# Urn problem: Probability of choosing balls with unique colors (nonuniform number of balls per color)

combinatoricsprobability

I am looking for a formula to efficiently compute the probability of choosing $$I$$ balls with distinct colors from an urn with $$N$$ balls and $$K$$ colors, where $$K$$ might be larger than $$I$$, without replacement.

As a concrete example: Let's assume we have 10 blue, 139 white, 44 green, and 1 red balls (so $$N = 194$$ and $$K = 4$$). Let's assume that we want to draw 2 balls with distinct colors (so $$I=2$$). For this example, it is tractable to enumerate all possibilities. For example, I computed the probability of first choosing a blue ball and then either a white, green, or red one by computing $$\frac{10}{194} * \frac{139+44+1}{193}=0.049$$. By computing also the probabilities for the other combinations and summing them all up, I obtained an overall probability of about $$43\%$$.

In the actual case I want to compute this for, I have a significantly larger number of $$K$$ and $$N$$. Also, $$I$$ is about $$K/2$$, making it infeasible to naively enumerate the different probabilities. Is there a more efficient way to compute this?

Just use generating functions , lets say that one of the color has $$f$$ balls in a color , so the generating function of this color is $$\binom{f}{1}x$$

Lets assume that we have $$K$$ distinct color of balls and they have $$f_1 ,f_2,f_K$$ balls.Moreover , we want to choose $$I$$ balls with distinct colors , so find the coefficient of $$x^I$$ in the expansion of $$\bigg[1+\binom{f_1}{1}x\bigg]\bigg[1+\binom{f_2}{1}x\bigg]..\bigg[1+\binom{f_K}{1}x\bigg]$$.

Moreover, we know that the denominator is $$\binom{N}{I}$$ ,because all balls are seem distinct when computing probability , so the answer is $$\frac{[x^I]\bigg[\bigg(1+\binom{f_1}{1}x\bigg)\bigg(1+\binom{f_2}{1}x\bigg)..\bigg(1+\binom{f_K}{1}x\bigg)\bigg]}{\binom{N}{I}}$$

$$\mathbf{EDITION:}$$ For $$I=2$$

• Generating function for $$10$$ blue ball : $$1+\binom{10}{1}x$$

• Generating function for $$139$$ white ball : $$1+\binom{139}{1}x$$

• Generating function for $$44$$ green ball : $$1+\binom{44}{1}x$$

• Generating function for $$1$$ red ball : $$1+\binom{1}{1}x$$

where $$1$$ represent $$x^0$$ ,i.e, we did not make selection , so by expansion

In link , we see that the coefficient of $$x^2$$ is $$8139$$ , so the answer is $$\frac{8139}{C(194,2) =18,721}= 0,434752417$$