Compute the probability that the person does not have the disease. Please Check my work A certain disease occurs in $36$ % of the population. A test for the disease is fairly accurate: it misclassifies people with the disease as healthy $10$ % of the time and reports that a healthy person is diseased just $7$% of the time. Suppose that a person tests positive for the disease. Compute the probability that the person does not have the disease. Round your answer to two decimal places. My answer: # $P$(Positive∣ Disease)= $(0.90*0.36)/(0.90*0.36+0.07-0.64)$ # =$0.324/0.3688$ # =$0.88$ Please check My work

I'm not sure why you computed $P(\text{Positive | Disease})$. The question asks for $P(\text{No disease | Positive})$.

We have

$$\begin{align*} P(\text{No disease | Positive}) &=\frac{P(\text{No disease } \cap \text{ Positive})}{P(\text{Positive})}\\\\\\\ &=\frac{0.64\cdot0.07}{0.64\cdot0.07+0.36\cdot0.9}\\\\\\\ &\approx 0.1215 \end{align*}$$