Complete question is;
In a certain community, 8% of all people above 50 years of age have diabetes. A health service in this community correctly diagnoses 95% of all person with diabetes as having the disease, and incorrectly diagnoses 10% of all person without diabetes as having the disease. Find the probability that a person randomly selected from among all people of age above 50 and diagnosed by the health service as having diabetes actually has the disease.
Answer:
P(has diabetes | positive) = 0.442
Step-by-step explanation:
Probability of having diabetes and being positive is;
P(positive & has diabetes) = P(has diabetes) × P(positive | has diabetes)
We are told 8% or 0.08 have diabetes and there's a correct diagnosis of 95% of all the persons with diabetes having the disease.
Thus;
P(positive & has diabetes) = 0.08 × 0.95 = 0.076
P(negative & has diabetes) = P(has diabetes) × (1 –P(positive | has diabetes)) = 0.08 × (1 - 0.95)
P(negative & has diabetes) = 0.004
P(positive & no diabetes) = P(no diabetes) × P(positive | no diabetes)
We are told that there is an incorrect diagnoses of 10% of all persons without diabetes as having the disease
Thus;
P(positive & no diabetes) = 0.92 × 0.1 = 0.092
P(negative &no diabetes) =P(no diabetes) × (1 –P(positive | no diabetes)) = 0.92 × (1 - 0.1)
P(negative &no diabetes) = 0.828
Probability that a person selected having diabetes actually has the disease is;
P(has diabetes | positive) =P(positive & has diabetes) / P(positive)
P(positive) = 0.08 + P(positive & no diabetes)
P(positive) = 0.08 + 0.092 = 0.172
P(has diabetes | positive) = 0.076/0.172 = 0.442
