Question

Among U.S. residents ages 65 years and older, 28% had diabetes in 2012. A special glucose test correctly diagnoses the presence of diabetes 97% of the time and correctly diagnoses its absence 95% of the time. (hint: a tree diagram is helpful). Find the probability that a person whose test results are negative actually has diabetes. In other words, given that a person had a negative test result, what is the chance they actually have diabetes? Ch 3.7   

Answer 1

let X be the event of having diabetes

then, P(X)=0.28, P( NOT X)=0.72

let A be the event of correctly diagnosing presence of diabetes (TRUE POSITIVE RESULT)

let B be the event of correctly diagnosing absence of diabetes (TRUE NEGATIVE RESULT)

then, P( A | X)=0.97(POSITIVE RESULT),

P(NOT A | X)=0.03(NEGATIVE RESULT),

P(B | NOT X)=0.95(NEGATIVE RESULT),

P(NOT B | NOT X)=0.05(POSITIVE RESULT)

SO, P(NEGATIVE RESULT) = P(NOT A | X).P(X) + P(B | NOT X).P( NOT X) = 0.03x0.28 + 0.95x0.72 = 0.6924

P(POSITIVE RESULT) =P( A | X).P(X) + P(NOT B | NOT X).P( NOT X) = 0.97x0.28 + 0.05x0.72 = 0.3076

So, given that a person had a negative test result, the chance they actually have diabetes = P( X | NEGATIVE RESULT) = P(NOT A | X).P(X) / P(NEGATIVE RESULT) = 0.03x0.28 / 0.6924 = 0.012132 (final answer)

tree diagram :