A diagnostic test for the presence of a virus yields a positive results 55% of t

Question

A diagnostic test for the presence of a virus yields a positive results 55% of the time. It is known that the probability that a person in the general population of the town carries thevirus is very close to 0.50. We also know that the probability that a person carries the virusgiven that she/he has tested positive is 0.8.
-a- Find the probability that a randomly selected person tests positive and actually carries the virus.

-b- Findtheprobabilitythatarandomlyselectedpersontestsnegativeanddoesnotactuallycarry the virus.
-c- Find the probability that a person does not carry the virus given that she/he has testedpositive.

Explanation / Answer

here let probabilty of positive result =P(P)=0.55

probabilty of negative result =P(N) =1-P(P) =0.45

and probabilty of carrying virus =P(CV) =0.5=probabilty of not carrying virus P(NCV)

probability that a person carries the virusgiven that she/he has tested positive =P(CV|P) =0.8

probability that a person not carries the virusgiven that she/he has tested positive =P(NCV|P) =1-P(CV|P) =0.2

a) probability that a randomly selected person tests positive and actually carries the virus =P(P)*P(CV|P) =0.55*0.8 =0.44

b)as probabilty of not carrying virus =P(NCV) =0.5 =P(P)*P(NCV|P)+P(N)*P(NCV|N)

0.5 =0.5*0.2+P(N)*P(NCV|N)

P(N)*P(NCV|N) =0.5-0.1 =0.4

hence probabilitythatarandomlyselectedpersontestsnegativeanddoesnotactuallycarry the virus =0.4

c) probability that a person does not carry the virus given that she/he has testedpositive= P(NCV|P) =0.2

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