2.5% of a population are infected with a certain disease. There is a test for the disease, however the test is not completely accurate. 90.4% of those who have the disease test positive. However 4.1% of those who do not have the disease also test positive (false positives). A person is randomly selected and tested for the disease. What is the probability that the person has the disease given that the test result is positive?
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Question:
2.5% of a population are infected with a certain disease. There is a test for the disease, however the test is not completely accurate. 90.4% of those who have the disease test positive. However 4.1% of those who do not have the disease also test positive (false positives). A person is randomly selected and tested for the disease. What is the probability that the person has the disease given that the test result is positive?
Answers
Answer:[tex]P(disease/positivetest) = 0.36116[/tex]Step-by-step explanation:This is a conditional probability exercise.Let's name the events :I : ''A person is infected''NI : ''A person is not infected''PT : ''The test is positive''NT : ''The test is negative''The conditional probability equation is : Given two events A and B : P(A/B) = P(A ∩ B) / P(B)[tex]P(B) >0[/tex]P(A/B) is the probability of the event A given that the event B happenedP(A ∩ B) is the probability of the event (A ∩ B)(A ∩ B) is the event where A and B happened at the same timeIn the exercise : [tex]P(I)=0.025[/tex][tex]P(NI)= 1-P(I)=1-0.025=0.975\\P(NI)=0.975[/tex][tex]P(PT/I)=0.904\\P(PT/NI)=0.041[/tex]We are looking for P(I/PT) :P(I/PT)=P(I∩ PT)/ P(PT)[tex]P(PT/I)=0.904[/tex]P(PT/I)=P(PT∩ I)/P(I)0.904=P(PT∩ I)/0.025P(PT∩ I)=0.904 x 0.025P(PT∩ I) = 0.0226P(PT/NI)=0.041P(PT/NI)=P(PT∩ NI)/P(NI)0.041=P(PT∩ NI)/0.975P(PT∩ NI) = 0.041 x 0.975P(PT∩ NI) = 0.039975P(PT) = P(PT∩ I)+P(PT∩ NI)P(PT)= 0.0226 + 0.039975P(PT) = 0.062575P(I/PT) = P(PT∩I)/P(PT)[tex]P(I/PT)=\frac{0.0226}{0.062575} \\P(I/PT)=0.36116[/tex]
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