Question

In: Statistics and Probability

A certain virus infects 1 in every 600 people. A test used to detect the virus...

A certain virus infects 1 in every 600 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. Let A be the event that the 'person is infected' and B be the event that the 'person tests positive'

A: Find the probability that the person has the virus given that they tested positive

B: Find the probability that the person does not have the virus given that they tested negaitive.

Solutions

Expert Solution

P(A) = 1/600 = 0.0017

P(B) = 0.9 * 0.0017 + 0.1 * (1 - 0.0017) = 0.1014

A) P(has the virus | tested positive) = P(tested positive | has the virus) * P(has the virus)/P(tested positive)

                                                         = 0.9 * 0.0017/0.1014 = 0.0151

B) P(does not have the virus | tested negative) = P(tested negative | does not have the virus) * P(does not have the virus)/P(tested negative)

                                                                        = (1 - 0.1) * (1 - 0.0017)/(1 - 0.1014) = 0.9999

orchestra answered 2 years ago
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