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.
$P({\left(\text{positive}{\mid}\text{disease}\right)}) = 0.6$
$P(\text{no disease}) = 0.9$
$P(\text{disease}) = 0.1$
$P{\left(\text{positive}{\mid}\text{no disease}\right)} = 0.02$
${P}{\left(\text{no disease}{\mid}\text{positive}\right)}=\frac{{{P}{\left(\text{no disease}\right)}{P}{\left(\text{positive}{\mid}\text{no disease}\right)}}}{{{P}{\left(\text{no disease}\right)}{P}{\left(\text{positive}{\mid}\text{no disease}\right)}+{P}{\left(\text{disease}\right)}{P}{\left(\text{positive}{\mid}\text{disease}\right)}}}$
Best Answer
My standard method: Imagine a population of 10000 people. Since 10% of the population are sick, there are 1000 sick people, 9000 healthy people. 0.002(900)= 18 of the healthy people test positive and 0.60(1000)= 600 of the sick people test positive, a total of 618 people who test positive.
Of the 618 people who test positive, 18/618= 0.029, 2.9% are actually healthy.