Probability... Urn problem with multiple colors, multiple draws and a given condition

Question

I have a real life probability problem, but I will break it down to an urn problem.

• The situation: There are $30$ balls in total. They have $5$ different colors. There are two balls from each color #1 through #4. The other $22$ balls are all of color #5.
• The question: What is the probability of having at least $1$ ball from color 1# through 4# after $n$ draws (without putting the balls back into the urn)?

Sadly real life seems to me more complex than stuff I did in school :( Could you explain how to approach this problem?

Answer 1

You can do this using inclusion-exclusion.

The probability to have $k_i$ particular balls of colour $i$ with $1\le i\le4$ after $n$ draws is

$$ \frac{\binom{30-\sum_ik_i}{n-\sum_ik_i}}{\binom{30}n}\;, $$

where, contrary to convention, binomial coefficients with negative lower index are taken to be zero.

Then by inclusion-exclusion the desired probability is

$$ \sum_{k_1,k_2,k_3,k_4=1}^2(-1)^{\sum_ik_i}\prod_i\binom2{k_i}\frac{\binom{30-\sum_ik_i}{n-\sum_ik_i}}{\binom{30}n}\\ =\binom{30}n^{-1}\left(16\binom{26}{n-4}-32\binom{25}{n-5}+48\binom{24}{n-6}-8\binom{23}{n-7}+\binom{22}{n-8}\right)\;. $$