# Removing all red balls from an urn

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

• This is wrong. Your answer was the probability that every red ball is removed before even a single blue or green ball is removed. You should be finding the probability that the last red ball is removed before the last blue ball and before the last green ball is removed, but it is perfectly acceptable for some blue balls and/or some green balls to be removed before the last red ball so long as it is not the case that all blue balls are removed before the last red ball, etc... Commented Nov 3, 2023 at 14:45
• Your problem already provides a good hint. Rather than thinking about pulling the balls in order from start to finish... try thinking about pulling the balls in the opposite order from last to first. Commented Nov 3, 2023 at 14:47
• The hint is not that helpful unfortunately, since exhausting red first is not the same as having a different color in last place... Commented Nov 3, 2023 at 14:52
• As for giving you a full answer... you have the image labeled "Midterm - I". Although you did share your attempt (flawed though it may be), this seems suspiciously like you are asking for help on your midterm exam which surely is against your institution's academic integrity policy and would be considered cheating. After enough time has passed, I could easily explain the full solution but I will opt not to for a few days at least in order to not help facilitate obvious cheating. Commented Nov 3, 2023 at 14:52
• @Surge no, it is incredibly helpful. The final answer will be easily expressible as the sum of products of two simple fractions. The punchline is that if the last ball is green, all you have to worry about then is whether when considering only the reds and blues if the blue is the last or not among those. Commented Nov 3, 2023 at 14:53

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 $$\dfrac{bg}{r+b+g}\left(\dfrac{1}{r+b}+\dfrac{1}{r+g}\right)$$

• In other words, in reverse order, it's the probability that red is the last color to appear.
– bof
Commented Nov 7, 2023 at 4:23