In a certain community, eight percent of all adults over age 50 have diabetes. If a health service in this community correctly diagnosis 95% of all persons with diabetes as having the disease and incorrectly diagnoses ten percent of all persons without diabetes as having the disease, find the probabilities that:

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

Using formula:

[tex]P(\text{diabetes diagnosis})\\[/tex]:

[tex]=\text{P(having diabetes and have been diagnosed with it)}\\ + \text{P(not have diabetes and yet be diagnosed with diabetes)}[/tex]

[tex]=0.08 \times 0.95+(1-0.08) \times 0.10 \\\\=0.08 \times 0.95+0.92 \times 0.10 \\\\=0.076+0.092\\\\=0.168[/tex]

[tex]\text{P(have been diagnosed with diabetes)}[/tex]:

[tex]=\frac{\text{P(have diabetic and been diagnosed as having insulin)}}{\text{P(diabetes diagnosis)}}[/tex]

[tex]=\frac{0.08\times 0.95}{0.168} \\\\=\frac{0.076}{0.168} \\\\=0.452\\[/tex]

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