Q. 2.2 - A First Course in Probability

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?

Step-by-Step Solution

Answer

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

Step 1 Given Information.

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

Step 2 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 $\left|X\right|$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 $20! · 10!$elements$A$, so

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