# Probability of drawing less than k black balls without replacement

I am dealing with a problem that resembles the following modeling:

Given a bag of $n$ balls constitute of $b$ black balls and $r$ red balls, we are choosing $\alpha$ many balls uniformly without replacement. What is the probability of having at most $k$ many black balls?

P.S: I know that the problem is broken down into the following formula. What we are seeking for is a very simplified version of the following formula to decide on $\alpha$ given $b$, $r$, $n$, and a target $p$.

$p = Pr(num_{black} < k) = \frac{\sum_{i=0}^{k-1} \binom{b}{i} \times \binom{r}{\alpha-i}}{\binom{n}{\alpha}}$

• The formula is the prob. of less than k black balls, not at most k black balls. And, the second choice should be from $n-b$, not $b$ Sep 14, 2018 at 8:38
• @gunes Thanks, edited, however, the second choice is correct I believe. Sep 14, 2018 at 8:46
• On the contrary, you've corrected the second choice from $b$ to $r = n-b$, so it is now correct. But, you still ask "What is the probability of having at most k many black balls?", which is not equal to the probability formula you wrote. You need $P(num_{black} \leq k)$ Sep 14, 2018 at 8:55

It depends a bit on your values for the parameters, but the normal approximation should work well here. The expected value and variance is

$$E(num_{black})=\alpha \frac{b}{n}=\mu$$ $$var(num_{black})=\alpha \frac{br}{n^2}\frac{n-\alpha}{n-1}=\sigma^2$$

You approximate the probaability as

$$Pr(num_{black}<k)=Pr(\frac{num_{black}-\mu}{\sigma}<\frac{k-\mu}{\sigma})\approx Pr(Z<\frac{k-\mu}{\sigma})=\Phi\left(\frac{k-\mu}{\sigma}\right)$$

(Note...$Z$ is a standard normal random variable). Now you want to set this equal to $p$ and you will get... $$p=\Phi\left(\frac{k-\mu}{\sigma}\right)$$ $$\Phi^{-1}(p)=\left(\frac{k-\mu}{\sigma}\right)$$

This is a non-linear function of $\alpha$. But you can get a good estimate by first assuming that $n>>\alpha$ which leaves you with quadratic formula...I.e. $\alpha=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a,b,c$ will be functions of your parameters $n,r,b,p$.

You can check this solution by manually plugging in the $\alpha$ into your exact formula. If it works you should get around $p$ as your answer. Then you might find a small line search is needed.

• @Price of Parse, if you're not seeking an exact formula, this is the best answer. Sep 14, 2018 at 9:10

This is actually Hypergeometric Distribution's CDF. Its closed form will not be as easy as you want, i.e. will include Generalized Hypergeometric Function.