C4, L1, S1 Chapter 2 Probability. C4, L1, S2 I am offered two lotto cards: –Card 1: has numbers...

C4, L1, S1

Chapter 2Probability

C4, L1, S2

I am offered two lotto cards:

– Card 1: has numbers

– Card 2: has numbers

Which card should I take so that I have the greatest chance of winning lotto?

Lotto

C4, L1, S3

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.

Roulette

C4, L1, S4

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 ?

C4, L1, S5

Coin Tossing

• Option A– 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.

C4, L1, S6

• 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?

TV Game Show

C4, L1, S7

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:• http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

C4, L1, S8

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 Goat

Lose Goat Car Goat

Lose Goat Goat Car

P(WIN) = 1/3

C4, L1, S9

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 Goat

Win Goat Car Goat

Win Goat Goat Car

Monty will show either B or C.

You switch to the one not shown

and lose.

Monty will show door C, you switch to B and win.

Monty will show door B, you switch to C and win.

P(WIN) = 2/3 !!!!

C4, L1, S10

Matching Birthdays• In a room with 23 people what is the

probability that at least two of them will have the same birthday?

• Answer: .5073 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!

C4, L1, S11

Probability

What is Chapter 6 trying to do? – Introduce us to basic ideas about probabilities:

• what they are and where they come from• simple probability models (genetics)• conditional probabilities• independent events• Baye’s Rule

Teach us how to calculate probabilities:• tables of counts and using properties of

probabilities such as independence.

C4, L1, S12

ProbabilityI 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 patient at random and observe whether they are successfully treated.

What are the possible outcomes?

“Success” and “Failure”

What is the probability of successful treatment?

?????

What factors influence this probability? ?????

C4, L1, S13

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

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Where do probabilities come from?

• Probabilities from models:The probability of getting a four when a fair dice is rolled is

1/6 (0.1667 or 16.7% chance)

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• Probabilities from data

or Empirical probabilities

What is the probability that a randomly selected patient is successfully treated?– In a clinical trial n = 67 patients are “randomly”

selected.– 40 of these patients are successfully treated.– The estimated probability that a randomly chosen

patient will have a successful outcome is

40/67 (0.597 or 59.7% chance)


C4, L1, S16

• Subjective Probabilities– The probability that there will be another

outbreak of ebola in Africa within the next year is 0.1.

– The probability of rain in the next 24 hours is very high. Perhaps the weather forecaster might say a there is a 70% chance of rain.

– A doctor may state your chance of successful treatment.


C4, L1, S17

For equally likely outcomes, and a given event A:

Simple Probability Models

“The probability that an event A occurs”

is written in shorthand as P(A).

P(A) =Number of outcomes in A

Total number of outcomes

C4, L1, S18

1. Heart Disease

In 1996, 6631 Minnesotans died from coronary heart disease. The numbers of deaths classified by age and gender are:

Sex

Age Male Female Total

< 45 79 13 92

45 - 64 772 216 988

65 - 74 1081 499 1580

> 74 1795 2176 3971

Total 3727 2904 6631

C4, L1, S19

Let

A be the event of being under 45B be the event of being maleC be the event of being over 64

1. Heart Disease

Sex


< 45 79 13 92

45 - 64 772 216 988

65 - 74 1081 499 1580

> 74 1795 2176 3971

Total 3727 2904 6631

C4, L1, S20

Find the probability that a randomly chosen member of this population at the time of death was:

a) under 45 P(A) = 92/6631 = 0.0139

1. Heart Disease

Sex


< 45 79 13 92

45 - 64 772 216 988

65 - 74 1081 499 1580

> 74 1795 2176 3971

Total 3727 2904 6631

C4, L1, S21

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”

C4, L1, S22

“The probability of event A occurring given that event B has already occurred”

is written in shorthand as P(A|B)

Conditional Probability

C4, L1, S23

P(A|B) =__________ , P(B) > 0

Conditional Probability and Independence

P(A and B) P(B)

Two events A and B are said to be independent if

P(A|B) = P(A) and P(B|A) = P(B)

i.e. knowing the occurrence of one of the events tells you nothing about the occurrence of the other.

C4, L1, S24

1. Heart Disease

Sex


< 45 79 13 92

45 - 64 772 216 988

65 - 74 1081 499 1580

> 74 1795 2176 3971

Total 3727 2904 6631


b) male assuming that the person was younger than 45.

C4, L1, S25

Sex


< 45 79 13 92

45 - 64 772 216 988

65 - 74 1081 499 1580

> 74 1795 2176 3971

Total 3727 2904 6631


b) male given that the person was younger than 45. P(B|A) = 79/92 = 0.8587

2. Heart Disease

P(B|A) = P(A and B)/P(A) = (79/6631)/(92/6631) = 79/92

C4, L1, S26

Sex


< 45 79 13 92

45 - 64 772 216 988

65 - 74 1081 499 1580

> 74 1795 2176 3971

Total 3727 2904 6631


c) male and was over 64.

