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6Probability
Chapter6 p126
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
Chapter6 p126
A∩B both A and B
A B ∪ either A or B
Ac not A, complement of A
Chapter6 p128
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
Chapter6 p128
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.
Chapter6 p128
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
Chapter6 p128
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) = ?
Chapter6 p128
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
Chapter6 p128
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
Chapter6 p128
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).
Chapter6 p138
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.
Chapter6 p138
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
Chapter6 p126
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.
Chapter6 p142
Chapter6 p126
Chapter6 p149
Chapter6 p126