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 $P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$, because the tests are independent.

By law of total probability we know that:

$P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$

Replacing, and assuming conditional independence given D, we have:

$P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$

This is the correct result, but now let's consider that:

$P({T}_{1},{T}_{2})=P({T}_{1}{)}^{2}$

We know that $P({T}_{1},{T}_{2})=P({T}_{1}{)}^{2}$ for all i because of symmetry, so we have $P({T}_{1},{T}_{2})=P({T}_{1}{)}^{2}$. Again, by law of total probability:

$P({T}_{1})=0.95\ast 0.01+0.05\ast 0.99\approx 0.059$

$P({T}_{1})=0.95\ast 0.01+0.05\ast 0.99\approx 0.059$

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.

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 $P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$, because the tests are independent.

By law of total probability we know that:

$P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$

Replacing, and assuming conditional independence given D, we have:

$P({T}_{1},{T}_{2})={0.95}^{2}\ast 0.01+{0.05}^{2}\ast 0.99=0.0115$

This is the correct result, but now let's consider that:

$P({T}_{1},{T}_{2})=P({T}_{1}{)}^{2}$

We know that $P({T}_{1},{T}_{2})=P({T}_{1}{)}^{2}$ for all i because of symmetry, so we have $P({T}_{1},{T}_{2})=P({T}_{1}{)}^{2}$. Again, by law of total probability:

$P({T}_{1})=0.95\ast 0.01+0.05\ast 0.99\approx 0.059$

$P({T}_{1})=0.95\ast 0.01+0.05\ast 0.99\approx 0.059$

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.