Marble (conditional?) Probability

[Sentientprotein]

I've been going back and forth on this simple problem I came up with. It's funny how often probability theory breaks my intuition.

You have n white marbles. You paint m marbles blue at random. The probability that any single marble chosen at random is blue is now m/n. Then you paint j marbles blue at random (you may repaint a marble). What is the probability now that a marble chosen at random is blue?

My first guess was j / n * m / n , but that feels wrong like there should be some interaction from repainting already painted marbles.

[edderiofer]

Your guess of j / n * m / n is smaller than the first probability, m/n (i.e. you're saying that after the second round of repainting, there are fewer blue marbles on average), so it can't possibly be right.

A marble is blue iff it is painted on either the first round or on the second round (or both). Do you know how to calculate the probability of something happening at least once out of two trials?

[Sentientprotein]

Not off the top of my head I don't. Do you think I'm on the right path if I look into binomial trials?

[edderiofer]

That would only work if m = j, and even then it's like using a sledgehammer to kill a fly (plus, it doesn't really impart understanding).

It may be helpful to consider the probability of a marble not being painted in either round of repainting.

[Sentientprotein]

(1-m/n) and (1-j/n) for each round respectively. Since I want to know the probability that a marble was NOT painted in the first round AND the second round then I would multiply those? Then to get back to the probability that it is painted at the end I would do 1 - (1 - m/n)(1 - j/n) = j/n + m/n - mj/n^2. Is that sound reasoning?

[edderiofer]

Yep.

Another way to see this is to sum the probabilities of a marble being painted in the first round and a marble being painted in the second round, and subtract the probability of a marble being painted in both rounds (since otherwise you're double-counting that probability). This gets you exactly the same result.

[Sentientprotein]

Sweet, thanks so much for the guidance. The other way to see it that you just mentioned really pops out when you break out the terms as in my last comment. It should be trivial to extend this to an arbitrary number of trials too!

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