Question

It is estimated that approximately 8.17% Americans are afflicted with diabetes. Suppose that a certain diagnostic evaluation for diabetes will correctly diagnose 98% of all adults over 40 with diabetes as having the disease and incorrectly diagnoses 3% of all adults over 40 without diabetes as having the disease. a) Find the probability that a randomly selected adult over 40 does not have diabetes, and is diagnosed as having diabetes (such diagnoses are called "false positives"). b) Find the probability that a randomly selected adult of 40 is diagnosed as not having diabetes. c) Find the probability that a randomly selected adult over 40 actually has diabetes, given that he/she is diagnosed as not having diabetes (such diagnoses are called "false negatives"). (Note: it will be helpful to first draw an appropriate tree diagram modeling the situation)

Answer 1

Solution:

Let

TP = Test Positive

TN = Test Negative

D = having diabetes

ND = Not having diabetes

Thus we have:

8.17% Americans are afflicted with diabetes.

That is:

P(D) = 0.0817

P(ND) = 1 - P(D)

P(ND) = 1 - 0.0817

P(ND) = 0.9183

A certain diagnostic evaluation for diabetes will correctly diagnose 98% of all adults over 40 with diabetes as having the disease

P( TP | D) = 0.98

and

incorrectly diagnoses 3% of all adults over 40 without diabetes as having the disease

P( TP | ND)= 0.03

Thus Tree diagram is:

Part  a) Find the probability that a randomly selected adult over 40 does not have diabetes, and is diagnosed as having diabetes

P( Does not have diabetes, and is diagnosed as having diabetes) = ........?

P( ND and TP ) =............?

Using: P( TP | ND)= 0.03 and conditional probability formula:

Part b) Find the probability that a randomly selected adult of 40 is diagnosed as not having diabetes.

That is find:

P( TN) = ..........?

Part c) Find the probability that a randomly selected adult over 40 actually has diabetes, given that he/she is diagnosed as not having diabetes

P(D | TN) =.............?

Using Bayes rule: