Question

In a certain region of the country it is known from past experience that the probability of selecting an adult over 40 years of age with cancer is 0.05. If the probability of a doctor correctly diagnosing a person with cancer as having the disease is 0.78 and the probability of incorrectly diagnosing a person without cancer as having the disease is 0.06, what is the probability that an adult over 40 years of age is diagnosed as having cancer?

Answer 1

Answer: Our required probability is 0.406.

Step-by-step explanation:

Since we have given that

Probability of selecting an adult over 40 years of age with cancer = 0.05

Probability of a doctor correctly diagnosing a person with cancer as having the disease = 0.78

Probability of incorrectly diagnosing a person without cancer as having the disease = 0.06

Let A be the given event i.e. adult over 40 years of age with cancer. P(A) = 0.05.

So, P(A')=1-0.05 = 0.95

Let C be the event that having cancer.

P(C|A)=0.78

P(C|A')=0.06

So, using the Bayes theorem, we get that

$P(A|C)=\dfrac{P(A).P(C|A)}{P(A).P(C|A)+P(A')P(C|A')}\\\\P(A|C)=\dfrac{0.78\times 0.05}{0.78\times 0.05+0.06\times 0.95}\\\\P(A|C)=0.406$

Hence, our required probability is 0.406.