Probability of having disease given positive test result I am trying to solve a puzzle and have struck in some fundamental concepts of probability.
In an epidemic, its found that only one in 10000 people actually suffer from the rare disease. So  the probability that I am positive will be

P1= 1/10000

But there is a medical test also which has 99% accuracy. Unfortunately my test result is positive. As per this test, the probability that I am positive will be

P2= 99/100

Now I have two probabilities P1 and P2.  The first question that comes to my mind is how can we have two probabilities for a single event (i.e. being "positive")? Why they are different?
Secondly, which one should I trust and why ?
Or what is the actual probability of me being found positive?
Edit
The specific question is if 100000 people are tested, and the test says I am positive, then what is the probability that I am actually suffering from the disease given the test results?

The idea of these types of questions/puzzles is to apply some Bayesian analysis and use the test result to update the probability that you are sick.
The people that got a positive test result are

*

*either sick and got a correct test result

*or not sick and got an incorrect test result.

The ratio of those two cases are the odds for being sick.

For example. Say 1 000 000 people do the test and the outcome is:
$$\begin{array}{}
& \text{sick} & \text{not sick} & \text{total} \\
\text{positive test}& 99 & 9999 & 10098 \\
\text{negative test}& 1 & 989901 & 989902\\
\text{total} &100 &999 900 & 1000000
\end{array}$$
Here we see that among the sick people 99% test positive and among the not sick people 99% test negative.
If you have a positive test then you are one of the 10 098 people with a positive test, but only 99 among those are truly sick. So you have $\frac{99}{10098} \approx 0.98\%$ probability of being sick.

So your problem is to make that table. Below this is done stepwise

*

*Fill in the total amount of people $n$ (the actual value does not matter, it will drop from the equation eventually)$$\begin{array}{}
& \text{sick} & \text{not sick} & \text{total} \\
\text{positive test}&  &  &  \\
\text{negative test}&  & & \\
\text{total} & &  & n
\end{array}$$


*The prior distribution sick is $p=0.0001$, you can fill this into the totals $$\begin{array}{}
& \text{sick} & \text{not sick} & \text{total} \\
\text{positive test}&  &  &  \\
\text{negative test}&  & & \\
\text{total} & p n & (1-p)  n & n
\end{array}$$


*If you know the probability distribution of the test result/outcome given the state sick/not-sick, then you can fill in the rest of the table.
If a fraction $x$ of the sick people have a positive test $$\begin{array}{}
& \text{sick} & \text{not sick} & \text{total} \\
\text{positive test}& x p n &  &  \\
\text{negative test}& (1-x) p n & & \\
\text{total} & p n & (1-p)  n & n
\end{array}$$


*If a fraction $y$ of the non sick people have a negative test $$\begin{array}{}
& \text{sick} & \text{not sick} & \text{total} \\
\text{positive test}& x p n & (1-y) (1-p) n &  \\
\text{negative test}& (1-x) p n & y (1-p) n & \\
\text{total} & p n & (1-p) n & n
\end{array}$$


*We can complete the totals on the right $$\begin{array}{}
& \text{sick} & \text{not sick} & \text{total} \\
\text{positive test}& x p n & (1-y) (1-p) n & x p n + (1-y) (1-p) n \\
\text{negative test}& (1-x) p n & y (1-p) n & (1-x) p n + y (1-p) n\\
\text{total} & p n & (1-p) n & n
\end{array}$$
Then the probability to be sick given a positive test result is
$$\begin{array}{}
P(\text{sick} | \text{positive test}) &=& \frac{x p n }{x p n + (1-y) (1-p) n} \\
&=& \frac{xp }{xp + (1-y)(1-p) } \\
& = & \frac{x }{xp + (1-y)(1-p) }p
\end{array}$$

In your problem, it is only stated that the accuracy is 99%. This is not a very complete description. The detection can make two types of errors a sick person with a negative test, and a not sick person with a positive test. The ratios of these two can be different.
In some of these problem statements, it is plainly just stated that the accuracy is 99% (like in your problem/puzzle), and implied that the false positive and false negative rates are the same and 1%. This implication is not correct/accurate but if you ignore this error in the question then you can say $x = y = 0.99$ and solve the above equation
$$\begin{array}{}
P(\text{sick} | \text{positive test}) &=& \frac{x }{xp + (1-y)(1-p) }p \\
&=& \frac{0.99 }{0.99 \cdot 0.0001 + 0.01 \cdot 0.9999 } 0.0001\\
&=&  \frac{0.99 }{  0.010098 } 0.0001
& = & 0.009803922
\end{array}$$
