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Problem description

Assume we have a bag filled with marbles of two different colors, red and blue. Our goal is to be able to pick out at least one red marble by picking out the least amount of marbles. Given the numbers $R$ and $B$ (amounts of red and blue marbles respectively), how many marbles do we need to randomly pick out to be at least 95% sure that at least one of them is red?

Independent case

If we assume that the picking of each marble is an independent event, then the answer is easy. For $n$ picked marbles:
$P(\text{at least one marble is red}) = 1 - P(\text{all marbles are blue}) \\ P(\text{at least one marble is red}) = 1 - P(\text{one marble is blue})^n \\ P(\text{at least one marble is red}) = 1 - \left( \frac{B}{R + B} \right)^n $.
And then we simply solve for $P(\text{at least one marble is red}) \geq 0.95$.

Dependent case

However, the independent approximation breaks down in certain cases. For example, consider the case where $R = 4$ and $B = 1$. For $n = 1$, we get $P(\text{at least one marble is red}) = 0.20$, as expected. But for $n = 2$, we get $P(\text{at least one marble is red}) = 0.96$, which is obviously false: no matter which marble we pick first in this configuration, it is impossible-to-not pick at least one red marble, so the probability should be 1.0.

Obviously, there is something missing from that formula, which maybe hints at a kind of dependence between the choices. How does one model this dependence in this case? I know that the classic formula for two dependent events is $P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2)$ (and how it extends to $N$ events), but I'm not sure how to model $E_1, E_2,\ldots, E_n$ in this case. Is there some generalized formula for this problem? Are the events dependent or independent after all?

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  • $\begingroup$ If you are choosing $n$ balls without replacement then the probability that you choose $n$ Black Balls is $\binom Bn\big / \binom {B+R}n$. $\endgroup$
    – lulu
    Commented Jun 24 at 16:15
  • $\begingroup$ @lulu but the question is, what is the probability of choosing at least one, not exactly $n$. $\endgroup$
    – jimkokko5
    Commented Jun 24 at 16:18
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    $\begingroup$ Which is $1$ minus the probability of choosing only the other color (note: I accidentally changed Blue into Black, but that doesn't change anything). $\endgroup$
    – lulu
    Commented Jun 24 at 16:23
  • $\begingroup$ @lulu I see, thank you. So this means that the events are indeed dependent? $\endgroup$
    – jimkokko5
    Commented Jun 24 at 16:29
  • $\begingroup$ If you are choosing without replacement, then of course the choices are dependent. $\endgroup$
    – lulu
    Commented Jun 24 at 16:30

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