Let C represent the event that a person has cancer. Let D represent the event that a person is diagnosed with cancer. 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 0.08. The probability of a doctor correctly diagnosing a person with cancer as having the disease is P(D | C) = 0.84, and the probability of incorrectly diagnosing a person without cancer as having the disease is P(D | C 0 ) = 0.04. What is the probability that a person diagnosed as having cancer actually has the disease, i.e. find P(C | D)? Round all values to 4 decimals, if needed.

Answer:0.6462Step-by-step explanation:Let [tex]\bar{C}[/tex] be the complement of C. We have that P(C)=0.08 because we know from past experience that the probability of selecting an adult over 40 years of age with cancer is 0.08, then the probability of selecting a person over 40 years of age without cancer isP([tex]\bar{C}[/tex]) =0.92, P(D | C) = 0.84 P(D | [tex]\bar{C}[/tex]) = 0.04Using the Bayes' Formula we haveP(C | D) = [tex]\frac{P(D | C)P(C)}{P(D | C)P(C)+P(D | \bar{C})P(\bar{C})}[/tex] = [tex]\frac{(0.84)(0.08)}{(0.84)(0.08)+(0.04)(0.92)}[/tex] =0.6462We used the Bayes' Formula because  C and [tex]\bar{C}[/tex] are mutually exclusive events satisfying [tex]C\cup\bar{C}[/tex] = S (S the sample space)