6Probability

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6.1 Operations on events and probability

An event is the basic element to which probability can be applied.

NotationsEvent: A, BA∩B both A and BA B ∪ either A or B

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A∩B both A and B

A B ∪ either A or B

Ac not A, complement of A

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Definition of probability-Many definitions, text used the frequentist definition

Probability of AP(A) = m / nm and n denote the frequency of occurrence of A and the total number of repeated experiments

Example – Table 5.1P(a child survives his or her first year) = 99149 / 100000 = 0.99149

P(A ∪Ac) = 1

P(A ∩ Ac) = 0

P(Ac) = (n – m) / n The probability that a newborn does not survive the first year of life is1- 0.99149 = 0.00851

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Mutually exclusively or disjoint eventA ∩ B = ØP( A ∩ B) = 0

When two events are mutually exclusively, the additive rule of probability applied, P( A ∪ B) = P(A) + P(B) Examplethe probability a newborn’s birth weight is under 2000 grams is 0.025, and the probability that it is between 2000 and 2499 grams is 0.043 P( baby weight is under 2500 grams) = 0.025 + 0.043 = 0.068

n mutually exclusively events, A1∩A2 = Ø, ……. An-1∩An = ØP(A1 A∪ 2 …… A∪ n ) = P(A1) + P(A2) + ….. P(An)

When two events are NOT mutually exclusively, P( A ∪ B) = P(A) + P(B) – P(A ∩ B)

Figure 6.2 Venn diagram representing two mutually exclusively events.

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6.2 Conditional probability The probability that an event B will occur given that we know the outcome of event A, does the prior occurrence of a cause the probability of B to change ?Example Find the probability that a person will live to the age of 65 given that the person has reached the age of 60 conditional probability, P(B|A) represent the probability of the event B given that the probability that event A has occurred.Multiplicative rule of probability The probability that two events A and B will both occur = probability of A * probability of B given that A has already occurred P( A ∩ B) = P(A) P(B|A) since is arbitrary which event we call A which we call B, we can also writeP( A ∩ B) = P(B) P(A|B) ExampleEvent A = a person is alive at age 60, Event B = a person survives to 65A ∩ B = the event that the person is alive at age 60 and also at 65. What are P(A), P(A ∩ B) and P(B|A) ? P(A) = 85993 / 100000 = 0.85993P(A ∩ B) = 80145 / 100000 = 0.80145P(B|A) = 0.80145 / 0.85993 = 0.9320If a person lives to be 60, this person chance of surviving to age 65 is greater than it was at birth, P(B), i.e. 0.9320 > 0.80145

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6.2 Conditional probability

When we are concerned with two events such that the outcome of one event has no effect on the occurrence or nonoccurrence of the other, the events are said to be independent.

If A and B are independent event, P(A|B) = P(A)and P(B|A) = P(B)P(A ∩ B) = P(A) P(B)

Independent and mutually exclusive DO NOT mean the same thingIf A and B are mutually exclusive, and event A occurs, event B cannot occur. By definition, P(B|A) = 0

6.3 Bayes’ Theorem

6.3 Bayes’ Theorem

3 Defects7 Good

Given 10 films, 3 of them are defected. What is the probability two successive films are defective?

6.3 Bayes’ Theorem

Loyalty of managers to their employer.

6.3 Bayes’ Theorem

Probability of new employee loyalty

6.3 Bayes’ Theorem

Probability (over 10 year and loyal) = ?

Probability (less than 1 year or loyal) = ?

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6.3 Bayes’ Theorem

Let E1, E2 and E3

= a person is currently employed, unemployed, and not in the labor force respectivelyP(E1) = 98917 / 163157 = 0.6063P(E2) = 7462 / / 163157 = 0.0457P(E3) = 56778 / 163157 = 0.3480Let H = a person has a hearing impairment due to injury, what are P(H), P(H|E1), P(H|E2) and P(H|E3) ?

