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.


Expert's answer

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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