I'm quite happy to answer part 1. of the following problem.

"You are diagnosed with an uncommon disease. You know that there is only a 1% chance of getting it. Let D stand for the event 'you have the disease' and T stand for 'the test says so'. It is known that the test is imperfect: $\displaystyle P(T | D) = 0.98$ and $\displaystyle P(T^{c} | D^{c}) = 0.95.$

- Given that you test positive, what is the probability that you really have the disease?

- You obtain a second opinion: an independent repetition of the test. You test positive again. Given this, what is the probability that you really have the disease?"

So, for part 1. I use Bayes' theorem to get$\displaystyle P(D|T) = \frac{P(T|D)P(D)}{P(T|D)P(D) + P(T|D^{c})P(D^{c})} = \frac{0.98 \times 0.01}{0.98 \times 0.01 + 0.05 \times 0.99} = 0.165 $

which is correct, according to my book. However, when I apply similar reasoning to the second scenario I get an impossible answer.$\displaystyle P(D|T_{1} \cap T_{2}) = \frac{ P(T_{1} \cap T_{2} | D)P(D) }{ P(T_{1}\cap T_{2}) } = \ldots $

This is where I get stuck. I know that the independence of $\displaystyle T_{1}$ and $\displaystyle T_{2}$ doesn't necessarily imply their conditional independence (conditioned on $\displaystyle D$) but I have no idea how to proceed. If I just assume the conditional independence I get an answer greater than 1, which is obviously wrong.