P(B and C) = (1081 + 1795)/6631= 2876/6631=.434

1. Heart Disease

C4, L1, S27

Sex


< 45 79 13 92

45 - 64 772 216 988

65 - 74 1081 499 1580

> 74 1795 2176 3971

Total 3727 2904 6631


d) over 64 given they were female (not B).

1. Heart Disease

C4, L1, S28

Sex


< 45 79 13 92

45 - 64 772 216 988

65 - 74 1081 499 1580

> 74 1795 2176 3971

Total 3727 2904 6631

P(C|not B) = (499+2176)/2904 = .9211

1. Heart DiseaseFind the probability that a randomly chosen member of this population at the time of death was:

d) over 64 given they were female (not B).

C4, L1, S29

2. Hodgkin’s Disease

Type None Partial

Positive

Row

Totals

LD 44 10 18 72

LP 12 18 74 104

MC 58 54 154 266

NS 12 16 68 96

Column

Totals

126 98 314 n = 538

Response to Treatment

C4, L1, S30


C4, L1, S31


Type

None Partial Positive

Row

Totals

LD 44 10 18 72

LP 12 18 74 104

MC 58 54 154 266

NS 12 16 68 96

Column

Totals

126 98 314 n = 538


a) Had positive response to treatment

P(pos) = 314/538 = .584 or 58.4% chance

C4, L1, S32


Type

None Partial PositiveRow

Totals

LD 44 10 18 72

LP 12 18 74 104

MC 58 54 154 266

NS 12 16 68 96

Column

Totals

126 98 314 n = 538


b) Had at least some response to treatment

P(par or pos) = (98 + 314)/538 = 412/538

= .766 or 76.6% chance

C4, L1, S33

2. Hodgkin’s DiseaseType

None Partial PositiveRow

Totals

LD 44 10 18 72

LP 12 18 74 104

MC 58 54 154 266

NS 12 16 68 96

Column

Totals

126 98 314 n = 538


c) Had positive response to treatment given they have LP

P(pos|LP) = 74/104 = .7115 or 71.15%

C4, L1, S34


Type

None Partial Positive

Row

Totals

LD 44 10 18 72

LP 12 18 74 104

MC 58 54 154 266

NS 12 16 68 96

Column

Totals

126 98 314 n = 538


d) Had positive response to treatment given they have LD.

P(pos|LD) = 18/72= .25 or 25.0% chance

C4, L1, S35

Mosaic Plot with Conditional Probs.

C4, L1, S36

Adding Conditional Probs. in JMP

C4, L1, S37

3. Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991)

Brain Injury

No Brain Injury

Row Totals

No Helmet 97 1918 2015

Helmet Worn 17 977 994

ColumnTotals 114 2895 3009

BI = the event the motorcyclist sustains brain injury

NBI = no braininjury

H = the event themotorcyclist waswearing a helmet

NH = no helmet worn P(BI) = 114 / 3009 = .0379

What is the probability that a motorcyclist involved in a accident sustains brain injury?

C4, L1, S38


Brain Injury

No Brain Injury

Row Totals

No Helmet 97 1918 2015






NH = no helmet worn P(H) = 994 / 3009 = .3303

What is the probability that a motorcyclist involved in a accident was wearing a helmet?

C4, L1, S39


Brain Injury

No Brain Injury

Row Totals

No Helmet 97 1918 2015



What is the probability that the cyclist sustained brain injury given they were wearing a helmet? P(BI|H) = 17 / 994 = .0171




NH = no helmet worn

C4, L1, S40


Brain Injury

No Brain Injury

Row Totals

No Helmet 97 1918 2015



What is the probability that the cyclist not wearing a helmet sustained brain injury? P(BI|NH) = 97 / 2015

= .0481




NH = no helmet worn

C4, L1, S41


Brain Injury

No Brain Injury

Row

Totals

No Helmet 97 1918 2015


Column

Totals 114 2895 3009

How many times more likely is a non-helmet wearer to sustain brain injury?

.0481 / .0171 = 2.81 times more likely. This is called the relative risk or risk ratio (denoted RR).

C4, L1, S42

Example 3: 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. (recall RR = 2.81)

Motorcyclists not wearing a helmet are at three times the risk of suffering brain injury.

C4, L1, S43

Building a Contingency Table from a Story

4. HIV Example

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.

C4, L1, S44

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)

C NC

NI

I

4. HIV Example

Total

Total

Condom UsageInfectio

n Status

C4, L1, S45

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.

