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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?

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  • $\begingroup$ 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. $\endgroup$ Commented Jan 16, 2019 at 22:51
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    $\begingroup$ 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. $\endgroup$ Commented Jan 16, 2019 at 22:59
  • $\begingroup$ @yumer.bekir: Do you agree with my rewording of your problem? $\endgroup$ Commented Jan 16, 2019 at 23:26
  • $\begingroup$ yup. it made the problem easier to understand $\endgroup$ Commented Jan 17, 2019 at 11:46

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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 $\frac{20}{87} - \frac{10}{177} = \frac{890}{5133}$.

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  • $\begingroup$ How would the solution change, if we make the following change: What is the probability that after you've taken out the last red ball there is at least one blue ball and at least one green ball remaining $\endgroup$ Commented Jan 19, 2019 at 21:37
  • $\begingroup$ Look at (1) the probability that when we take the last red ball, no blue balls are left (2) the probability that when we take the last red ball, no green balls are left (3) the probability that both of these occur (so the last red ball is the last ball). $\endgroup$ Commented Jan 19, 2019 at 23:57

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