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Introduction to probability. Probability . Probability – the chance that an uncertain event will occur (always between 0 and 1) Symbols: P(event A) = “the probability that event A will occur” P(red card) = “the probability of a red card”

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Introduction to probability

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Introduction to probability

- Probability – the chance that an uncertain event will occur (always between 0 and 1)
Symbols:

P(event A) = “the probability that event A will occur”

P(red card) = “the probability of a red card”

P(~event A) = “the probability of NOT getting event A” [complement]

P(~red card) = “the probability of NOT getting a red card”

P(A & B) = “the probability that both A and B happen” [joint probability]

P(red card & ace) = “the probability of getting a red ace”

1. Theoretical probability—based on theory (a priori understanding of a phenomena)

e.g.: theoretical probability of rolling a 2 on a standard die is 1/6

theoretical probability of choosing an ace from a standard deck is 4/52

theoretical probability of getting heads on a regular coin is 1/2

2. Empirical probability—based on empirical data

e.g.: you toss an irregular die (probabilities unknown) 100 times and find that you get a two 25 times; empirical probability of rolling a two is 1/4

empirical probability of a major Earthquake (>=6.7 on the Richter scale) in Bay Area by 2032 is .62 (based on historical data)

empirical probability of a lifetime smoker developing lung cancer is 15 percent (based on empirical data)

- Sample space: the set of all possible outcomes.
For example, in genetics, if both the mother and father carry one copy of a recessive disease-causing mutation (d), there are three possible outcomes (the sample space):

- child is not a carrier (DD)
- child is a carrier (Dd)
- child has the disease (dd).

- Probabilities: the likelihood of each of the possible outcomes (always 0 P 1.0).
- P(genotype=DD)=.25
- P(genotype=Dd)=.50
- P(genotype=dd)=.25.

Note: mutually exclusive, exhaustive probabilities sum to 1.

Child’s outcome

Father’s allele

Mother’s allele

P(DD)=.5*.5=.25

P(♂D=.5)

P(♀D=.5)

P(♂d=.5)

P(Dd)=.5*.5=.25

P(♂D=.5)

P(dD)=.5*.5=.25

P(♀d=.5)

P(dd)=.5*.5=.25

______________

1.0

P(♂d=.5)

Mendel example: What’s the chance of having a heterozygote child (Dd) if both parents are heterozygous (Dd)?

Rule of thumb: in probability, “and” means multiply, “or” means add

Conditional Probability: Read as “the probability that the father passes a D allele given that the mother passes a d allele.”

Intuitively, we understand that the mother’s and father’s outcomes are independent. The mother’s and father’s alleles are segregating independently.

Intuitively, independence means that the probability of A happening does not change depending on whether B happens and vice versa.

It’s easy to see on the probability tree: the father’s probabilities are the same regardless of which maternal branch you’re on:

P(♂D/♀D)=.5 and P(♂D/♀d)=.5

Joint Probability: The probability of two events happening simultaneously.

Marginal probability: This is the probability that an event happens at all, ignoring all other outcomes.

Formal definition:

A and B are independent if and only if P(A&B)=P(A)*P(B)

Formally, P(DD)=.25=P(D♂)*P(D♀)

Conditional probability

Child’s outcome

Marginal probability: mother

Mother’s allele

Joint probability

P(DD)=.5*.5=.25

P(♀D=.5)

P(Dd)=.5*.5=.25

P(dD)=.5*.5=.25

P(♀d=.5)

P(dd)=.5*.5=.25

______________

1.0

Sum to get Marginal probability for the father

Father’s allele

P(♂D/ ♀D )=.5

P(♂d=.5)

P(♂D=.5)

P(♂d=.5)

- What is the probability of picking 2 aces from a full deck?

P(Ace=3/51)

P(Ace=4/52)

P(~Ace=48/51)

P(Ace=4/51)

P(~Ace=48/52)

P(~Ace=47/51)

Outcome

Draw 2

Draw 1

______________

1.0

Outcome

______________

1.0

Are the events (draw ace on the first and draw ace on the second) independent?

Intuitively Probabilities change depending on the outcome of the first draw.

Mathematically

P(Ace1)=.0045+.0723 =.0768 (marginal)

Outcome

______________

1.0

Are the events (draw ace on the first and draw ace on the second) independent?

Intuitively Probabilities change depending on the outcome of the first draw.

