Question

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?

Short answer

There is a one in$300$ million chance of that happening.

Step-by-step solution

Given Information.

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

Explanation.

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

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