Removing all red balls from an urn

Question

An urn contains $r$ red balls, $b$ blue balls, and $g$ green balls, $r,g,b \geq 1$. The balls are removed one at a time without replacement. What is the probability that the red balls are the first to be removed completely? (That is, the probability that all the red balls are removed and there are still some blue balls and some green balls in the urn.) Hint: Consider all ways of removing all the balls and condition on the color of the final ball.

Solution

Answer 1

Consider the last ball drawn. It will be green with probability $\dfrac{g}{r+b+g}$, noting that the probability for the last ball is the same as the probability for the first. Similarly, the probability the last ball is blue will be $\dfrac{b}{r+b+g}$

Given that the last ball is green, the probability that there is at least one blue after the last red is going to be the probability the last ball if we were considering only the red and blue balls and ignoring the greens will be $\dfrac{b}{r+b}$. A similar argument is made in the other case.

Our probability is then:

$$\left(\dfrac{g}{r+b+g}\times \dfrac{b}{r+b}\right) + \left(\dfrac{b}{r+b+g}\times \dfrac{g}{r+g}\right)$$

or if desired to rearrange, could write it as something like $\dfrac{bg}{r+b+g}\left(\dfrac{1}{r+b}+\dfrac{1}{r+g}\right)$