diseaseTest = function (hasPos = .8, notHasPos = .1, actuallncidence = .01) {#P(A) probability of having disease (actuallncidence) {#P(B) probability of testing positive numerator = hasPos* actuallncidence # P(BIA)*P(A) denominator = actuallncidence * hasPos + (1-actuallncidence) * notHasPos numerator/denominator } What does the result of the function tell us? O P(A|B). O P(BIA). O P(A). O P(B).
diseaseTest = function (hasPos = .8, notHasPos = .1, actuallncidence = .01) {#P(A) probability of having disease (actuallncidence) {#P(B) probability of testing positive numerator = hasPos* actuallncidence # P(BIA)*P(A) denominator = actuallncidence * hasPos + (1-actuallncidence) * notHasPos numerator/denominator } What does the result of the function tell us? O P(A|B). O P(BIA). O P(A). O P(B).
Essentials of Business Analytics (MindTap Course List)
2nd Edition
ISBN:9781305627734
Author:Jeffrey D. Camm, James J. Cochran, Michael J. Fry, Jeffrey W. Ohlmann, David R. Anderson
Publisher:Jeffrey D. Camm, James J. Cochran, Michael J. Fry, Jeffrey W. Ohlmann, David R. Anderson
Chapter15: Decision Analysis
Section: Chapter Questions
Problem 16P: Suppose that you are given a decision situation with three possible states of nature: s1, s2, and...
Related questions
Question
a medical test has some probability of being positive if the patient has the disease (hasPos) and another probability of testing positive if the person does not have the disease (notHasPos). a random member of the entire population has a real problem of having the disease (actual incidence). Based on the attached information what does the result of the function?
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