A certain disease has a prevalence of 1 % in the population. A diagnostic test for the disease has a sensitivity of 98 % and a specificity of 95 % . If a person from this population tests positive, then the probability that they actually have the disease is...

Step-by-step solution

Let D be the event that the person has the disease, and T be the event that the test is positive. We have P(D) = 0.01 and P(D^c) = 1 - 0.01 = 0.99 . The sensitivity is P(T|D) = 0.98 . The specificity is P(T^c|D^c) = 0.95 , which gives P(T|D^c) = 1 - 0.95 = 0.05 . Using Bayes' theorem, the probability that the person has the disease given a positive test is: P(D|T) = P(T|D)P(D) P(T|D)P(D) + P(T|D^c)P(D^c) Substituting the values: P(D|T) = 0.98 0.01 0.98 0.01 + 0.05 0.99 P(D|T) = 0.0098 0.0098 + 0.0495 P(D|T) = 0.009