P(H) = 947 / 163157 = 0.0058P(H|E1) = 552 / 98917 = 0.0056P(H|E2) = 27 / 7462 = 0.0036P(H|E3) = 368 / 56778 = 0.0065

Employment status Population Impairments

Currently employed 98917 552

Currently unemployed 7462 27

Not in the labor force 56778 368

Total 163157 947

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6.3 Bayes’ Theorem

H = a person has a hearing impairment due to injury,

What is P(H)?May be expressed as the union of three mutually exclusively events, i.e. E1∩H, E2∩H, and E3∩ HH = (E1∩H)∪(E2∩H)∪(E3∩ H) Apply the additive ruleP(H) = P(E1∩H) + P(E2∩H) + P(E3∩ H) Apply the Bayer’ theoremP(H) = P(E1) P(H|E1) + P(E2) P(H|E2) + P(E3) P(H|E3)

Event P(Ei) P(H | Ei) P(Ei) P(H | Ei)

E1 0.6063 0.0056 0.0034

E2 0.0457 0.0036 0.0002

E3 0.3480 0.0065 0.0023

P(H) 0.0059

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6.3 Bayes’ Theorem

The more complicate methodP(H) = P(E1) P(H|E1) + P(E2) P(H|E2) + P(E3) P(H|E3) ………………. (1)is useful when we are unable to calculate P(H) directly.

How about we want to compute P(E1|H) ?The probability that a person is currently employed given that he or she has a hearing impairment.The multiplicative rule of probability states thatP(E1∩H) = P(H) P(E1 | H) P(E1 | H) = P(E1∩ H) / P(H)

Apply the multiplicative rule to numerator, we haveP(E1 | H) = P(E1) P(H | E1) / P(H) ……………………………………..(2)Substitute (1) into (2), we have the expression for Bayes’ Theorem

)E|P(H )P(E )E|P(H )P(E )E|P(H )P(E

)E | P(H )P(E H)|E P(

332211

111

Chapter6 p128

6.3 Bayes’ Theorem

)E|P(H )P(E )E|P(H )P(E )E|P(H )P(E

)E | P(H )P(E H)|E P(

332211

111

Event P(Ei) P(H | Ei) P(Ei) P(H | Ei)

E1 0.6063 0.0056 0.0034

E2 0.0457 0.0036 0.0002

E3 0.3480 0.0065 0.0023

P(H) 0.0059

= (0.6063)(0.0056) / [(0.6063)(0.0056)+(0.0457)(0.0036)+(0.3480)(0.0065)]= 0.583

P(E1 | H) = 552 /947= 0.583

Employment status Population Impairments

Currently employed 98917 552

Currently unemployed 7462 27

Not in the labor force 56778 368

Total 163157 947

6.3 Bayes’ Theorem Exercise

A box contains 10 balls, 3 black in color and 7 white in color.. Balls are drawn from the box without return.

(a) Calculate the probability that the second ball is black in color.

(b) Given that the second ball is black in color, determine the probability that the first is also black in color, that is compute P (first black | second black).

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6.4 Diagnostic test

Screening is the application of a test to individuals who have not yet exhibited any clinical symptoms in order to classify them with respect to their probability of having a particular disease.

Test positive likely to have the disease further diagnostic

Bayes’ theorem allows us to use probability to evaluate the associated uncertainties

Probability of false negative P(test negative | cancer)Probability of true positive P(test positive | cancer)Probability of false positive P(test positive | no cancer)

6.4.1 Sensitivity and Specificity

6.4.1 Sensitivity and Specificity

Correlation coefficient (CC)

Sp

))()()((

)()(

NFTNFPTPFPTNFNTP

FPFNTNTPCC

Correlation coefficient (CC) ranges from -1 to 1, where a value of 1 means prefect prediction, a value of -1 indicates zero correction prediction.

6.4.2 Applications of Bayes’ Theorem

Notation : D1 is the event that an individual has a particular diseaseD2 is the event that an individual does not has a particular diseaseT+ denotes a positive screening test resultT- denotes a negative screening test resultExample

Cervical cancer, Pap smear (子宮頸抹片) screening testProbability of false negative P(test negative | cancer) = P(T- | D1)= 0.1625Probability of true positive P(test positive | cancer) P(T+ | D1)= 1-0.1625 = 0.8375Probability of false positive P(test positive | no cancer) = P(T+ | D2 )= 0.1864P(D1) = 0.000083, rate of cervical cancer in 1983-1984 was 8.3 per 100,000 prevalence of the diseaseP(D2) = 1 - P(D1)

)|()()|()(

)|()(

)(

)()|(

2211

1111 DTPDPDTPDP

DTPDP

TP

TDPTDP

000373.0%)64.18%9917.99(%)75.83%0083.0(

%75.83%0083.0

P(D1|T+) is called the predictive value of a positive test

For every 1,000,000 women with positive Pap smear, only 373 represent true cases of cervical cancer.

