Answer to Question #127600 in Statistics and Probability for Harsha

Question

A certain “Covid 19” disease occurs in 1% of the population. A simple screening procedure is
Available and in 8 out of 10 cases where the patient has the disease, it produces a positive result. If
The patient does not have the disease there is still a 0.05 chance that the test will give a positive result.
Find the probability that a randomly selected individual:
(a) Does not have the “Covid 19” disease but gives a positive result in the screening test
(b) Gives a positive result on the test
(c) Ton has taken the test and her result is positive. Find the probability that he has the
“Covid 19” disease.
Let C represent the event “the patient has the disease” and S represent the event “the screening test gives a
positive result”. ”.

Answer 1

a) P(C'⋂ S)=0.99*0.05=0.0495

b) P(S)=P(C ⋂ S)+P(C' ⋂ S)=0.01*0.8+ 0.99*0.05=0.0575

c)P(C|S)=P(C ⋂ S)/P(S)=0.01*0.8/0.0575=0.139