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ProbabilityProbability

Prof. Khaled MahmoudProf. Khaled MahmoudAssociate ProfessorAssociate Professor

Department of community, Environmetnal Department of community, Environmetnal and occupational medicine, and occupational medicine,

Faculty of Medicine, Ain Shams UniversityFaculty of Medicine, Ain Shams University

The probability of an event is a quantitative The probability of an event is a quantitative expression of the likelihood of its occurrenceexpression of the likelihood of its occurrence

Pr (A)= Pr (A)= number of times A does occurnumber of times A does occur

total no. of times A can occurtotal no. of times A can occur

Example: in food outbreak. 158 people Example: in food outbreak. 158 people attendedattended a banquet. The probability of a banquet. The probability of illness for a person selected at random isillness for a person selected at random is

Pr (illness)= Pr (illness)= 99 = 99 = 0.63 or 63%0.63 or 63%

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Probability can be expressed in fractions, Probability can be expressed in fractions, decimals, decimal fractionsm must fall in the decimals, decimal fractionsm must fall in the range of 0 to 1 so that range of 0 to 1 so that

Pr (illness)= Pr (illness)= 99 = 99 = 0.63 or 63%0.63 or 63%

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Probability can be expressed in fractions, decimals, Probability can be expressed in fractions, decimals, decimal fractions must fall in the range of 0 to 1 decimal fractions must fall in the range of 0 to 1 so thatso that

Pr (event A does not occur)=1-Pr (event A occurs)Pr (event A does not occur)=1-Pr (event A occurs)

Pr (illness does not occur) = 1- 0.63 = 0.37 or 37%Pr (illness does not occur) = 1- 0.63 = 0.37 or 37%

Conditional probabilityConditional probability الشرطى الشرطى األحتمال األحتمال

conditional probability measures the conditional probability measures the probability of an event given that (by probability of an event given that (by assumption, presumption, assertion or assumption, presumption, assertion or evidence) another event has occurredevidence) another event has occurred

In food poisoning example the probability of a In food poisoning example the probability of a given person becomes ill was 0.63given person becomes ill was 0.63

The probability of illness can be modified if The probability of illness can be modified if we knew the food the person ate.we knew the food the person ate.

Pr (A| B)= Pr (A| B)= number of times A and B occur jointlynumber of times A and B occur jointly

total no. of times B occurtotal no. of times B occur

Pr (Illness| turkey)= Pr (Illness| turkey)= number who ate turkey and became illnumber who ate turkey and became ill

number of people who ate turkeynumber of people who ate turkey

= = 97 = 97 = 0.73 or 73%0.73 or 73%

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So the probability of getting illness if the subject consumes So the probability of getting illness if the subject consumes turkey is 73%turkey is 73%

Complex eventsComplex events Events expressed as specific combination A Events expressed as specific combination A

and B, Expressed as specified alternatives and B, Expressed as specified alternatives A or BA or B

Pr (A and B)= probability that A and B occur Pr (A and B)= probability that A and B occur jointlyjointly

Pr (A or B)= probability that A occurs or B Pr (A or B)= probability that A occurs or B occurs= probability that at least one of the occurs= probability that at least one of the stated alternatives will occurstated alternatives will occur

Multiplication RuleMultiplication Rule Pr ( A and B) = Pr (A) X Pr (B)Pr ( A and B) = Pr (A) X Pr (B)

Example Example

Side effects of certain drug occurs in 10% of all Side effects of certain drug occurs in 10% of all patients who take it. A physician has 2 patients patients who take it. A physician has 2 patients who are taking the drug. What is the probability who are taking the drug. What is the probability that both will experience side effectsthat both will experience side effects

The events are independent, that is the occurrence The events are independent, that is the occurrence of side effects in one patient does not affect the of side effects in one patient does not affect the likelihood of side effects in the other patient.likelihood of side effects in the other patient.

