Suppose that the reliability of an HIV test i | vedclass

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(A) Let $E$ denote the event that the person selected actually has $HIV$ and $A$ be the event that the person's $HIV$ test is diagnosed as positive. W

Vedclass · Accepted Answer

$0.083$

Vedclass · Answer

(A) Let $E$ denote the event that the person selected actually has $HIV$ and $A$ be the event that the person's $HIV$ test is diagnosed as positive. We need to find $P(E|A)$. Also,$E^{\prime}$ denotes the event that the person selected does not have $HIV$. Clearly,$\{E, E^{\prime}\}$ is a partition of the sample space. We are given: $P(E) = 0.1\% = \frac{0.1}{100} = 0.001$ $P(E^{\prime}) = 1 - P(E) = 0.999$ $P(A|E) = P(	ext{Person tested as } HIV 	ext{ positive given that he/she has } HIV) = 90\% = 0.9$ $P(A|E^{\prime}) = P(	ext{Person tested as } HIV 	ext{ positive given that he/she does not have } HIV) = 1\% = 0.01$ By Bayes' theorem: $P(E|A) = \frac{P(E) P(A|E)}{P(E) P(A|E) + P(E^{\prime}) P(A|E^{\prime})}$ $P(E|A) = \frac{0.001 	imes 0.9}{(0.001 	imes 0.9) + (0.999 	imes 0.01)}$ $P(E|A) = \frac{0.0009}{0.0009 + 0.00999} = \frac{0.0009}{0.01089} = \frac{90}{1089} \approx 0.083$ Thus,the probability that the person actually has $HIV$ is approximately $0.083$.

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