Bayesian theory: Is this difference legit or I'm missing something?

Question

Say 2 groups. Healthy and sick people.
Sick people are more likely to vomit. 
Healthy people are less likely to vomit. 
4 million healthy vs 1000 sick.
healthy person vomiting: 0.05/10
sick person vomiting: 5/10

you see a person vomiting. is he sick?
2 million * 0.05/10 = 10,000 people who would vomit but are not sick.
1000 * 5/10 = 500 people who vomit and are sick

10000:500 ratio shows that person is not likely to be sick. 
20:1

But what about odds of YOU encountering such person.
10,000 / (2mil + 1000) = A
1000 / (2mil + 1000) = B 
10000:1000
10:1.
A > B

That's weird. Your chances of encountering Healthy person is 10 to 1. Yet person's chance of being healthy is 20 to 1.

Why the difference?

Answer 1

The reason you find a difference is because you are not being consistent with the sample size and proportions of people who are sick and healthy. This answer assume the original total population $N=4,001,000$ individuals with $4$ million healthy, and $1$ thousand sick. This gives yield to a Bernoulli trial, where we observe one person from the sample and whether or not they are sick with probabilities determined by the known make up of the population. Then, further, we add in the event that they may vomit which is dependent on the events of being sick or healthy. Finally, the chance of observing a person who is sick given that they are vomiting can be computed by Bayes' theorem.

Let $H$ denote healthy, $S$ denote sick, $V$ denote vomiting. You wish to find $P(S|V)$.

You have given us $$P(H)=\frac{4,000,000}{4,001,000}= 0.99975006248$$ $$P(S)=\frac{1,000}{4,001,000}= 0.00024993751$$ (this can be interpreted as a Bernoulli trial, sampling one person from the $N=4,001,000$ and success being healthy).

You have also given that, $$P(V|H)=0.05/10=0.005$$ and $$P(V|S)=5/10=0.5.$$

Hence, the probability of vomiting is given by the law of total probability $$P(V)=P(V|H)P(H)+P(V|S)P(S) = 0.005123719$$

We can finally compute $P(S|V)$ by Bayes' law: $$P(S|V)=\frac{P(V|S)P(S)}{P(V)}=0.02439024$$

Answer 2

If you take the sick people amount as 1000, you also need to take the healthy population 4 million while calculating probability of a random person you met being sick.

Edit: Alright, apart from pointing out why I think the second part of the solution is wrong, let me try to answer to the last part as well:

Bayesian inference calculates the probibality of the underlying system based on the observations made so far. This means that a Bayesian probability is only a inference given what have already happened, that is, it will not match the exact intrinsic properties of the system. It is more like a long term educated guess.

In this system, knowing a person have puked updates our a priori model of the system, where we learn that their probability of being sich is changed. This is a result of different sample spaces: Those who puked will have a different probabilistic model than of general public. This investigation is called Bayesian statistics.