Mathematically

P(Ace1)=.0045+.0723 =.0768 (marginal)

P(Ace2) = .0045 + .0723 = .0768 (marginal)

Outcome

______________

1.0

- Are the events (draw ace on the first and draw ace on the second) independent?
Intuitively Probabilities change depending on the outcome of the first draw.

Mathematically

- P(Ace1)=.0045+.0723 =.0768 (marginal)
- P(Ace2) = .0045 + .0723 = .0768 (marginal)
- P(Ace1)* P(Ace2)=.0768*.0768 = .0059
- P(ace,ace)=.0045 (joint)
- NOT independent!

If HIV has a prevalence of 3% in San Francisco, and a particular HIV test has a false positive rate of .001 and a false negative rate of .01, what is the probability that a random person selected off the street will test positive?

Joint probability of being + and testing +

Conditional probability: the probability of testing + given that a person is +

Marginal probability of carrying the virus.

P(test +)=.99

P(+)=.03

P(test - )= .01

P(test +) = .001

P(-)=.97

P(test -) = .999

Marginal probability of testing positive

P (+, test +)=.0297

P(+, test -)=.003

P(-, test +)=.00097

P(-, test -) = .96903

______________

1.0

P(test +)=.0297+.00097=.03067

P(+&test+)P(+)*P(test+)

.0297 .03*.03067 (=.00092)

Dependent!

One of these has to be true (mutually exclusive, collectively exhaustive). They sum to 1.0.

- What is the probability of rolling a total of 8 on two dice?
- 1/36
- 8/36
- 1/6
- 5/36
- 4/6

- What is the probability of rolling a total of 8 on two dice?
- 1/36
- 8/36
- 1/6
- 5/36
- 4/6

36 ways to roll two die (6x6).

5 ways to get an 8:

5-3, 3-5, 4-4, 2-6, 6-2

- What is the probability of rolling two sixes in a row?
- 1/36
- 8/36
- 1/6
- 5/36
- 4/6

- What is the probability of rolling two sixes in a row?
- 1/36
- 8/36
- 1/6
- 5/36
- 4/6

P(6&6) = P(6)*P(6)

by independence

So P(6&6) = 1/6*1/6 = 1/36

- The probability of getting a straight flush from a fair deck of 52 cards is an example of:
- An empirical probability
- A theoretical probability
- None of the above
- Both of the above

- The probability of getting a straight flush from a fair deck of 52 cards is an example of:
- An empirical probability
- A theoretical probability
- None of the above
- Both of the above

- Which of the following is an example of statistically independent events?
- P(A)=.2; P(B)=.5; P(A&B) = .07
- P(A)=.2; P(B)=.5; P(A&B) = .10
- P(A)=.2; P(B)=.5; P(A&B) = .70
- P(A)=.2; P(B)=.5; P(A&B) = 0

- Which of the following is an example of statistically independent events?
- P(A)=.2; P(B)=.5; P(A&B) = .07
- P(A)=.2; P(B)=.5; P(A&B) = .10
- P(A)=.2; P(B)=.5; P(A&B) = .70
- P(A)=.2; P(B)=.5; P(A&B) = 0

The “law of total probability” gives you:

- A conditional probability
- A marginal probability
- A joint probability
- All of the above
- None of the above

The “law of total probability” gives you:

- A conditional probability
- A marginal probability
- A joint probability
- All of the above
- None of the above

Bayes’ rule

- Definition:
Let A and B be two events with P(B) 0. The conditional probability of A given B is:

The idea: if we are given that the event B occurred, the relevant sample space is reduced to B—eg, the P(B)=1 because we know B is true—and the probability of A is simply the proportion of B that’s also A (joint probability).

40% chance of failing both

50% chance of failing the midterm

50% chance of failing the class

A

A&B

B

Sample space

A&B

B

Sample space is reduced to this 50%.

A&B

B

What’s the probability of A now that we’re restricted to B? It’s the A&B proportion of B:

can be re-arranged to:

and, since also:

From the “Law of Total Probability”

OR

- Why do we care??
- Why is Bayes’ Rule useful??
- It turns out that sometimes it is very useful to be able to “flip” conditional probabilities. That is, we may know the probability of A given B, but the probability of B given A may not be obvious. An example will help…

- If HIV has a prevalence of 3% in San Francisco, and a particular HIV test has a false positive rate of .001 and a false negative rate of .01, what is the probability that a random person who tests positive is actually infected (also known as “positive predictive value”)?

