red, yellow, or orange?
P(red or yellow or orange) = P(red) + P(yellow) + P(orange)
= = 0.5

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

26. Example: college students
Suppose 56% of all students live on campus, 62% of all students purchase a campus meal plan and 42% do both.
Question: what is the probability that a randomly selected student either lives OR eats on campus.
L = {student lives on campus}
M = {student purchases a meal plan}
P(a student either lives or eats on campus)
= P(L or M) = P(L) + P(M) - P(L and M)
= – 0.42
= 0.76

27. THE RELATIONSHIP BETWEEN ODDS AND PROBABILITIES
World Series Odds
The odds at the above link are the odds against a team winning the World Series, though the author claims they’re “odds for winning the World Series”
Odds are frequently a source of confusion. Odds for? Odds against?
From probability to odds
From odds to probability

28. From Probability to Odds
If event A has probability P(A), then the odds in favor of A are P(A) to 1-P(A). It follows that the odds against A are 1-P(A) to P(A)
If the probability of an earthquake in California is .25, then the odds in favor of an earthquake are .25 to .75 or 1 to 3. The odds against an earthquake are .75 to .25 or 3 to 1

29. From Odds to Probability
If the odds in favor of an event E are a to b, then
P(E)=a/(a+b)
in addition,
P(E’)=b/(a+b)
If the odds in favor of UNC winning the NCAA’s are 3 (a) to 1 (b), then
P(UNC wins)=3/4
P(UNC does not win)=
1/4

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

31. 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

32. Dice
You toss two dice. What is the probability of the outcomes summing to 5?
This is S:
{(1,1), (1,2), (1,3), ……etc.}
There are 36 possible outcomes in S, all equally likely (given fair dice). Thus, the probability of any one of them is 1/36.
P(the roll of two dice sums to 5) =
P(1,4) + P(2,3) + P(3,2) + P(4,1) = 4 / 36 = 0.111

33. We Need Efficient Methods for Counting Outcomes

34. 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?

35. 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)

36. 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

37. 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.

38. 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

39. 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?

40. 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

41. 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

42. Factorials with calculators and Excel
non-graphing: x ! (second function)
graphing: bottom p. 9 T I Calculator Commands
(math button)
Excel:
Insert function: Math and Trig category, FACT function

43. 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
20! is roughly the age of the universe in seconds

44. 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

45. Permutations (cont.)

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

47. 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

48. Combinations (cont.)

49. ST 311 Powerball Lottery From the numbers 1 through 20,
choose 6 different numbers.
Write them on a piece of paper.

50. And the numbers are ... 16 11 2 10 8 4
wow
scream

51. Chances of Winning?

52. Example: Illinois State Lottery

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

54. The Forrest Gump Visualization of 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

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

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

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

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

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

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

61. Chances of Winning NC Powerball Lottery?
Remember: one of the bills you put down is a $100 bill; all others are $1 bills.
Put on a blindfold and begin walking along the trail of bills.
Your chance of winning the lottery is the same as your chance of selecting the $100 bill if you stop at a random location along the trail and pick up a bill .

62. Virginia State Lottery

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

64. Probability Tree Example: probability of playing professional baseball
6.1% of high school baseball players play college baseball. Of these, 9.4% will play professionally.
Unlike football and basketball, high school players can also go directly to professional baseball without playing in college…
studies have shown that given that a high school player does not compete in college, the probability he plays professionally is .002.
Question 1: What is the probability that a high school baseball player ultimately plays professional baseball?
Question 2: Given that a high school baseball player played professionally, what is the probability he played in college?

65. Question 1: What is the probability that a high school baseball player ultimately plays professional baseball?
.061*.094=
.939*.002=
P(hs bb player plays professionally)
= .061* *.002 = =

66. Question 2: Given that a high school baseball player played professionally, what is the probability he played in college?
.061*.094=
.061*.094=
P(hs bb player plays professionally) = =
.939*.002=

67. Example: AIDS Testing V={person has HIV}; CDC: Pr(V)=.006
P : test outcome is positive (test indicates HIV present)
N : 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

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

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

70. Probability Tree Multiply clinical reliability branch probs

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

72. 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

73. Question 2: If your test comes back positive, what is the probability that you have HIV?

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

75. 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:
Pr(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.

76. 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