First I must sorry in advance, because this question sounds really basic, but couldn't find anything on the internet as I don't know how to phrase this question into a simple google query.. if this is a duplicate I would highly appreciate if you could send a link to a similar question. Sorry again!
This is a theoretical question, so I am making this situation up.
Let's say that we have a company which consists of $10,000$ workers, we'll call this company $A$.
Then we have $6,000$ people from company $B$, then $2,000$ from $C$ and so on. The numbers don't need to follow a special rule, except:
- $A$ has the largest amount of workers (In this case $10,000$)
- The sum of the workers of all the companies $B, C, D, \dots $ must be larger than $A$'s.
We enclose all of the workers of all these companies in a huge room.
Now, "special sense" of probability and statistics (AKA intuition) says that if we take big tweezers and choose a random worker, the probability of him being from company $A$ is the greatest.
However, it doesn't make sense if I think about it longer, because the question is if the worker is from $A$ or not $A$, and say the chances of him is $x$ then the chances of $\bar{x}$ is bigger as the sum of all the non-$A$ workers is larger...
Which way of thinking is correct? It got me very confused
In order to showcase those intuitions I will present an example.
Say for example, we have this set of data:
- $A$ has $10,000$ workers
- $B$ has $5,000$
- $C$ has $6,000$
- $D$ has $7,000$
Clearly it follows the rule that $\#A > \#\text{not}A$ and the sum of all the workers from all the companies which are not $A$ is larger than $10,000$.
What would be the probability to choose a person from $A$?
$$P(A) = \frac{10000}{10000+5000+6000+7000} = \frac{10}{28} \\ P(\bar{A}) = 1 - \frac{10}{28} = \frac{18}{28}$$
Clearly: $$ P(A) < P(\bar{A})$$
So, which statement is correct? Statistically speaking if we choose a random worker, it would be $A$'s as there are more workers from this company than any other company, but the probability of choosing not $A$ proved to be larger, from the example above..