Balls are randomly removed from an urn initially containing $20$red and $10$blue balls. What is the probability that all of the red balls are removed before all of the blue ones have been removed?

Balls are randomly removed from an urn initially containing $20$red and $10$blue balls.

The outcome space $S$contains every possible order (permutation) of $30$differentiable balls.

If all events $S$are considered equally likely, the probability of an event $A\subseteq S$is:

$P\left(A\right)=\frac{\left|A\right|}{\left|S\right|}$

where localid="1649489216671" $\left|X\right|$denotes the number of elements in $X$

It is known that the number of permutations of $30$different elements is $30!=\left|S\right|$

Let's name $A$the event that all red balls are removed before the blue ones.

The $20$red balls can be the first $20$drawn in$20!$permutations. Whichever permutation of$20$balls is in the beginning the remaining$10$(blue) balls can be rearranged in $10!$ways By the basic principle of counting, there are localid="1649489058748" $20!·10!$elements$A$, so

$P\left(A\right)=\frac{20!·10!}{30!}=\frac{1}{30045015}\approx 3.33·{10}^{-8}$.