Conditional Probability and Independence nonsense in a problem The statement:
Suppose that a patient tests positive for a disease affecting $1$% of the population. For a patient who has the disease, there is a $95$% chance of testing positive, and for a patient who doesn't has the disease, there is a $95$% chance of testing negative. The patient gets a second, independent, test done, and again tests positive. Find the probability that the patient has the disease.
The problem:
I can solve this problem, but I'm unable to understand what is wrong with the following:
Let $T_i$ be the event that the patient tests positive in the $i$-th test, and let $D$ be the event that the patient has the disease. 
The problem says that $P(T_1, T_2) = P(T_1)P(T_2)$, because the tests are independent.
By law of total probability we know that: 
$$P(T_1, T_2) = P(T_1, T_2\,|\,D)P(D) + T(T_1, T_2\,|\,D^c)P(D^c)$$
Replacing, and assuming conditional independence given $D$, we have:
$$P(T_1, T_2) = 0.95^2*0.01 + 0.05^2*0.99 = 0.0115$$
This is the correct result, but now let's consider that:
$$P(T_1, T_2) = P(T_1)P(T_2)$$
We know that $T_1 = T_i$ for all $i$ because of symmetry, so we have $P(T_1, T_2) = P(T_1)^2$. Again, by law of total probability:
$$P(T_1) = P(T_1\,|\,D)P(D) + P(T_1|D^c)P(D^c)$$
$$P(T_1) = 0.95*0.01 + 0.05*0.99 \approx 0.059$$
So we have:
$$P(T_1, T_2) = P(T_1)^2 \approx 0.059^2 \approx 0.003481$$
The second approach is wrong, but it seems legitimate to me, and I'm unable to find what's wrong.
Thank's for your help, you make self studying easier.
 A: It is not the case that $T_1$ and $T_2$ are independent. They are assumed conditionally independent given $D$ and conditionally independent given $D^c$ but this does not imply (and in fact usually prohibits) that they are unconditionally independent.
Intuitively if $T_1$ is positive, it is more likely they have the disease which makes it more likely for $T_2$ to be positive. So they are not independent.
A: 
The patient gets a second, independent, test done ... 

The word "independent" as used in the question does not refer to statistical independence. It refers to the second test being done without any direct causal influence from the first test. The lack of a causal relationship between the two tests does not imply the lack of a correlation between them (in fact, the results of the two tests are quite positively correlated with each other). 
Incidentally, the $^{\text{(fallacious)}}$ idea that correlation implies causation is equivalent (by contrapositive) to the $^{\text{(also fallacious)}}$idea that a lack of causation implies a lack of correlation. 
Perhaps the folks over at the English Language & Usage Stack Exchange can clarify the non-mathematical use of the word "independent" better than I can. 

We know that for any two events, regardless of independence, the following is true:
$$P(T_1 \cap T_2) = P(T_1) P(T_2 \,|\, T_1)$$
Suppose, for contradiction, that $T_1$ and $T_2$ are statistically independent. By definition,  $$P(T_1 \cap T_2) = P(T_1)P(T_2) $$
In order for the first and second equations above to both be satisfied, it must be the case that $$P(T_2\,|\,T_1)=P(T_2)$$
This is a contradiction. If the first test is positive, the second test clearly must be more likely to also be positive. Therefore, $$P(T_2 \,|\, T_1) \neq P(T_2) \,\,\implies \,\,P(T_1 \cap T_2) \neq P(T_1)P(T_2) \,\, \implies \,\, T_1 \text{ and } T_2 \text{ are not independent }$$

The original problem, which asks for the value of $P(D\,|\,T_1, T_2)$, can be solved using Bayes' theorem, and no assumption about the statistical independence between $T_1$ and $T_2$ is necessary.
