Bayes, two tests in a row

As an aside, I believe the proper value for $P(D|T)$ is exactly $.88 = \frac{8}{10}\frac{1}{100}\frac{110}{1}$

We have $P(T)$, the probability of the test showing a positive regardless of disease state as $\frac{1}{110}$. This has to be the conditional probability of a positive given diseased plus the conditional probability of a positive given disease-free. In other words: $$ \begin{align} P(T) &= P(T\cap D) + P(T\cap \neg D)\\ &= P(T|D)P(D) + P(T|\neg D)P(\neg D)\\ \frac{1}{110} &=\frac{8}{10}\frac{1}{100} + P(T|\neg D)\frac{99}{100}\\ P(T|\neg D) &=\frac{2}{1815} \end{align} $$

Next: $$ \begin{align} P(TT) &= P(TT|D)P(D) + P(TT|\neg D)P(\neg D)\\ &= \frac{64}{100}\frac{1}{100} + \frac{4}{3294225}\frac{99}{100}\\ &=\frac{21087}{3294225} = \frac{213}{33275} \approx 0.006401202 \end{align} $$ Now $$ \begin{align} P(D|TT) &= \frac{P(TT|D)P(D)}{P(TT)}\\ &= \frac{64}{100}\frac{1}{100}\frac{33275}{213}\\ &= \frac{5324}{5325} \approx 0.999812207 \end{align} $$

So, after two tests, we are really sure this person is diseased.

Update

In general, though, with Bayesian estimation, one can use the previous posterior as the current prior-- see slides 3 and 4. This will follow through as well here. Let $P(D^*)$ be the new prior (after one test). Now we live back in one test world, as one test after one test is the same as two tests after no tests. So $P(D^*)$ is $0.88$ from above. $P(T|D^*)$ remains the same as does $P(T|\neg D^*)$. So, all we need is: $$ \begin{align} P(TT) &= P(T|D^*)P(D^*) + P(T|\neg D^*)P(\neg D^*)\\ &= 0.8\cdot.88 + \frac{2}{1815}\cdot0.12\\ &= \frac{426}{605} \approx 0.704132231 \end{align} $$

Note that $P(TT)$ in the $D^*$ world is much greater than $P(TT)$ in the $D$ world. It stands to reason since $TT$ in $D^*$ is actually $T$ (one test) after already knowing a positive test. $TT$ in $D$ is a priori two tests knowing nothing. Now, as per before: $$ \begin{align} P(D|TT) &= \frac{P(TT|D)P(D)}{P(TT)}\\ &=\frac{8}{10}\frac{88}{100}\frac{605}{426}\\ &=\frac{5324}{5325} \approx 0.999812207 \end{align} $$

You compute $P(TT)$ the same way you computed $P(T)$ - using the Law of total probability: $$P(TT)=P(TT|D)P(D)+P(TT|\neg D)P(\neg D)=0.8^2\times 0.01 + P(T|\neg D)^2\times 0.99$$

Alas, I cannot quite figure out what $P(T|\neg D)$ in your problem statement is.

This is an interesting one. It seems like you can't carry the independence across the conditions. What it means is, if you tested positive, then the next test is also more likely to be positive (can you explain why?).

Thus, to find $P(TT)$, you need to condition first, like sds did in his answer.

To find $P(T|\neg D)$, we can use $P(T) = 1/110$ and $P(T | D) = 0.8$ Then, $$P(T, \neg D) = P(T) - P(TD) = P(T) - P(T|D)P(D) \approx 0.00909 - 0.008 $$ et cetera.

Also, it can be discussed whether test errors are truly independent from one test to another on the same person (since they might depend on certain chemicals in the body, they are likely not), but this discussion is beyond this simple problem.