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I am solving another problem and it boils down to the following.

Say we have a bag with identical balls numbered 1,2,...,100 and we draw 5 of them on each attempt.
Say we make N draws/attempts and after each draw we put back the 5 drawn ones and we shuffle the bag.

Event A = {Ball i was drawn at least once after the N draws}
Event B = {Ball j was drawn at least once after the N draws}

Here $i \ne j$

Are these two events independent?

My intuition tells me they are but I am not absolutely sure and somehow I fail to convince myself. But who knows, maybe my intuition is misleading me here.

peter.petrov
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    They are not, because knowing that event A happens makes it less likely that B happens. This is obvious with $N=1$. For larger $N$, just consider what happens to the event space, and count accordingly. – Calvin Lin Apr 07 '20 at 19:55
  • Say $N=1$. If Event A happens, what is the probability event B happens? – Paul Apr 07 '20 at 19:55
  • Guys, if I can answer/prove these myself I would not be asking this question. @CalvinLin But why does it make it less likely? Say in each draw, you draw the 5 balls at once i.e. simultaneously. Would that make a difference? This is where I am getting confused. – peter.petrov Apr 07 '20 at 19:57
  • @Paul I am not sure, that is the thing. Aren't they both ${99 \choose 4} / {100 \choose 5} = 0.05$ ? – peter.petrov Apr 07 '20 at 20:22
  • By asking this, I am just trying to understand if my solution here is correct https://math.stackexchange.com/a/3614566/116591 Now I am getting more and more convinced that it is actually correct. – peter.petrov Apr 07 '20 at 20:43

1 Answers1

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Two events $A,B$ are said to be independent if and only if $P(A\cap B) = P(A)P(B)$. Suppose $N=1$. Then:

$$P(A) = \dfrac{\dbinom{99}{4}}{\dbinom{100}{5}} = \dfrac{1}{20} \\ P(B) = \dfrac{\dbinom{99}{4}}{\dbinom{100}{5}} = \dfrac{1}{20} \\ P(A\cap B) = \dfrac{\dbinom{2}{2}\dbinom{98}{3}}{\dbinom{100}{5}} = \dfrac{1}{495} \neq P(A)P(B)$$

The two events you list are not independent.

SlipEternal
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  • OK, thanks... So after all my solution to the other problem is incorrect, even though numerically I seem to be getting the correct value. Hm... unfortunately but nice... Thanks. – peter.petrov Apr 07 '20 at 21:15
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    It is a very small difference between your calculation and the actual solution. This problem (when $N=1$) was fairly far apart. The greater $N$ is, the closer the two events become to appearing to be independent. While it is not a true equality, you can say that the events are "approximately independent" when $N$ is large. – SlipEternal Apr 07 '20 at 21:17
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    Yes, I was just going to write the same comment. I guess that's why in that other problem I am getting the correct numerical value. Because we make 100 draws and so the events are "almost independent" :) In fact this is what my intuition was: after 100 lottery draws these two events A and B (intuitively) seem to be really independent... But actually they are "almost independent" (roughly speaking). – peter.petrov Apr 07 '20 at 21:18
  • Right but in the other problem, it was pure luck that I got 114.1 and not 113.8. In fact, now I am not sure what is the ideal answer there for the integer N, is it 114 or 115? We cannot be sure how far away my approx. number 114.1 is from the truly correct answer, unless we develop a really precise math solution to that other problem, right? That other problem is quite intriguing, I would say. – peter.petrov Apr 07 '20 at 21:22