P (+, test +)=.0297

P(test +)=.99

P(+)=.03

P(test - = .01)

P(+, test -)=.003

P(test +) = .001

P(-, test +)=.00097

P(-)=.97

P(-, test -) = .96903

P(test -) = .999

______________

1.0

A positive test places one on either of the two “test +” branches.

But only the top branch also fulfills the event “true infection.”

Therefore, the probability of being infected is the probability of being on the top branch given that you are on one of the two circled branches above.

An insurance company believes that drivers can be divided into two classes—those that are of high risk and those that are of low risk. Their statistics show that a high-risk driver will have an accident at some time within a year with probability .4, but this probability is only .1 for low risk drivers.

- Assuming that 20% of the drivers are high-risk, what is the probability that a new policy holder will have an accident within a year of purchasing a policy?
- If a new policy holder has an accident within a year of purchasing a policy, what is the probability that he is a high-risk type driver?

Assuming that 20% of the drivers are of high-risk, what is the probability that a new policy holder will have an accident within a year of purchasing a policy?

Use law of total probability:

P(accident)=

P(accident/high risk)*P(high risk) +

P(accident/low risk)*P(low risk) =

.40(.20) + .10(.80) = .08 + .08 = .16

P(accident, high risk)=.08

P(accident/HR)=.4

P(high risk)=.20

P( no acc/HR)=.6

P(no accident, high risk)=.12)

P(accident/LR)=.1

P(accident, low risk)=.08

P(low risk)=.80

P( no accident/LR)=.9

P(no accident, low risk)=.72

______________

1.0

If a new policy holder has an accident within a year of purchasing a policy, what is the probability that he is a high-risk type driver?

P(high-risk/accident)=

P(accident/high risk)*P(high risk)/P(accident)

=.40(.20)/.16 = 50%

Or use tree:

P(high risk/accident)=.08/.16=50%

In epidemiology, the association between a risk factor or protective factor (exposure) and a disease may be evaluated by the “risk ratio” (RR) or the “odds ratio” (OR).

Both are measures of “relative risk”—the general concept of comparing disease risks in exposed vs. unexposed individuals.

1:1

3:1

1:9

1:99

Note: An odds is always higher than its corresponding probability, unless the probability is 100%.

Exposed

Disease-free cohort

Not Exposed

Disease

Disease-free

Target population

Disease

Disease-free

TIME

Exposure (E)

No Exposure (~E)

Disease (D)

a

b

No Disease (~D)

c

d

a+c

b+d

risk to the exposed

risk to the unexposed

The Risk Ratio, or Relative Risk (RR)

Normal BP

Congestive Heart Failure

High Systolic BP

No CHF

400

400

1500

3000

1100

2600

- This risk ratio seems like the perfect measure of relative risk. Why not stop here? Why introduce the more complicated odds ratio??
- We cannot calculate a risk ratio from a case-control study. Case-control studies are a popular study design in epidemiology, because they are useful for studying rare diseases.
- In a case-control study, we sample conditional on disease status, so we cannot calculate risk of disease.

Disease

(Cases)

Exposed in past

Not exposed

Target population

Exposed

No Disease

(Controls)

Not Exposed

Hep C +

Hep C -

Cases: Liver cancer

90

10

Controls

30

70

The Odds Ratio (OR)

100

100

What are P(D/E) and P(D/~E) here?

We can’t tell, because, by design, we have fixed the proportion of liver cancer cases in this sample at 50% simply by selecting half controls and half cases.

All these data give us is: P(E/D) and P(E/~D).

Hep C +

Hep C -

Cases: Liver cancer

90

10

Controls

30

70

Unfortunately, our sampling scheme precludes calculation of the marginals: P(E) and P(D), but turns out we don’t need these if we use an odds ratio because the marginals cancel out!

The Odds Ratio (OR)

100

100

by Bayes’ Rule…

Luckily, P(E/D) [=the quantity you have]

Odds of disease in the exposed

Odds of exposure in the cases

Odds of disease in the unexposed

Odds of exposure in the controls

The Odds Ratio (OR)

This expression is mathematically equivalent to:

Backward from what we want…

The direction of interest!

Proof:

Odds of exposure in the cases

Odds of exposure in the controls

Bayes’ Rule

Odds of disease in the exposed

What we want!

