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?

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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