# What is the probability that after removing all red balls, you have exactly 1 blue and at least 1 green ball remaining?

You have a box with 10 red, 20 blue, and 30 green balls. You take out balls from the box, without returning them.

What is the probability that after you've taken out the last red ball there is exactly one blue ball and at least one green ball remaining?

• Very unclear. So the person keeps selecting balls until she takes the 10th red ball, and you want to know after that that she takes exactly (?) one blue and at least one green? How many selections is she allowed after the last red has been chosen? Please re-write your question carefully. Jan 16, 2019 at 22:51
• The way I understand the problem is, that the moment you have no more red balls remaining in the box, you have 1 blue ball in the box, and at least 1 green ball. What is the likelihood of this event to occur. Jan 16, 2019 at 22:59
• @yumer.bekir: Do you agree with my rewording of your problem? Jan 16, 2019 at 23:26
• yup. it made the problem easier to understand Jan 17, 2019 at 11:46

First let's count the probability that after you take out all the red balls, there is exactly one blue ball left. For this part, we can ignore the green balls completely, and just pretend that there are $$10$$ red balls and $$20$$ blue balls.
If there is exactly one blue ball left, that means that if we take out all $$30$$ red or blue balls, the last ball drawn is blue and the next-to-last ball is red. By symmetry, the probability of this happening is the same as the probability that the first ball we draw is blue and the second red, which is $$\frac{20}{30} \cdot \frac{10}{29} = \frac{20}{87}$$.
From this, we should subtract the probability that (in the original setting where green balls do exist) once we take out all the red balls, there is exactly one blue ball but no green balls left. Equivalently, once we take out all $$60$$ balls, the last is blue and the second-to-last is red. Equivalently, by symmetry, the first ball is blue and the second is red. This happens with probability $$\frac{20}{60} \cdot \frac{10}{59} = \frac{10}{177}.$$
So the overall probability we want is $$\frac{20}{87} - \frac{10}{177} = \frac{890}{5133}$$.