In a certain region of the country it is known from past experiencethat the probability of selecting an adult over 40 years of agewith cancer is 0.05. If the probability of a doctor correctlydiagnosing a person with cancer as having the disease is 0.78 andthe probability of incorrectly diagnosing a person without canceras having the disease is 0.06, what is the probability that aperson is diagnosed as having cancer?

**EXPERT ANSWER**

Let A be “having cancer”, A’ be “not having cancer”,

B be “be diagnosed with cancer” and “not be diagnosed withcancer”.

We are given that:

P(A)=0.05

P(A’)=1-P(A)=0.95

P(B/A)=0.78

P(B/A’)=0.06

We are looking for P(B).

From total probability rule:

P(B)=P(B/A)*P(A)+P(B/A’)*P(A’)=0.78*0.05+0.06*0.95=0.096