# bayes' theorem, sick people exercise

I have been practicing probabilities for quite some time now and i came across bayes' theorem. I've been sitting on this exercise for quite some time now and i am not completely sure if this is correct way to do it. i would love some more insight if the displayed solution is not correct

if a person is sick, the probability to diagnose the disease is 0.6. Probability for a healthy person to test positive for a disease is 0.02. Let's say 10% of population are sick people. What is the probability for a person to be healthy if he was diagnosed sick.

$$P({\left(\text{positive}{\mid}\text{disease}\right)}) = 0.6$$

$$P(\text{no disease}) = 0.9$$

$$P(\text{disease}) = 0.1$$

$$P{\left(\text{positive}{\mid}\text{no disease}\right)} = 0.02$$

$${P}{\left(\text{no disease}{\mid}\text{positive}\right)}=\frac{{{P}{\left(\text{no disease}\right)}{P}{\left(\text{positive}{\mid}\text{no disease}\right)}}}{{{P}{\left(\text{no disease}\right)}{P}{\left(\text{positive}{\mid}\text{no disease}\right)}+{P}{\left(\text{disease}\right)}{P}{\left(\text{positive}{\mid}\text{disease}\right)}}}$$

• That will work when you substitute the numbers Mar 23 '20 at 17:23
• thanks for reconfirming Mar 23 '20 at 17:25

• I think there would be $180$ healthy people who test positive rather than $18$ Mar 23 '20 at 18:01