# Conditional binomial problem

An urn contains black balls and white balls.

We know that if we draw randomly with replacement, for example $N = 100$ balls, we have a probability greater than $P$ (example $P= 0.60$) to get more than $k$ (example $k = 33$) black balls.

What is the minimum initial percentage $R$ of the number of black balls contained in the urn by the total number of balls?

Is there an explicit formula $R(N, k, p)$ giving $R$ as a function of $N,\;k,$ and $p$?

Same questions if we perform the draw of the $N$ balls without replacement

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The question seems not very well posed to me. The "probability greater than" bit should be "probability equal to". Further, the relation with the title is not clear. – leonbloy May 30 '11 at 19:14

## 1 Answer

If you draw randomly with replacement $N$ balls and assume the probability of getting one black ball in one draw is $r$ then the probability of drawing at least $k$ black balls is $N$ draws is: $$P=\sum_{t=k}^{N}\left(\binom{N}{t}\cdot r^t \cdot (1-r)^{N-t}\right)$$ (You get it by setting $t$ to be the actual number of black balls you draw ($t\geq k$), choosing the $t$ appearances of the black balls, multiplying by the probabilities and dividing by total number of possibilities).
From here, for given $N,k,P$ you can find $r$, but it's not that nice and not explicit... :(

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Are you sure about /2^N. It seems to me that we do not have to divide by 2^N . – Jean-Pierre Jun 1 '11 at 0:29
@Jean-Pierre: Thanks, you are right – Dennis Gulko Jun 2 '11 at 15:07