Refer to the table which summarizes the results of testing for a certain disease. $$\begin{array} { l c c }
& \begin{array} { c }
\text { Positive Test } \
\text { Result }
\end{array} & \begin{array} { c }
\text { Negative Test } \
\text { Result }
\end{array} \
\text { Subject has the disease } & 87 & 9 \
\text { Subject does not have the disease } & 27 & 312
\end{array}$$ A test subject is randomly selected and tested for the disease. What is the probability the subject has the disease given that the test result is negative?
A) 0.094
B) 0.972
C) 0.028
D) 0.221

Question

Refer to the table which summarizes the results of testing for a certain disease. $$\begin{array} { l c c } 
& \begin{array} { c } 
\text { Positive Test } \
\text { Result }
\end{array} & \begin{array} { c } 
\text { Negative Test } \
\text { Result }
\end{array} \
\text { Subject has the disease } & 87 & 9 \
\text { Subject does not have the disease } & 27 & 312
\end{array}$$ A test subject is randomly selected and tested for the disease. What is the probability the subject has the disease given that the test result is negative?
A) 0.094
B) 0.972
C) 0.028
D) 0.221

Quizplus · Accepted Answer

To find the conditional probability of having the disease given a negative test result, we need to use Bayes' theorem: 
$$P(	ext{disease}\,|\,	ext{negative})=\dfrac{P(	ext{negative}\,|\,	ext{disease})\cdot P(	ext{disease})}{P(	ext{negative})}$$ 
We can fill in the values from the table: 
$$P(	ext{disease}\,|\,	ext{negative})=\dfrac{P(	ext{negative}\,|\,	ext{disease})\cdot P(	ext{disease})}{P(	ext{negative}\,|\,	ext{disease})\cdot P(	ext{disease})+P(	ext{negative}\,|\,	ext{no disease})\cdot P(	ext{no disease})}$$ 
$$P(	ext{disease}\,|\,	ext{negative})=\dfrac{(9/96)\cdot (96/435)}{(9/96)\cdot (96/435)+(312/339)\cdot (339/435)}=0.028$$ 
Therefore, the correct answer is choice (C).

Refer to the Table Which Summarizes the Results of Testing