A certain virus infects five in every 400 people. A test used to detect the viru

Question

A certain virus infects five in every 400 people. A test used to detect the virus ina person is positive 85% of the time when the person has the virus and 8% of the time when the person does not have the virus. Let A be the event "the person is infected" and B be the event "the person tests positive." (20 pts) Using Bayes' Theorem, when a person tests positive, determine the probability that the person is infected Using Bayes' Theorem, when a person tests negative, determine the probability that a person is not infected a. b.

Explanation / Answer

Here, we are given that:

P( infected ) = 5/400 = 0.0125

P( positive | infected ) = 0.85, therefore, we get here: P( negative | infected ) = 0.15
P( positive | not infected ) = 0.08, therefore, we get here: P( negative | not infected ) = 0.92

a) Using law of total probability, we get here:

P( positive ) = P( positive | infected )P( infected ) + P( positive | not infected ) P( not infected )

P( positive ) = 0.85*0.0125 + 0.08*(1 - 0.0125) = 0.089625

Given that a test is positive probability that the person is infected is computed using bayes theorem as:

P( infected | positive ) = P( positive | infected )P( infected ) / P( positive )
P( infected | positive ) = 0.85*0.0125 / 0.089625 = 0.1185

Therefore 0.1185 is the required probability here.

b) Here, we first compute:

P( negative ) = 1 - P( positive ) = 1 - 0.089625 = 0.910375

Using bayes theorem, we get:

P( not infected | negative ) = P( negative | not infected )*P(not infected ) / P( negative )

P( not infected | negative ) = 0.92*(1 - 0.0125) / 0.910375 = 0.9979

Therefore 0.9979 is the required probability here.

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