Use Bayes' theorem to find the indicated probability.
-Refer to the table which summarizes the results of testing for a certain disease. $$\begin{array} { l | c c }
& \text { Positive Test Result } & \text { Negative Test Result } \
\hline \text { Subject has the disease } & 93 & 10 \
\text { Subject does not have the disease } & 27 & 93
\end{array}$$ A test subject is randomly selected and tested for the disease. What is the probability the subject does not have the disease given that the test result is positive?
A) 0.141
B) 0.65
C) 0.225
D) 0.775

Question

Use Bayes' theorem to find the indicated probability.
-Refer to the table which summarizes the results of testing for a certain disease. $$\begin{array} { l | c c } 
& \text { Positive Test Result } & \text { Negative Test Result } \
\hline \text { Subject has the disease } & 93 & 10 \
\text { Subject does not have the disease } & 27 & 93
\end{array}$$ A test subject is randomly selected and tested for the disease. What is the probability the subject does not have the disease given that the test result is positive?
A) 0.141
B) 0.65
C) 0.225
D) 0.775

Quizplus · Accepted Answer

We want to find the probability that the subject does not have the disease given that the test result is positive. Let A be the event that the subject does not have the disease, and B be the event that the test result is positive. Then, we want to find $P(A|B)$. By Bayes' theorem, we have:
$$P(A|B) = \dfrac{P(B|A)P(A)}{P(B)}$$
We can find each of these probabilities from the table:
\begin{align*}
P(B|A) &= \dfrac{27}{120}\
P(A) &= \dfrac{120}{240} = \dfrac{1}{2}\
P(B) &= P(B|A)P(A) + P(B|A^c)P(A^c)\
&= \dfrac{27}{120} \cdot \dfrac{1}{2} + \dfrac{93}{120} \cdot \dfrac{1}{2} = \dfrac{60}{120} = \dfrac{1}{2}
\end{align*}
where $A^c$ is the complement of event A, i.e. the event that the subject has the disease. Putting it all together, we have:
$$P(A|B) = \dfrac{P(B|A)P(A)}{P(B)} = \dfrac{\frac{27}{120} \cdot \frac{1}{2}}{\frac{1}{2}} = \boxed{0.225}$$

Use Bayes' Theorem to Find the Indicated Probability A Test Subject Is Randomly Selected and Tested for the the Table