1. Probability

2. Lotto I am offered two lotto cards: Card 1: has numbers
Which card should I take so that I have the greatest chance of winning lotto?

3. Roulette
In the casino I wait at the roulette wheel until I see a run of at least five reds in a row.
I then bet heavily on a black.
I am now more likely to win.

4. Coin Tossing I am about to toss a coin 20 times.
What do you expect to happen?
Suppose that the first four tosses have been heads and there are no tails so far. What do you expect will have happened by the end of the 20 tosses ?

5. Coin Tossing Option A Option B
Still expect to get 10 heads and 10 tails. Since there are already 4 heads, now expect to get 6 heads from the remaining 16 tosses. In the next few tosses, expect to get more tails than heads.
Option B
There are 16 tosses to go. For these 16 tosses I expect 8 heads and 8 tails. Now expect to get 12 heads and 8 tails for the 20 throws.

6. TV Game Show In a TV game show, a car will be given away.
3 keys are put on the table, with only one of them being the right key. The 3 finalists are given a chance to choose one key and the one who chooses the right key will take the car.
If you were one of the finalists, would you prefer to be the 1st, 2nd or last to choose a key?

7. Let’s Make a Deal Game Show
You pick one of three doors
two have booby prizes behind them
one has lots of money behind it
The game show host then shows you a booby prize behind one of the other doors
Then he asks you “Do you want to change doors?”
Should you??! (Does it matter??!)
See the following website:

8. Game Show Dilemma
Suppose you choose door A. In which case Monty Hall will show you either door B or C depending upon what is behind each.
No Switch Strategy ~ here is what happens
Result A B C
Win
Car
Goat
Lose
P(WIN) = 1/3

9. Game Show Dilemma
Suppose you choose door A, but ultimately switch. Again Monty Hall will show you either door B or C depending upon what is behind each.
Switch Strategy ~ here is what happens
Result A B C
Lose
Car
Goat
Win
P(WIN) = 2/3 !!!!

10. Matching Birthdays
In a room with 23 people what is the probability that at least two of them will have the same birthday?
Answer: or 50.73% chance!!!!!
How about 30?
.7063 or 71% chance!
How about 40?
.8912 or 89% chance!
How about 50?
.9704 or 97% chance!

11. Probability In our discussion of probability
Introduce the basic ideas about probabilities:
what they are and where they come from
simple probability models (genetics)
conditional probabilities
independent events
Baye’s Rule
Examine how to calculate probabilities:
Using counting methods, tables of counts (not in text) and using properties of probabilities such as independence.

12. What is the probability it will turn up heads?
I toss a fair coin (where fair means ‘equally likely outcomes’)
What are the possible outcomes?
Head and tail ~ This is called a “dichotomous experiment” because it has only two possible outcomes. S = {H,T}.
What is the probability it will turn up heads?
1/2
I choose a duck nest at random and observe whether it gets predated or not.
Predated or Not Predated (“Success” and “Failure”)
What is the probability of predation?
?????
What factors influence this probability? ?????

13. What are Probabilities?
A probability is a number between 0 & 1 that quantifies uncertainty.
A probability of 0 identifies impossibility
A probability of 1 identifies certainty

14. Where do probabilities come from?
Probabilities from models:
The probability of getting a four when a fair dice is rolled is
1/6 ( or 16.7% chance)
The probability of winning craps on a “Pass Line” bet is .493
The probability of winning the jackpot in a Powerball lottery is

15. Where do probabilities come from?
Probabilities from data
or Empirical probabilities
What is the probability that a randomly selected starling is female?
A random sample n = 67 starlings was taken.
40 of these starlings are female.
The estimated probability that a randomly chosen starling will be a female is
40/67 (0.597 or 59.7% chance)

16. Where do probabilities come from?
Subjective Probabilities
The probability that there will be another outbreak of ebola in Africa within the next year is 0.1.
The probability of snow in the next 24 hours is very high. Perhaps the weather forecaster might say a there is a 80% chance of snow.
A doctor may state your chance of successful treatment, e.g. 70% chance of remission.

17. Simple Probability Models
“The probability that an event A occurs”
is written in shorthand as P(A).
For equally likely outcomes, and a given event A:
P(A) =
Number of outcomes in A
Total number of outcomes

18. Example 1: Hodgkin’s Disease
Response to Treatment
Type
None
Partial
Positive
Row
Totals LD 44 10 18 72 LP 12 74
104 MC 58 54
154
266 NS 16 68 96
Column
126 98
314
n = 538

19. Example 1: Hodgkin’s Disease
Response to Treatment
Type
None
Partial
Positive
Row
Totals LD 44 10 18 72 LP 12 74
104 MC 58 54
154
266 NS 16 68 96
Column
126 98
314
n = 538
Had positive response to treatment
P(Pos) = 314/538 = .584 or 58.4% chance

