Question #345876

Q: A certain virus infects one in every 20 people. A test used to detect the virus in a person delivers positive outcome at 85% accuracy for infected persons. Moreover, it provides negative outcome for healthy persons at 95% accuracy. Compute followings:

 

a)     Find the probability that a person has the virus given that his test outcome is positive.

b)     Find the probability that a person does not have the virus given that his test outcome is negative.

c)     Find the probability that a person has the virus given that his test outcome is negative.


1
Expert's answer
2022-06-01T11:18:00-0400

Let AA denote the event "infected person". Let VV denote the event "test outcome is positive".

Given P(A)=0.05,P(VA)=0.85,P(VCAC)=0.95.P(A)=0.05, P(V|A)=0.85, P(V^C|A^C)=0.95.

a)


P(AV)=P(VA)P(A)P(VA)P(A)+P(VAC)P(AC)P(A|V)=\dfrac{P(V|A)P(A)}{P(V|A)P(A)+P(V|A^C)P(A^C)}

=0.85(0.05)0.85(0.05)+(10.95)(10.05)=0.4722=\dfrac{0.85(0.05)}{0.85(0.05)+(1-0.95)(1-0.05)}=0.4722

b)


P(ACVC)=P(VCAC)P(AC)P(VCAC)P(AC)+P(VCA)P(A)P(A^C|V^C)=\dfrac{P(V^C|A^C)P(A^C)}{P(V^C|A^C)P(A^C)+P(V^C|A)P(A)}

=0.95(10.05)0.95(10.05)+(10.85)(0.05)=0.9918=\dfrac{0.95(1-0.05)}{0.95(1-0.05)+(1-0.85)(0.05)}=0.9918

c)


P(AVC)=P(VCA)P(A)P(VCA)P(A)+P(VCAC)P(AC)P(A|V^C)=\dfrac{P(V^C|A)P(A)}{P(V^C|A)P(A)+P(V^C|A^C)P(A^C)}

=0.05(10.85)0.05(10.85)+0.95(10.05)=0.0082=\dfrac{0.05(1-0.85)}{0.05(1-0.85)+0.95(1-0.05)}=0.0082


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