Odds of disease in the unexposed

=

Hep C +

Hep C -

Cases: Liver cancer

90

10

Controls

30

70

The odds ratio

100

100

- Interpretation: there is a 21-fold higher odds of stroke in those who are Hep C +.

- The odds ratio will always be bigger than the corresponding risk ratio if RR >1 and smaller if RR <1 (the harmful or protective effect always appears larger)
- The magnitude of the inflation depends on the prevalence of the disease.

1

1

When a disease is rare:

P(~D) = 1 - P(D) 1

Odds ratio

Odds ratio

Odds ratio

Risk ratio

Risk ratio

Odds ratio

Risk ratio

Risk ratio

Rare Outcome

1.0 (null)

Common Outcome

1.0 (null)

- A cross-sectional study on risk factors for wrinkles found that heavy smoking significantly increases the risk of prominent wrinkles.
- Adjusted OR=3.92 (heavy smokers vs. nonsmokers)
- Interpretation: heavy smoking increases risk of prominent wrinkles nearly 4-fold?
- The prevalence of prominent wrinkles in non-smokers is roughly 45%. So, it’s not possible to have a 4-fold increase in risk!
- In fact, though the OR=3.92, the RR is closer to 2. Still a huge absolute increase, but not a 4-fold increase!

Raduan et al. J Eur Acad Dermatol Venereol. 2008 Jul 3.

- If the outcome has a 10% prevalence in the unexposed group, the maximum possible RR=10.0.
- For 20% prevalence, the maximum possible RR=5.0
- For 30% prevalence, the maximum possible RR=3.3.
- For 40% prevalence, maximum possible RR=2.5.
- For 50% prevalence, maximum possible RR=2.0.
- The prevalence of the outcome in the unexposed group is often NOT given! So, you have to estimate or consult other data.

Hep C +

Hep C -

Cases: Liver cancer

10

50

Controls

10

100

What is the odds ratio from this case-control study of liver cancer and Hep C?

- 2.0
- 1.5
- 0.5
- 3.0
- Cannot calculate

Hep C +

Hep C -

Cases: Liver cancer

10

50

Controls

10

100

What is the odds ratio from this case-control study of liver cancer and Hep C?

- 2.0
- 1.5
- 0.5
- 3.0
- Cannot calculate

Hep C +

Hep C -

Cases: Liver cancer

10

50

Controls

10

100

What is the risk ratio from this case-control study of liver cancer and Hep C?

- 2.0
- 1.5
- 0.5
- 3.0
- Cannot calculate

Hep C +

Hep C -

Cases: Liver cancer

10

50

Controls

10

100

What is the risk ratio from this case-control study of liver cancer and Hep C?

- 2.0
- 1.5
- 0.5
- 3.0
- Cannot calculate

If a patient has a 10% chance of a bad outcome, what are her odds of a bad outcome?

- 1/10
- 1/9
- 1/11
- 1/100
- Cannot calculate

If a patient has a 10% chance of a bad outcome, what are her odds of a bad outcome?

- 1/10
- 1/9
- 1/11
- 1/100
- Cannot calculate

If the true risk ratio for the chance of developing breast cancer for exercisers vs. sedentary women is .43, what is a possible value for the true OR?

- .41
- .45
- 1.0
- 2.0
- Cannot determine

If the true risk ratio for the chance of developing breast cancer for exercisers vs. sedentary women is .43, what is a possible value for the true OR?

- .41
- .45
- 1.0
- 2.0
- Cannot determine

- You are on the Monty Hall show. You are presented with 3 doors (A, B, C), only one of which has something valuable to you behind it. You do not know what is behind any of the doors. You choose door A; Monty Hall opens door B and shows you that there is nothing behind it. Then he gives you the option of sticking with A or switching to C. Do you stay or switch? Does it matter?

- http://query.nytimes.com/gst/fullpage.html?res=9D0CEFDD1E3FF932A15754C0A967958260&sec=&spon=&pagewanted=all
- http://www.nytimes.com/2008/04/08/science/08tier.html?_r=1&em&ex=1207972800&en=81bdecc33f60033e&ei=5087%0A&oref=slogin
- http://www.nytimes.com/2008/04/08/science/08monty.html#

- Problem set 2
- Read the assigned journal article, and fill out a review sheet.

- Who wants to lead the journal article discussion next week?