Answer to Question #345876 in Statistics and Probability for Toilet Gamer

Question

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.

Answer 1

Let $A$ denote the event "infected person". Let $V$ denote the event "test outcome is positive".

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

a)

$$P(A|V)=\dfrac{P(V|A)P(A)}{P(V|A)P(A)+P(V|A^C)P(A^C)}$$

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

b)

$$P(A^C|V^C)=\dfrac{P(V^C|A^C)P(A^C)}{P(V^C|A^C)P(A^C)+P(V^C|A)P(A)}$$

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

c)

$$P(A|V^C)=\dfrac{P(V^C|A)P(A)}{P(V^C|A)P(A)+P(V^C|A^C)P(A^C)}$$

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