There are 10 red balls, 20 green balls, 30 blue balls in an urn. We take them all out one by one without replacement. What is the probability that by the time we take all the blue balls out there is at least one red ball and one green ball left in the urn?


By symmetry, a particular order is just as likely as its reversal. And in the reversal, we pick at least one red and one green ball before we pick a blue one. But the probability of this is easy to calculate:

The probability that red is the first colour to be picked, followed by green, is

$$\frac{10}{60}\times \frac{20}{50}=\frac{1}{15}$$

And the probability that green is the first colour to be picked, followed by red, is

$$\frac{20}{60}\times \frac{10}{40}=\frac{1}{12}$$

So the answer is $\frac{1}{15}+\frac{1}{12}=\frac{3}{20}$.

  • $\begingroup$ A nice use of symmetry. $\endgroup$ – N. F. Taussig May 22 at 23:54

I'm thinking there's an easier way to attack this problem, but I don't see it at the moment. Here's one way to attack it.

Enumerate the states just before you draw the last blue ball. (There are a lot of them.) One of them is $4$ red, $6$ green, and $1$ blue.

What's the probability that you arrive at this state? In other words, what is the probability of drawing $10-4=6$ red balls, $20-6=14$ green balls, and $30-1=29$ blue balls, in any order? Then, what's the probability that the next ball you draw is the final blue one?

Then you rinse and repeat for every pair of possible red and green ball numbers.

Very tedious, but it should get you to the correct answer.

If you're allowed to express your answer with summations, then just figure out the probability $P(m,n)$ for one of the states in general (there are $m$ red balls and $n$ green balls after you draw the final blue ball).

Then your total probability is just the sum of all of these:

$$P = \sum_{m=1}^{10}\sum_{n=1}^{20}P(m,n).$$


There are alltogether $$ N=\binom {60}{10,20,30} $$ ways to arrange the balls.

From them $$ M=\sum_{i=0}^{9}\sum_{j=0}^{19}\binom {i+j+29}{i, j,29}\binom {30-i-j}{10-i, 20-j} $$ ways lead to success (the last 30-th blue ball has the number $i+j+30$), so that the probability in question is $$\frac MN=\frac 3 {20} .$$

  • $\begingroup$ How do you propose to evaluate that $200$-element sum? $\endgroup$ – TonyK May 22 at 20:23

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