20. Example 1: Hodgkin’s Disease
Response to Treatment
Type
None
Partial
Positive
Row
Totals LD 44 10 18 72 LP 12 74
104 MC 58 54
154
266 NS 16 68 96
Column
126 98
314
n = 538
Had at least some response to treatment
P(Par and Pos) = ( )/538 = 412/538
= .766 or 76.6% chance

21. Example 1: Hodgkin’s Disease
Response to Treatment
Type
None
Partial
Positive
Row
Totals LD 44 10 18 72 LP 12 74
104 MC 58 54
154
266 NS 16 68 96
Column
126 98
314
n = 538
Had LP and positive response to treatment
P(LP and Pos) = 74/538 = .138 or 13.8%

22. Example 1: Hodgkin’s Disease
Response to Treatment
Type
None
Partial
Positive
Row
Totals LD 44 10 18 72 LP 12 74
104 MC 58 54
154
266 NS 16 68 96
Column
126 98
314
n = 538
Had LP or NS as their histological type.
P(LP or NS) = ( )/538 = .372 or 37.2% chance

23. Conditional Probability
We wish to find the probability of an event occuring given information about occurrence of another event. For example, what is probability of developing lung cancer given that we know the person smoked a pack of cigarettes a day for the past 30 years.
Key words that indicate conditional probability are:
“given that”, “of those”, “if …”, “assuming that”

24. Conditional Probability
“The probability of event A occurring given that event B has already occurred”
is written in shorthand as P(A|B)

25. Independence Events A and B are said to be independent if
P(A|B) = P(A) and P(B|A) = P(B)
i.e. knowing something about the occurrence of B tells you nothing about the occurrence of A.

26. Independence

27. Example 1: Hodgkin’s Disease
Response to Treatment
Type
None
Partial
Positive
Row
Totals LD 44 10 18 72 LP 12 74
104 MC 58 54
154
266 NS 16 68 96
Column
126 98
314
n = 538
Conditional Probs
P(Pos|LD) = ?
P(Pos|LD) = 18/72 = .25
or a 25% chance

28. Example 1: Hodgkin’s Disease
Response to Treatment
Type
None
Partial
Positive
Row
Totals LD 44 10 18 72 LP 12 74
104 MC 58 54
154
266 NS 16 68 96
Column
126 98
314
n = 538
Conditional Probs
P(Pos|LP) = ?
P(Pos|LP) = 74/104 = .712
or a 71.2% chance

29. Example 1: Hodgkin’s Disease
Response to Treatment
Type
None
Partial
Positive
Row
Totals LD 44 10 18 72 LP 12 74
104 MC 58 54
154
266 NS 16 68 96
Column
126 98
314
n = 538
Conditional Probs
P(Pos|MC) = ?
P(Pos|MC) = 154/266 = .579
or a 57.9% chance

30. Example 1: Hodgkin’s Disease
Response to Treatment
Type
None
Partial
Positive
Row
Totals LD 44 10 18 72 LP 12 74
104 MC 58 54
154
266 NS 16 68 96
Column
126 98
314
n = 538
Conditional Probs
P(Pos|NS) = ?
P(Pos|NS) = 68/96 = .708
or a 70.8% chance

31. Example 1: Hodgkin’s Disease
See the tutorial “Bivariate Displays for Categorical Data …” for a full details on this example and how to construct this plot in JMP. Under JMP Tutorials, not the new JMP Videos page.

32. Example 1: Hodgkin’s Disease

33. Example 1: Hodgkin’s Disease
Right-click on a colored panel in the mosaic plot and select Cell Labeling > Show Percents.

34. Example 2: Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991)
Brain Injury (BI)
No Brain Injury
(NBI)
Row
Totals
No Helmet (NH) 97
1918
2015
Helmet Worn (H) 17
977
994
Column
114
2895
3009
BI = the event the motorcyclist sustains brain injury
NBI = no brain injury
H = the event the motorcyclist was wearing a helmet
NH = no helmet worn
What is the probability that a motorcyclist involved in a accident sustains brain injury?
P(BI) = 114 / 3009 = .0379

35. Example 2: Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991)
Brain Injury (BI)
No Brain Injury (NBI)
Row
Totals
No Helmet (NH) 97
1918
2015
Helmet Worn (H) 17
977
994
Column
114
289
3009
BI = the event the motorcyclist sustains brain injury
NBI = no brain injury
H = the event the motorcyclist was wearing a helmet
NH = no helmet worn
What is the probability that a motorcyclist involved in a accident was wearing a helmet?
P(H) = 994 / 3009 = .3303

