Answered by AI, Verified by Human Experts
Note that theprobabilitythat a person selected at random would test positive for this disease is0.06894.What is the rationale for the above answer?We are given that the detection of a certain disease gives apositive result5 percent of the time for people who do not have the disease.This means:P(positive results | do not have disease) = 0.05We also know that the test gives a negative result 0.3 percent of the time for people who have the disease. What this means is that the test gives a positive result 99.7 percent of the time for people who have the disease.Thus,P(positive results | have disease) = 0.997;Then, we know thatlarge-scale studies have shown that the disease occurs in about 2 percent of the population. That is P(have disease) = 0.02, P(do not have disease) = 0.98Thus, theprobabilitythat a person selected randomly would test positive is:P(positive results) = P( Positive results ∩ have disease) + P(Positive results ∩ do not have disease)= P(have disease) P (Positve results | have disease) + P(do not have disease) P(Positive results) | do not have disease)= (0.02 x 0.0997) + (0.98 x 0.05)= 0.06894Thus, it is right to state that the likelihood that a person selected at random would test positive for this disease is0.06894.Learn more aboutProbability:brainly.com/question/30034780#SPJ1...