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6.4.2 Applications of Bayes’ Theorem

Calculate the predictive value of a negative test.

)|()()|()(

)|()(

)(

)()|(

1122

2222 DTPDPDTPDP

DTPDP

TP

TDPTDP

999983.0%)25.16%0083.0(%)36.81%9917.99(

%36.81%9917.99

For every 1,000,000 women with negative Pap smear, 999,983 do not have cervical cancer

Figure 6.3 illustrates the results of the entire diagnostic testing process. All numbers have been rounded to the nearest integer.

6.4.3 Receiver Operation Characteristic (ROC) curve - Sensitivity and Specificity

6.4.3 Receiver Operation Characteristic (ROC) curve - Sensitivity and Specificity

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6.4.3 ROC curve

2.9 mg% as an indicator of imminent rejection, the test has a sensitivity of 0.303 and a specificity of 0.909.To increase the sensitivity, we could lower the arbitrary cutoff poing that distinguishes a positive test result from a negative one, if we use 1.2 mg%, for example, a much greater proportion of the results will reject the organ. At the same time we would increase the probability of a FP result, thereby decreasing the specificity.

6.5 The relative risk and the odds ratio

Relative Risk (RR) – want to compare the probabilities of disease in two different groups.

)exp|(

)exp|(

osedundiseaseP

oseddiseasePRR

Example – breast cancer studyExposed – a woman first gave birth at age 25 of older Sample – 4540 (1628) women who gave birth to their first child before the age of 25 (or older), 65 (31) developed breast cancer

33.14540/65

1628/31RR

Women who first gave birth at a later age are 33% more likely to develop breast cancer. In chapter 15, we will explain how to determine whether this is an important difference.

6.5 The relative risk and the odds ratio

Odds Ratio (機會 ,勝算 ), or relative odds (OR) – measure of the relative probabilities of disease. If an event takes place with probability p, the odds in favor of the event are p/(1– p) to 1. If p = ½, the odds are (1/2)/(1/2) = 1 the event is equally likely either to occur or not to occur. For every 100,000 individuals there are 9/3 cases of tuberculosis, the odds of randomly selected person’s having the disease are

1__00009301.0000,100/7.99990

000,100/3.9

1to

p

p

OR is defined as the odds of disease among exposed individuals divided by the odds of disease among the unexposed, or

)]|(exp1/[)|(exp

)]|(exp1/[)|(exp

__

)]exp|(1/[)exp|(

)]exp|(1/[)exp|(

dnondiseaseosedPdnondiseaseosedP

diseasedosedPdiseasedosedPOR

definitionequivalentanother

osedundiseasePosedundiseaseP

oseddiseasePoseddiseasePOR

6.5 The relative risk and the odds ratio

Risk factors for breast cancer - use of oral contraceptives A case-control study – examine the effects of the use of oral contraceptives. Determine whether the exposure in question was present or absent for each individual 989 women who has breast cancer, 273 had used oral contraceptives and 716 had not9901 women who did not have breast cancer, 2641 had used oral contraceptives and 7260 had not Subjects with and without the disease are chosen, therefore, the probability of disease in the exposed and unexposed group cannot be determined. We can determine the probability of exposure for both cases and controls

05.17260/2641

716/273

)9901/26411/(9901/2641

)989/2731/()989/273(

)]|(exp1/[)|(exp

)]|(exp1/[)|(exp

dnondiseaseosedPdnondiseaseosedP

diseasedosedPdiseasedosedPOR

Women who have used oral contraceptives have an odds of developing breast cancer that is only 1.05 times the odds of nonusers. Chapter 15 will interpret this result.

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