1. 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”  indep.
“male” and “taller than 6 ft”  dep.
“male” and “pregnant”  dep.
This is completely different from the idea of events that are disjoint.

2. 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 = 1/2. If 1st frog is male, P(2nd frog is male) = 9/19.  Here, successive picks are dependent.
Survey of a whole county with thousands of frogs (half males, half females).
P(1st frog is male) = 1/2
If 1st frog is male, P(2nd frog is male) = 499/999 ≈ 1/2.  Here, successive picks are “nearly” indep.

3. 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.

4. 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=5/11≈.46
P(HI | B) ≠ P(HI ), therefore HI and B are dependent.

5. 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 .34
A) 0.03 B) C) 0.34 D) E) 0.97
Are “Smoker” and “Lung cancer” independent? No because cond. prob. above are diff.
A) Yes B) No
Another approach: P(L)=.06 but P(L|S)=.34. therefore L and S are dependent.

6. 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).

7. 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)

8. 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) =(10/20)*(9/19) = .237
Blood donation center
Unrelated visitors are independent.
Probability that the next two unrelated visitors are both type O:
P(O and O) = .45*.45 = .2025

9. 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) = .15
P(trunk) = .41
P(limbs) = .44
% in each group who are women:
P(woman | head) = .56
P(woman | trunk) = .37
P(woman | limbs) = .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

10. Skin cancer example) P(H) = .15; P(H|M) = ??
Bayes’s Theorem:
𝑃 𝐴 𝑗 𝐵 = 𝑃 𝐵∩ 𝐴 𝑗 𝑃(𝐵) = 𝑃( 𝐴 𝑗 )𝑃(𝐵| 𝐴 𝑗 ) 𝑖=1 𝑘 𝑃( 𝐴 𝑖 )𝑃(𝐵| 𝐴 𝑖 )
where A1, A2, … , Ak nonzero mutually exclusive & exhaustive events, B any other event w/ P(B)≠0, P(B)≠1 but P(B) not given directly. P(Aj) prior probability; 𝑃( 𝐴 𝑗 |𝐵) posterior probability.
Skin cancer example) P(H) = .15; P(H|M) = ??
Q) Mutually exclusive and exhaustive events that include H?
H, T, L
Q) Role of M is Event B. How to compute P(M) = P(M and H) + P(M and T) + P(M and L) = P(H)*P(M|H) + P(T)*P(M|T) + P(L)*P(M|L)
P(M) = .15* * *.2 = .4123
P(H|M) = P(H and M)/P(M) = [P(H)*P(M|H)]/P(M) = [.15*.44]/.4123 = .16
≠ P(M|H) = .44

11. Example) The proportion of people in a given community who have a certain disease is A test is available to diagnose the disease. When a person has the disease, the probability that the test will produce a positive signal is .99. When a person does not have the disease, the probability that the test will produce a positive signal is .01. When a person tests positive, what is the probability that the person actually has the disease?
Answer) Let D represent the event that the person actually has the disease, and let + represent the event that the test gives a positive signal. We wish to find P(D|+). We are given the following probabilities:
𝑃 𝐷 =.005, 𝑃 + 𝐷 = ??, 𝑃 + 𝐷 𝑐 =??
Using the Bayes’ rule,
P 𝐷 + 𝐴= 𝑃 + 𝐷 𝑃(𝐷) 𝑃 + 𝐷 𝑃 𝐷 +𝑃 + 𝐷 𝑐 𝑃( 𝐷 𝑐 ) = ?? (??) ?? (??)+ ?? (??) =.332

12. Example) An automobile dealer has kept records on the customers who visited his showroom. Forty percent of the people who visited his dealership were women. Furthermore, his records show that 37% of the women who visited his dealership purchased an automobile, while 21% of the men who visited his dealership purchased an automobile. Suppose a customer visited the showroom and purchased a car. What is the probability that the customer was a woman?
Answer) Define the events: A = customer is a woman, Ac = customer is a man,
B = customer purchased.
“40% of the people who visited were women.” => 𝑃 𝐴 =??, 𝑃 𝐴 𝑐 =??
“37% of the women who visited purchased.” =>𝑃 𝐵 𝐴 =??
“21% of the men who visited purchased.” =>𝑃 𝐵 𝐴 𝑐 =??
 We need to compute 𝑃 𝐴 𝐵 .
𝑃 𝐴 𝐵 = 𝑃 𝐵 𝐴 𝑃 𝐴 𝑃 𝐵 𝐴 𝑃 𝐴 +𝑃 𝐵 𝐴 𝑐 𝑃 𝐴 𝑐 = (??)(??) (??)(??)+(??)(??) = ?? ??+?? ≈ .54

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 prevalence
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*.0060 =
PPV = P(disease | positive) = P(disease and positive) / P(positive)
= P(true positive) / P(true or false positive)
= /( ) = .627

16. What is the rate of prostate cancer among mature men
What is the rate of prostate cancer among mature men? P(cancer)= (3+1)/100
=P(high PSA and cancer)
+P(normal PSA and cancer) = (10/100)*(3/10) + (90/100)*(1/90) = 4/100.
How “sensitive” is the PSA test? P(+ | cancer)=
=P(+ and Cancer)/P(Cancer) = (.1*.3)/.4 = 3/4
How “specific” is the PSA test? P(– | no cancer)=
P(- and no cancer)/P(no cancer)= .90*(89/90)/.96
= 89/96
If a man gets an elevated PSA test (+), what is the probability that he has prostate cancer? PPV = P(cancer | +)= .3