Mr. Bayes will help you:
This is a probability problem on a past exam for my university. It asks....
A test correctly identifies a disease in 95% of people who have it. I correctly identifies no disease in 94% of people who don't have it. In the population, 5% of people have this disease. A person is tested at random.
1. What is the probability that they will test positive?
2. what is the probability that they have the disease given that they tested positive.
I'm useless at probability but I have a feeling that this has something to do with the multiplication law. I just can't seem to figure out how this would be done. I know that the second part would be like...
P(person actually has the disease | person tested positive)
Could anyone point me in the right direction??
Ok, I had a belt at this, here's what i have.
A = probability the the person has the disease (.05).
B = probability that the person hasn't got the disease (.95).
C = probability of detecting the disease if present (.95).
D = probability of detecting no disease if not present (.94).
P(person tests positive) = (0.05 * 0.95) + (0.95 * 0.06 (ŹD)) = 0.1045
So a person will test positive about 10% of the time, is this right or at least close to it??
And thank you Colby, you defo pointed me in the right direction .