Compute the probability that the person does not have the disease. Please Check my work

Question

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

Answer 1

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

We have

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

Answer 2

well,I think that you have made some mistakes. Let A＝"no disease" , B＝‘positive’ we need to derive P(A|B) according to bayes, it equals to 0.07*(1-0.36)/(0.07*(1-0.36)+0.36*0.9)＝0.12 Actually, I am also not so sure about that. So, you had better do it yourself again.