Drawing from an urn - probability of drawing maximum 2 of each color.

Question

I have an urn with 100 balls; 10 colors with each 10 balls.

Now I draw 10 balls from the urn without replacement - I want to calculate the probability of getting 3 or more of any color.

How do I calculate this? My problem is that for example the event "more than 3 greens" and "more than 3 yellows" are not exclusive, there are possible draws which fit into both categories. So I cant the hypergeometric distribution and multiply it by 10.

I feel like this question is school level.... I got a math degree, so the answer doesnt need to be super basic, I just suck at stochastics. Help greatly appreciated.

Answer 1

There are $\frac{100!}{90!}$ possible combinations. To get an answer compute the number of combinations $(C)$where there are at most $2$ of any color. The combinations can be broken down into $6$ sets depending on how many balls appear twice. These are $A_k=90^k\times 10^{10-2k}$ and The number of permutations for each $k$ are $B_k=\frac{10!}{(k!)^2(10-2k)!}$ $C=\sum_{k=0}^5 A_kB_k$, giving $Q=\frac{90!C}{100!}$ as the probability of no more than $2$ of any color and the desired result is $P=1-Q$

Calculating $Q$ i s tedious but not impossible.

Answer 2

The probability you are looking for can be found by inversion: it is one minus the probability of drawing two or less of each color, which is much easier to calculate since a draw has to satisfy the condition for each color rather than at least for some.

What combinations can you have that would count in? There can be 10 balls of 10 different colors $(1,1,\ldots,1)$, or 2 of the same color + 8 uniquely colored ones $(2,1,\ldots,1,0)$, or $(2,2,1,\ldots,1,0,0)$, etc. Write down the probability of each case, denoting each by the number $i \in \{0, \ldots, 5\}$ of same-color couples: \begin{align} p_i &= \frac{90!}{100!} \cdot 10^{10-i} \cdot 9^{i} \cdot (10-i)! \cdot \frac{10!}{2^i}. \end{align}

Here the first term is the probability of each particular 10-ball draw (with the order of drawing kept important), the second term reflects the probability of drawing a new color the first time, and the third term, the second time; the fourth term is for all the ways to assign a particular color the number of occurrences in the draw, and the rest account for the fact that the order of drawing is actually irrelevant.

Important: I haven't done combinatorics in a while, so my exact formula for $p_i$ may be flawed, but I stand for the overall logic.

Then the probability you're looking for is $1 - \sum_{i=0}^{5} p_i$.

Answer 3

Observe that the event you are looking for is complementary to the event "there are no more than two balls of the same color". In terms of the generating functions the number of combinations corresponding to the latter event is: $$ N_2=[x^{10}]\left(1+\binom{10}1x+\binom{10}2x^2\right)^{10}, $$ where the function $[x^n]$ extracts the coefficient at $x^n$ in the following expression.

Hence the probability in question is $$ 1-\frac{N_2}{\binom{100}{10}}\approx0.53. $$

EDIT:

In general case we can assume that the balls are of $m$ colors, and there are $N_i$ balls of $i$-th color. The question is: how many combinations of $n$ drawn balls contain no more than $k_i$ balls of the $i$-th color ($k_i$ are given for each $i$, they must not be the same). The answer to the question is: $$ [x^n] \prod_{i=1}^m \sum_{k=0}^{k_i}\binom{N_i}k x^k. $$