Asked by Cathy Wilson
A certain virus infects one in every 500 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 10% of the time if the person does not have the virus.
Find the probability that a person does not have the virus given that they have tested negative.
Find the probability that a person does not have the virus given that they have tested negative.
Answers
Answered by
Joe
A=infected
~A=not infected
(+)=positive diagnosis
(-)= negative diagnosis
We seek P(~A|(-))
We are given:
P(A) = 1/500 = 0.002
P((+)|A) = 0.9
P((+)|~A) = 0.1
We infer P(~A) = 1 - P(A) = 499/500 = 0.998 and that P((-)|~A) = 1 - P((+)|~A) = 0.9
We now find that:
P(~A|(-)) = P(~A) × P((-)|~A) =
0.998 × 0.9 = 0.898
which is the desired result.
QED
~A=not infected
(+)=positive diagnosis
(-)= negative diagnosis
We seek P(~A|(-))
We are given:
P(A) = 1/500 = 0.002
P((+)|A) = 0.9
P((+)|~A) = 0.1
We infer P(~A) = 1 - P(A) = 499/500 = 0.998 and that P((-)|~A) = 1 - P((+)|~A) = 0.9
We now find that:
P(~A|(-)) = P(~A) × P((-)|~A) =
0.998 × 0.9 = 0.898
which is the desired result.
QED
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