C NC

NI

I

4. HIV Example

Total

Total


n Status

470293

3

C4, L1, S46

Of the 177 couples who did not always use a condom, 20 of the non-infected partners became infected with the virus.

C NC

NI

I

4. HIV Example

Total

Total


n Status

470293

3 20

177

290 15723

447

C4, L1, S47

a) What proportion of the couples in this study always used condoms?

C NC

NI

I

Total

Total

Condom UsageInfection

Status

470293

3 20

177

290 15723

447

4. HIV Example

P(C )

C4, L1, S48

a) What proportion of the couples in this study always used condoms?

C NC

NI

I

Total

Total


Status

470293

3 20

177

290 15723

447

4. HIV Example

P(C ) = 293/470 (= 0.623)

C4, L1, S49

b) If a non-infected partner became infected, what is the probability that he/she was one of a couple that always used condoms?

4. HIV Example

C NC

NI

I

Total

Total


Status

470293

3 20

177

290 15723

447

P(C|I ) = 3/23 = 0.130

C4, L1, S50

4. HIV Example

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 = .0102 or 1.02%• What is relative risk of infection associated

with not wearing a condom?

RR = P(I|NC) / P(I|C) = 11.08 times more likely to become infected.

C4, L1, S51

4. HIV Example

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.

The risk of infection is 11 time higher in no condom group

C4, L1, S52

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?

C4, L1, S53


The ODDS for an event A are defined as

Odds for A = _______P(A)

1 – P(A)

For example suppose we roll a single die the odds for a 3 are:

Odds for 3 = P(3)/(1 – P(3)) = = (1/6)/(1 – (1/6)) = 1/5

1 three for every 5 rolls that don’t result in a six.

(Odds for a 3 are 1:5 and odds against are 5:1)

C4, L1, S54

Relative Risk (RR) and Odds Ratio (OR)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)

_______________________

The Odds Ratio gives us the multiplicative increase in odds associated with having the “risk factor”.

Odds for disease amongst those with risk factor present

Odds for disease amongst those without the risk factor.

C4, L1, S55


Age at 1st Pregnancy Case Contro

l

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 !

C4, L1, S56



l

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%

C4, L1, S57


Age at 1st Pregnancy Cas

eContr

ol

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

C4, L1, S58



l

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.

C4, L1, S59


Odds Ratio

The ratio of dark to light shading is 3.37 times larger for the cervical cancer group than it is for the control group.

C4, L1, S60

e) 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 statusOdds for Disease for 1st Preg. Age < 25 = .171/(1 - .171) = .207Odds for Disease for 1st Preg. Age > 25 = .058/(1 - .058) = .061


C4, L1, S61

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.

C4, L1, S62


Risk Factor

Status Case Control

Risk Factor Present

a b

Risk Factor Absent

c d

Disease Status

OR = _____a X d

b X c

Much easier computational formula!!!

C4, L1, S63

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 and relative risk are similar.

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.

C4, L1, S64



l

Row Totals

Age < 25

a42

b203 245

Age > 25

c7

d114 121

Column

Totals 49 317 n = 366

OR = (42 X 114)/(7 X 203) = 3.37 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.

C4, L1, S65

More About RR and OR• The most commonly cited advantage of the RR over the OR

is that the former is the more natural interpretation. The relative risk comes closer to what most people think of when they compare the relative likelihood of events.

e.g. suppose there are two groups, one with a 25% chance of mortality and the other with a 50% chance of mortality. Most people would say that the latter group has it twice as bad. But the odds ratio is 3, which seems too big.

RR = .50/.25 = 2.00 OR = P(death|high mortality)/P(survive|high mortality) P(death|low mortality)/P(survive|low mortality)

= .50/(1 - .50) = 3.00 .25/(1 - .25)

C4, L1, S66

More About RR and OREven more extreme examples are possible. A change

from 25% to 75% mortality represents a relative risk of 3, but an odds ratio of 9. A change from 10% to 90% mortality represents a relative risk of 9 but an odds ratio of 81.

RR = .90 /.10 = 9.00

OR = P(death|high mortality)/P(survive|high mortality)

P(death|low mortality)/P(survive|low mortality)

= .90/(1 - .90) = 81.00

.10/(1 - .10)

C4, L1, S67

More About RR and OR

• OR’s arise as part of logistic regression which we will study later in the course.

• Despite their pitfalls OR’s are really the only option when case-control studies are used.

• Any study of risk needs to adjust for potential confounding factors which is typically done using logistic regression.

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C4, L1, S1 Chapter 2 Probability. C4, L1, S2 I am offered two lotto cards: –Card 1: has numbers...

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