It’s been a while since I did statistics, and I only did mediocre, but I’m going to give it a try. I didn’t read everything above, it was just too much, but none of the solutions I did read stated what my calulations did. Of course this means there is a higher likelihood I am wrong, but who gives a smeg. This is the intarweb.

I did use google, but I knew what I had to do: find P(B|A) while knowing P(A|B), P(B) and P(A) and didn’t remember the formula. If I had my formula book available, I would have used it so I don’t consider this cheating.

On to my calculations.

Chance of having breast cancer

P© = 1%

Chance of recieving positive result, having breast cancer

P(T|C) = 80% (read as T given B)

Change of recieving a false positive

P(T|^C) = 9,6%

Chance of recieveing positive result, both having and not

P(T) = P(T|C) + P(T|^C) = 89,6%

Question stated: A woman in this age group had a positive mammography in a routine screening. What is the probability that she actually has breast cancer?

So what is probability of cancer, given a positive result? : P(C|T)

Bayes theorem says P(B|A) = P(A|B)*(P(B)/P(A)), so for my notation i’m using P(C|T) = P(T|C)*(P©/P(T)) which gives me

P(C|T) = 80%*(1%/89,6%) = 0,89%

Yes. Having been given a positive mammogram result there is a 0,89% chance that one actually does have breast cancer.

It sounds unintuitive to me, but both my statistics book said so at the time and Jeff said so himself in this article.