36. Example 2: Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991)
Brain Injury (BI)
No Brain Injury (NBI)
Row
Totals
No Helmet (NH) 97
1918
2015
Helmet Worn (H) 17
977
994
Column
114
2895
3009
BI = the event the motorcyclist sustains brain injury
NBI = no brain injury
H = the event the motorcyclist was wearing a helmet
NH = no helmet worn
What is the probability that the cyclist sustained brain injury given they were wearing a helmet?
P(BI|H) = 17 / 994 = .0171

37. Example 2: Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991)
Brain Injury (BI)
No Brain Injury (NBI)
Row
Totals
No Helmet (NH) 97
1918
2015
Helmet Worn (H) 17
977
994
Column
114
2895
3009
BI = the event the motorcyclist sustains brain injury
NBI = no brain injury
H = the event the motorcyclist was wearing a helmet
NH = no helmet worn
What is the probability that the cyclist not wearing a helmet sustained brain injury?
P(BI|NH) = 97 / 2015 = .0481

38. Example 2: Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991)
Brain Injury (BI)
No Brain Injury
(NBI)
Row
Totals
No Helmet (NH) 97
1918
2015
Helmet Worn (H) 17
977
994
Column
114
2895
3009
How many times more likely is a non-helmet wearer to sustain brain injury?
.0481 / = 2.81 times more likely. This is called the relative risk or risk ratio (denoted RR).

39. Example 2: Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991)
The shading for Brain Injury for the No Helmet group is roughly three times higher than the shading for Brain Injury for the Helmet Worn group.
Motorcyclists not wearing a helmet are at three times the risk of suffering brain injury.

40. Building a Contingency Table from a Story
Example 3: HIV and Condom Use (not in notes)
A European study on the transmission of the HIV virus involved 470 heterosexual couples. Originally only one of the partners in each couple was infected with the virus. There were 293 couples that always used condoms. From this group, 3 of the non-infected partners became infected with the virus. Of the 177 couples who did not always use a condom, 20 of the non-infected partners became infected with the virus.

41. Example 3: HIV and Condom Use (not in notes)
Let C be the event that the couple always used condoms. (NC be the complement)
Let I be the event that the non-infected partner became infected. (NI be the complement)
Condom Usage
Infection Status C NC
Total NI I
Total

42. Example 3: HIV and Condom Use (not in notes)
A European study on the transmission of the HIV virus involved 470 heterosexual couples. Originally only one of the partners in each couple was infected with the virus. There were 293 couples that always used condoms. From this group, 3 of the non-infected partners became infected with the virus.
Condom Usage
Infection Status C NC
Total NI I 3
Total
293
470

43. Example 3: HIV and Condom Use (not in notes)
Of the 177 couples who did not always use a condom, 20 of the non-infected partners became infected with the virus.
Condom Usage
Infection Status C NC
Total NI I 3 20 23
290
157
447
Total
293
177
470

44. Example 3: HIV and Condom Use (not in notes)
What proportion of the couples in this study always used condoms?
P(C ) C NC NI I
Total
Condom Usage
Infection Status
470
293 3 20
177
290
157 23
447

45. Example 3: HIV and Condom Use (not in notes)
What proportion of the couples in this study always used condoms?
P(C ) = 293/470 (= 0.623) C NC NI I
Total
Condom Usage
Infection Status
470
293 3 20
177
290
157 23
447

46. Example 3: HIV and Condom Use (not in notes)
If a non-infected partner became infected, what is the probability that he/she was one of a couple that always used condoms?
P(C|I ) = 3/23 = 0.130 C NC NI I
Total
Condom Usage
Infection Status
470
293 3 20
177
290
157 23
447

47. Example 3: HIV and Condom Use (not in notes)
c) In what percentage of couples did the non-HIV partner become infected amongst those that did not use condoms?
P(I|NC) = 20/177 = .113 or 11.3%
Amongst those that did where condoms?
P(I|C) = 3/293 = or 1.02%
What is relative risk of infection associated with not wearing a condom?
RR = P(I|NC) / P(I|C) = times more likely to become infected.

48. Example 3: HIV and Condom Use (not in notes)
The percentage of couples where the non-HIV partner became infected in the non-condom user group is 11 times higher than that for condom group.

49. Relative Risk (RR) and Odds Ratio (OR)
Example: Age at First Pregnancy and Cervical Cancer
A case-control study was conducted to determine whether there was increased risk of cervical cancer amongst women who had their first child before age 25. A sample of 49 women with cervical cancer was taken of which 42 had their first child before the age of 25. From a sample of 317 “similar” women without cervical cancer it was found that 203 of them had their first child before age 25.
Q: Do these data suggest that having a child at or before age 25 increases risk of cervical cancer?

50. Odds Ratio (OR) The ODDS for an event A are defined as
P(A)
1 – P(A)
For example suppose we roll a single die the odds for a 6 are:
Odds for 6 = P(6)/(1 – P(6)) = = (1/6)/(1 – (1/6))
= 1/5 (1:5 odds for or 5:1 odds against)
i.e. 1 six for every 5 rolls that don’t result in a six.

