# Question on how to apply Bayes Theorem realistically

So I'm new to Bayes Theorem and am trying to understand it. For simplicity I'll refer to the usual example of testing for cancer.

Let's assume:
P(Cancer) = 0.001

there is a 5% probability that a patient who takes a test for cancer gets a positive result, given he doesn't have it:
P(+|no C) = 0.05
therefore,
P(-|no C) = 0.95

also there is a 1% probability that a patient who takes a test for cancer gets a negative result, given he does have cancer:
P(-|C) = 0.01
P(+|C) = 0.99


So we now what to know what is the probability that he really has cancer if he received a positive result to the test.

My understanding is that we have a "Prior" probability P(C) that gets updated with the new conditional information provided to us by the theorem.

So out "posterior" is calculated as:

P(C|+) = P(+|C)P(C) / [P(+|C)P(C) + P(+|no C)P(no C)] = 0.02


Now here is my question, let's say he took the test and received a positive result, he goes and takes the test a 2nd time, but now we know that he already took the test once, and got a positive, so our prior would now be 0.02 instead of 0.001. Correct? And with each new test he takes, we update the prior to the previous posterior generated in the last test he got.

My question is, instead of recalculating the Bayes theorem everytime he takes the test, and updating the prior accordingly, let's say a patient comes to me and tells me I took the test three times already, and each time received a positive result. Can I calculate P(C|+) in one go, as opposed to doing it three times sequentially?

1- Is there a way to update the unconditional probability once (3 consecutive +ve results), as opposed to doing the whole calculation three times?

2- Also realistically, there usually isn't only one variable that plays a role in the conditional probability. Sometimes u can have a person that says "I took the test, and it was positive, but I also am vomiting", and we know that people with cancer vomit 40% of the time, and people who don't have cancer vomit 4% of the time. No only that, u can also add a 3rd similar variable and a 4th, and so on. Would this information be integrated into calculating the posterior probability? If so, how?

For your first question, the answer is that you can do it in one go. What would change in your posterior calculation? P(C) would not change, but P(+|C) would become P(+++|C), which you can calculate if you know P(+|C) (0.99) and assume that the tests are independent. Similarly for P(+++|noC).
P(x|y)=P(x,y)/P(y)
P(x|y,z) = P(x,y,z) / P(y,z)