Multiplication RuleMultiplication Rule Pr ( A and B) = Pr (A) X Pr (B)Pr ( A and B) = Pr (A) X Pr (B)

Pr of both patients experience side effectsPr of both patients experience side effects Pr (A and B)= 0.1 X0.1 =0.01 Pr (A and B)= 0.1 X0.1 =0.01

So the probability of both patients A and B So the probability of both patients A and B will experience side effects is 1%will experience side effects is 1%

Addition RuleAddition Rule Pr ( A or B) = Pr (A) + Pr (B)- Pr ( A and B)Pr ( A or B) = Pr (A) + Pr (B)- Pr ( A and B)

Example Example What is the probability that at least one of the What is the probability that at least one of the

patients experiences side effects.patients experiences side effects.

Pr (A or B)= 0.1 +0.1-0.01= 0.19 or 19%Pr (A or B)= 0.1 +0.1-0.01= 0.19 or 19%So the probability that at least one patient will So the probability that at least one patient will

experience side effects is 19%experience side effects is 19%

Bayes TheoremBayes Theorem Application of conditional probability based Application of conditional probability based

on additional information that is later on additional information that is later obtained. Dealing with sequential eventsobtained. Dealing with sequential events

Priori probabilityPriori probability: is an initial probability : is an initial probability originally obtained before any additional originally obtained before any additional information is obtainedinformation is obtained

Posterior probabilityPosterior probability: is the probability value : is the probability value that has been revised by using additional that has been revised by using additional information that is later obtainedinformation that is later obtained

Diagnosis a problemDiagnosis a problem

Application of certain clinical test to Application of certain clinical test to diagnose cancer cervix .diagnose cancer cervix .

The test is 90% sensitive The test is 90% sensitive The prevalence rate of the disease is 1% in The prevalence rate of the disease is 1% in

the same age group as the patientthe same age group as the patient The test has a false positive 10%The test has a false positive 10%

Cancer

NO Cancer

0.01

0.99

Cancer

NO Cancer

0.01

0.99

+ve

-ve

0.9

0.1

Cancer

NO Cancer

0.01

0.99

+ve

-ve

0.9

0.1

+ve

-ve

0.1

0.9

Cancer

NO Cancer

0.01

0.99

+ve

-ve

0.9

0.1

+ve

-ve

0.1

0.9

Probability of having test positive results



Pr (B)Pr (B)

Pr (cancer| +)= Pr (cancer| +)= (0.9) X (0.01) (0.9) X (0.01)

(0.01)(0.9) + (0.99)(0.1)(0.01)(0.9) + (0.99)(0.1)


Pr (cancer| +)= Pr (cancer| +)= (0.009) (0.009) (0.009) + (0.099)(0.009) + (0.099)

= 0.009/0.108 = 9/108 = 0.083 = 0.009/0.108 = 9/108 = 0.083 =8.3%=8.3%So the probability that the subject has a breast So the probability that the subject has a breast

cancer after doing the test and being positive is cancer after doing the test and being positive is only 8.3%only 8.3%

Diagnosis a problemDiagnosis a problem

Application of certain clinical test to diagnose Application of certain clinical test to diagnose breast cancer .breast cancer .

The test is 90% sensitive The test is 90% sensitive The prevalence rate of the disease is 1/5000 The prevalence rate of the disease is 1/5000

(0.0002)in the same age group as the patient(0.0002)in the same age group as the patient The test has a false positive 0.005 5 positive The test has a false positive 0.005 5 positive

results in every 1000 women who don’t have the results in every 1000 women who don’t have the disease.disease.

Breast Cancer

NO Cancer

0.0002

0.9998

+ve

-ve

0.9

0.1

+ve

-ve

0.005

0.995

Probability of having test positive results


Pr (B)Pr (B)

Pr (cancer| +)= Pr (cancer| +)= (0.9) X (0.0002) (0.9) X (0.0002)

0.9X0.0002 + (0.9998)(0.005)0.9X0.0002 + (0.9998)(0.005)

Pr (cancer|+)= 0.00018/0.005179= 0.0348Pr (cancer|+)= 0.00018/0.005179= 0.0348

=3.48%=3.48%

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