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

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

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.

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 = 0.5. 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) ≈ 0.5.  Here, successive picks are “nearly” independent.

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.

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.

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 0.34. The probability that a person will get lung cancer if the person doesn’t smoke is 0.03. The probability P(Lung cancer | Smoker) is A) 0.03 B) 0.0394 C) 0.34 D) 0.583 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

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 0.34. 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).

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)

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

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

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)

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

HIV-AIDS: Suppose that about 1% of a large population has HIV-AIDS antibodies. The enzyme immuno-assay test has sensitivity .9985 and specificity .9940. 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)?

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) = .09985 P(false positive) = P(no disease and positive) = P(no disease)P(positive | no disease) = (.99)(.006) = .00594 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) = .09985 / (.09985 + .00594) ≈ 62.7%

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