I saw this question in a textbook and the answer it has, seems pretty weird to me:
Suppose we have 12 balls in a container, of which 8 are white and the 4 others are black. We choose 4 of the balls (After we pick up a ball, we record the result and put it back in the container so it can get selected again). What's the probability of the first and the third ball being white, given we know that exactly 3 of the 4 balls we picked were white?
This is what the book gave for the answer:
Suppose we show the white balls with a and the black balls with b. Let's say event B is selecting exactly 3 white balls:
B = {aaab,aaba,abaa,baaa}
If we display the event of the first and the third ball being white with A, then A∩B = {aaab, abaa} so P(A|B)=n(A∩B)/n(B)=1/2
I find this quite confusing as the values the question gave us (8 of the balls are white and 4 others are black) have been of no use whatsoever. Other than that, it doesn't make sense to get the same answer if we had some other number of balls; for example, we had 23243 balls and 800 of them were white and the rest were black)
Is there something I'm missing out here?
You are not missing out. The propability in question is completely independent from the distribution of colours in the container. The event of drawing is not considered, you just have a fixed set of outcomes.
It is confusing in the way, that the description of the container is part of question text, I suppose it is meant to be intentionally misleading to make this point.