TITLE: How to calculate Pr(Diseased | 2 Positive Tests)?
QUESTION [7 upvotes]: Diagnosed with a rare disease, you know that there is only a 1% chance of getting it.
Abbreviate D as the event "you have the disease" and T as "you test positive for the disease."
The test is imperfect: $\Pr(T | D) = 0.98$ and $\Pr(T^C | D^C) = 0.95.$
$\large{A.}$ Given that you test positive, what is the probability that you really have the disease?
$\large{B.}$  You obtain a second opinion: an independent repetition of the test. You test positive again. given two positive tests, what is the probability that you really have the disease?

So, first question I have is this: Does $\Pr(T | D^C) = 1 - 0.95 = 0.05$?

This would mean that $P(T) = P(T | D) + P (T | D^C) = 0.98 + 0.05 = 1.03$ ...? This can't be right...P(T) can't be 1.03! Is there an error in this question?

Then, here's my strategy for solving part A: Calculate $\Pr(D | T) = \dfrac{P(T \cap D) }{ P(T) } = \dfrac{ P(T | D) \times P(D) }{ P(T) } $

UPDATE: Here is a hint from the professor:
During office hours today, the following hint came up that I thought would be good to share with the entire class for part B above. 
Let's define two events T (first test positive) and S (second test positive). When you use Bayes' rule, you are going to need to figure out how to compute the total probability $\Pr(T \cap S)$. To do this, you should assume that these two tests are independent, and therefore you will get:
$\Pr(T \cap S | D) = P(T | D) * P(S | D) \quad$ and $\quad \Pr(T \cap S | D^c) = P(T | D^c) * P(S | D^c)$.
A point to remember here is that the rules of probability stay the same for conditional probability if the event on the right of the "|" stays constant. For instance, the complement rule looks like this: $P(A^c | B) = 1 - P(A | B)$. The other rules we have learned work out similarly.

REPLY [6 votes]: To get a handle on what Bayes’ rule is really doing in a simple case like your first problem, look at a little table:
$$\begin{array}{r|cc|c}
&T&\lnot T&\text{Total}\\ \hline
D&\color{red}{0.0098}&\color{red}{0.0002}&0.01\\
\lnot D&\color{red}{0.0495}&\color{red}{0.9405}&\color{blue}{0.99}\\ \hline
\text{Total}&\color{green}{0.0593}&\color{green}{0.9407}&\color{blue}{1}
\end{array}$$
What’s in black shows the basic setup: the number in row $D$ and column $T$ will be the probability that you have the disease and test positive, for instance. (I’ve used $\lnot D$, meaning $\text{not }D$, for the event of not having the disease.) I’ve also written in black the one number that you were explicitly given that goes in the table, namely, the probability that a randomly chosen person has the disease. Obviously the probability in the lower righthand corner must be $1$, and the probability that a randomly chosen person does not have the disease must be $1-0.01=0.99$, so I added those in blue.
Next, we know that $98$% of the $1$% who have the test will test positive, so the probability that a randomly chosen person both has the disease and tests positive is $0.98\cdot0.01=0.0098$, and the probability that such a person both has the disease and tests negative must be $0.01-0.0098=0.0002$; I’ve added these in red. 
Similar reasoning tells us that the probability that a randomly chosen person does not have the disease and tests negative is $0.95\cdot0.99=0.9405$, so the probability that such a person does not have the disease but tests positive is $0.99-0.9405=0.0495$; I’ve added these in red as well.
At this point we can complete the table by adding the $T$ and $\lnot T$ columns to get the missing totals on the bottom line.
Now, suppose that you test positive. That means that you’re one of the $14.75$% represented by the first column of the table, and you want to know your chance of being one of the $0.98$% who test positive and have the disease. That probability is $\dfrac{0.0098}{0.0593}\approx 0.1653$: you have almost one chance in six of having the disease.
Bayes’ rule boils all of this down to a single formula. We ended up calculating the ratio $$\frac{\mathrm{Pr}(D\cap T)}{\mathrm{Pr}(T)}\;,$$ but we got the numerator by calculating $\mathrm{Pr}(T|D)\mathrm{Pr}(D)$, so in fact we calculated
$$\mathrm{Pr}(D|T)=\frac{\mathrm{Pr}(T|D)\mathrm{Pr}(D)}{\mathrm{Pr}(T)}\;,$$
which is Bayes’ rule.
In this case we also had to work a bit to get the denominator; that’s typical.