law of total probability (thats the denominator) and bayes rule
There's a medical diagnostic test for a disease. If a person has the disease, the probability that the test gives a positive result is 0.95. If a person does not have a disease the probablilty of a negative result is 0.90. In the population 1% of people have the disease. What is the probability that a person tested has the disease, given the result is positive? Let T be the event of a positive test and D be the event of having the disease.
Bayes Rule: P(D|T) = [P(T|D) * P(D)]/[P(T|D) * P(D) + P(T|D') *P(D')
where D' is 'not D' and P(D') = 1 - P(D).
P(T|D) = 0.95
P(D) = 0.01
P(T|D') = 0.10
P(D') = 0.99
so P(D|T) is [0.95 *0.01]/[0.95 * 0.01 + 0.10 *0.99] = 0.0875.
Is that correct? Doesn't seem right to me ... would expect to get a higher number than that.
My answer is .087557604
I also had expected it to be higher, BUT you should also compare P(D|T) to P(D) which is only .01.
So this is 8 times higher than not knowing any info.
Also, do not get me started on doctors and false positives.
I've had eight surgeries already.