1. 10. General rules of probability
The Practice of Statistics in the Life Sciences
Third Edition
© 2014 W.H. Freeman and Company

2. Objectives (PSLS Chapter 10)
General rules of probability
Independent events
Conditional probability
Multiplication rule
Tree diagrams
Diagnosis tests

3. Independent events
Two events are independent if knowing that one event is true or has happened does not change the probability of the other event.
“male” and “get head when flipping a coin”  independent
“male” and “taller than 6 ft”  not independent
“male” and “high cholesterol”  it’s not obvious (I’m guessing independent)
“male” and “pregnant”  not independent
This is completely different from the idea of events that are disjoint.

4. Sampling without replacement
Pick one frog at random from your target population and don’t put it back. Then pick another frog, etc.
Artificial pond with 10 male and 10 female frogs. P(1st frog is male) = 10/20 = If 1st frog is male, P(2nd frog is male) = 9/19 = 0.47 (≠ 0.5).  Here, successive picks are not independent.
Survey of a whole county with thousands of frogs (half males, half females).
P(1st frog is male) = 0.5.
If 1st frog is male, P(2nd frog is male) ≈  Here, successive picks are “nearly” independent.

5. Conditional probability
Conditional probabilities reflect how the probability of an event can be different if we know that some other event has occurred or is true.
The conditional probability of event B, given event A is: (provided that P(A) ≠ 0)
When two events A and B are independent, P(B | A) = P(B).
No information is gained from the knowledge of event A.

6. Probabilities of hearing impairment and blue eyes among Dalmatian dogs.
HI = Dalmatian is hearing impaired
B = Dalmatian is blue eyed
Neither HI nor B
0.66
B and not HI
0.06
HI and B
0.05
HI and not B
0.23
P(HI and B) = .05 P(HI) = .28 P(B) = .11
P(HI | B) = P(HI and B) / P(B) = .05/.11≈ .45
P(HI | B) ≠ P(HI ), therefore HI and B are not independent.

7. Large-scale surveys indicate that 11% of the population smokes
Large-scale surveys indicate that 11% of the population smokes. Medical researchers know that the probability that a smoker will get lung cancer is The probability that a person will get lung cancer if the person doesn’t smoke is The probability P(Lung cancer | Smoker) is
A) 0.03 B) C) 0.34 D) E) 0.97
11% = P(Smoker)
0.34 = P(Lung cancer | Smoker)
0.03 = P(Lung cancer | Nonsmoker)
Notice how the second and third sentences are quite different grammatically but both describe a conditional probability: a probability not for the entire population but for a subgroup.
Smoker and Lung cancer are NOT independent, because P(Lung cancer | Smoker) ≠ P(Lung cancer | Nonsmoker).
Are “Smoker” and “Lung cancer” independent?
A) Yes B) No

8. Non-smoker and no Lung cancer
Large-scale surveys indicate that 11% of the population smokes. Medical researchers know that the probability that a smoker will get lung cancer is The probability that a person will get lung cancer if the person doesn’t smoke is 0.03.
Lung cancer
Non-smoker and no Lung cancer
Smoker
0.11
0.34 = P(Lung cancer | Smoker) is the fraction of smokers (the orange circle) that will get lung cancer (the blue dots that are within the orange circle).
This is why by definition P(Lung cancer | Smoker) = P(Lung cancer and Smoker)/P(Smoker).

9. Multiplication rule General multiplication rule:
The probability that any two events, A and B, both occur is:
P(A and B) = P(A)P(B|A)
Multiplication rule for independent events:
If A and B are independent, then:
P(A and B) = P(A)P(B)

10. Artificial pond with 10 male and 10 female frogs
Successive captures are not independent.
Probability of randomly capturing 2 male frogs in a row:
P(male and male) = P(1st is male) P(2nd is male | 1st is male)
= (10/20)(9/19) ≈ 0.237
Blood donation center
Unrelated visitors are independent.
Probability that the next two unrelated visitors are both type O:
P(O and O) = P(O) * P(O)
= (.45)(.45) ≈ .2025

11. Tree diagrams
Tree diagrams are used to represent probabilities graphically and facilitate computations.
Probabilities of skin cancer among men and women by body locations.
Man
Woman
In a random individual with skin cancer:
P(head) = 0.15
P(trunk) = 0.41
P(limbs) = 0.44
% in each group who are women:
P(woman | head) = 0.56
P(woman | trunk) = 0.37
P(woman | limbs) = 0.80
0.44
0.56
Head
Trunk
Limbs
0.15
0.44
0.63
0.37
Man
Woman
0.41
0.20
0.80
Man
Woman

12. Diagnostic tests and screening
Unknown medical status
Known test result
Test sensitivity
P(+|disease)
Disease
No disease
Individual
Positive
Negative
True positive
False negative
Rate of disease in target population
P(disease)
False positive
True negative
Test specificity
P(–|no disease)
If a person gets a positive test result, what is the probability that he/she actually has the disease?
This is the positive predictive value:
PPV = P(disease | positive test)

13. Diagnosis tests Diagnosis sensitivity Disease prevalence
No disease
Individual
Positive
Negative
True positive
False negative
False positive
True negative
If a person gets a positive test result, what is the probability that he/she actually has the disease?
This is the positive predictive value:
PPV = P(disease | positive test)
Diagnosis specificity

14. HIV-AIDS: Suppose that about 1% of a large population has HIV-AIDS antibodies.
The enzyme immuno-assay test has sensitivity and specificity
Diagnosis sensitivity
Disease incidence
Positive
True positive
.9985
HIV antibodies
.01
.0015
Negative
False negative
Random adult taking the test
Positive
False positive
.99
.006
No HIV antibodies
Incidence of HIV-AIDS in the general population
.994
Negative
True negative
Diagnosis specificity
HIV-test performance
If a person who takes this test gets a positive test result, what is the probability that he or she actually has HIV-AIDS, P(HIV-AIDS | positive test)?

15. PPV = P(disease | positive)
Diagnosis sensitivity
What’s the positive predictive value of the enzyme immuno-assay test for HIV-AIDS?
PPV = P(disease | positive)
Disease incidence
.9985
Positive
HIV antibodies
.01
.0015
Negative
False negative
Random adult taking the test
.0060
Positive
False positive
.99
No HIV antibodies
Incidence of HIV-AIDS in general population
.9940
Negative
Diagnosis specificity
HIV-test performance
P(true positive) = P(disease and positive) = P(disease)P(positive | disease)
= (.01)(.9985) =
P(false positive) = P(no disease and positive) = P(no disease)P(positive | no disease)
= (.99)(.006) =
This means that a random adult who gets a positive test result using this method actually has a ~63% probability of having HIV/AIDS.
PPV = P(disease | positive) = P(disease and positive) / P(positive)
= P(true positive) / P(all positives)
= / ( ) ≈ 62.7%

16. What is the rate of prostate cancer among mature men? P(cancer)= ?
How “sensitive” is the PSA test? P(+ | cancer)= ?
How “specific” is the PSA test? P(– | no cancer)= ?
If a man gets an elevated PSA test (+), what is the probability that he has prostate cancer? PPV = P(cancer | +)= ?
P(cancer) = (3+1)/100 = 0.04
P(+ | cancer) = 3/4 = 0.75
P(– | no cancer) = 89/96 = 0.927
PPV = P(cancer | +) = 3/10 = 0.30