51. Odds Ratio (OR) OR = _________________________ P(Disease|Risk Factor)
Odds for disease amongst those with risk factor present
The Odds Ratio (OR) for a disease associated with a risk factor is ratio of the odds for disease for those with risk factor and the odds for disease for those without the risk factor
OR = _________________________
P(Disease|Risk Factor)
_____________________
1 – P(Disease|Risk Factor)
P(Disease|No Risk Factor)
_______________________
1 – P(Disease|No Risk Factor)
Odds for disease amongst those without the risk factor.
The Odds Ratio gives us the multiplicative increase in odds associated with having the “risk factor”.

52. Relative Risk (RR) and Odds Ratio (OR)
Age at 1st Pregnancy
Case
Control
Row Totals
Age < 25 42
203
245
Age > 25 7
114
121
Column
Totals 49
317
n = 366
Cervical Cancer
a) Why can’t we calculate P(Cervical Cancer | Age < 25)?
Because the number of women with disease was fixed in advance and therefore NOT RANDOM !

53. Relative Risk (RR) and Odds Ratio (OR)
Age at 1st Pregnancy
Case
Control
Row Totals
Age < 25 42
203
245
Age > 25 7
114
121
Column
Totals 49
317
n = 366
Cervical Cancer
b) What is P(risk factor|disease status) for each group?
P(Age < 25|Case) = 42/49 = .857 or 85.7%
P(Age < 25|Control) = 203/317 = .640 or 64.0%

54. Relative Risk (RR) and Odds Ratio (OR)
Age at 1st Pregnancy
Case
Control
Row Totals
Age < 25 42
203
245
Age > 25 7
114
121
Column
Totals 49
317
n = 366
Cervical Cancer
c) What are the odds for the risk factor amongst the cases?
Amongst the controls?
Odds for risk factor cases = .857/(1-.857) = 5.99
Odds for risk factor controls = .64/(1- .64) = 1.78

55. Relative Risk (RR) and Odds Ratio (OR)
Age at 1st Pregnancy
Case
Control
Row Totals
Age < 25 42
203
245
Age > 25 7
114
121
Column
Totals 49
317
n = 366
Cervical Cancer
d) What is the odds ratio for the risk factor associated with being a case?
Odds Ratio (OR) = 5.99/1.78 = 3.37, the odds for having 1st child on or before age 25 are 3.37 times higher for women who currently have cervical cancer versus those that do not have cervical cancer.

56. Relative Risk (RR) and Odds Ratio (OR)
The ratio of dark to light shading is 3.37 times larger for the cervical cancer group than it is for the control group.

57. Relative Risk (RR) and Odds Ratio (OR)
Even though it is inappropriate to do so calculate P(disease|risk status).
P(case|Age<25) = 42/245 = .171 or 17.1%
P(case|Age>25) = 7/121 = .058 or 5.8%
Now calculate the odds for disease given the risk factor status
Odds for Disease for 1st Preg. Age < 25
= .171/( ) = .207
Odds for Disease for 1st Preg. Age > 25
= .058/( ) = .061

58. Relative Risk (RR) and Odds Ratio (OR)
f) Finally calculate the odds ratio for disease associated with 1st pregnancy age < 25 years of age.
Odds Ratio = .207/.061 = 3.37
This is exactly the same as the odds ratio for having the risk factor (Age < 25) associated with being in the cervical cancer group!!!!
Final Conclusion: Women who have their first child at or before age 25 have 3.37 times the odds of developing cervical cancer when compared to women who had their first child after the age of 25.

59. Relative Risk (RR) and Odds Ratio (OR)
Risk Factor
Status
Case
Control
Risk Factor Present a b
Risk Factor Absent c d
Disease Status
a X d
OR = _____
b X c
Much easier computational formula!!!

60. Relative Risk (RR) and Odd’s Ratio (OR)
When the disease is fairly rare, i.e. P(disease) < .10 or 10%, then one can show that the odds ratio (OR) and relative risk (RR) are similar in value.
OR is approximately equal to RR when
P(disease) < .10 or 10% chance.
In these cases we can use the phrase:
“… times more likely” when interpreting the OR.

61. Relative Risk (RR) and Odds Ratio (OR)
Age at 1st Pregnancy
Case
Control
Row Totals
Age < 25 a 42 b
203
245
Age > 25 c 7 d
114
121
Column
Totals 49
317
n = 366
OR = (42 X 114)/(7 X 203) = Because less than 10% of the population of women develop cervical cancer we can say: “Women who have their first child at or before age 25 are 3.37 times more likely to develop cervical cancer than women who have their first child after age 25.”