Given a pool of 30 balls (5 of each color). When drawing 8 balls without replacement, what is the probability of getting at least one of each color?

Related: Probability of drawing at least one red and at least one green ball.

When drawing more than 2 colors you need to exclude overlapping 'hands'. Thus when finding the probability of drawing no red, you can have a hand made up of blue, green, white, black and grey. But when you are determining the probability of drawing no blue you draw from red, green, white, black, grey. So you need to exclude all green, white, black, grey hands as they have already been counted. And the same for the other colors as well.

The other complexity of the problem is that since there are only 5 of each color, no draw will only include balls of the same color.

  • $\begingroup$ At least one in this case means exactly one. so we want the probability they are of different colours. So you want the probability they all have different colours. What is the probability the second ball has a different colour than the first? $\endgroup$ – André Nicolas Aug 15 '14 at 21:29
  • $\begingroup$ I apologize, the other condition was that it must be at least. I changed the question so that we are drawing 8 balls instead of 6 now. $\endgroup$ – Rovert Renchirk Aug 15 '14 at 21:31

The quickest approach is to count ways to get $6$ colors on $8$ balls. There are essentially two cases, $\langle 3,1,1,1,1,1\rangle$ and $\langle 2,2,1,1,1,1\rangle$. There are $\binom{6}{1}\binom{5}{3}\binom{5}{1}^5=187,500$ ways to get the first case. The $\binom{6}{1}$ counts the number of ways of choosing one color to get three balls, and the rest is the number of ways of choosing three balls from that one color and one for each of the other colors.

The second case has $\binom6 2\binom 5 2^2\binom 5 1^4=937,500$ different ways. There are $\binom 6 2$ ways to choose two colors to receive two balls, and the rest is the number of ways of choosing two balls from each of those colors and one ball from the others.

So the total cases are $1,125,000$, out of $\binom{30}{8}=5,852,925$. That gives a probability of about $0.1922$.

  • $\begingroup$ nice, simple and fast $\endgroup$ – Nikos M. Aug 15 '14 at 22:15

Edit: The question has changed. We leave the answer to the original question, which specified $6$ draws. We append a sketch for the new question.

We want the probability that the balls all have different colours. Imagine (it makes no difference) that we draw the balls one at a time.

Whatever the first ball was, the probability the second has a different colour is $\frac{25}{29}$.

Given that the second ball had a different colour from the first, the probability we get a new colour on the third is $\frac{20}{28}$.

Thus the probability that the first three balls have different colours is $\frac{25}{29}\cdot \frac{20}{28}$.

Given that the first three are of different colours, the probability of a new colour on the fourth is $\frac{15}{27}$. So the probability the first four balls are of different colours is $\frac{25}{29}\cdot \frac{20}{28}\cdot \frac{15}{27}$.

Continue. We are close to the end.

Drawing $8$: There are $\binom{30}{8}$ equally likely ways to draw $8$ balls. We count the number of favourable draws, in which we get all $6$ colours. Since $8$ is not much larger than $6$, a division into cases is efficient. For larger numbers, we would suggest using Inclusion/Exclusion.

These favourables are of two types.

(i) One each of $5$ colours, and $3$ of another. The abundant colour can be chosen in $\binom{6}{1}$ ways, and for each of these the actual balls can be chosen in $\binom{5}{3}$ ways. The rest of the balls can then be chosen in $\binom{5}{1}^5$ ways. Multiply.

(ii) One each of $4$ colours, and $2$ each of $2$ others. The lucky colours can be chosen in $\binom{6}{2}$ ways. For each way, the actual balls can be chosen in $\binom{5}{2}^2$ ways. And the remaining $4$ can be chosen in $\binom{5}{1}^4$ ways.

  • $\begingroup$ I apologize, I changed my question so that it is not a trivial answer. We are drawing 8 balls instead of 6 so this solution no longer works. I appreciate the answer though! $\endgroup$ – Rovert Renchirk Aug 15 '14 at 21:43
  • $\begingroup$ I like this - it is simpler than using Inclusion-Exclusion. $\endgroup$ – user84413 Aug 15 '14 at 22:03
  • $\begingroup$ @user84413: Well, I don't think that you should. Except in the case where the number of "extra" balls drawn is small, Inclusion/Exclusion is preferable. The division into cases approach rapidly becomes unwieldy. $\endgroup$ – André Nicolas Aug 15 '14 at 22:22

For the new version of the problem, we can use Inclusion-Exclusion:

If $E_i$ is the set of draws without at least one ball of color i, we have

$|E_1\cup\cdots\cup E_6|=\sum|E_i|-\sum|E_i\cap E_j|+\sum|E_i\cap E_j\cap E_k|-\cdots$


so the probability will be $\displaystyle1-\frac{\binom{6}{1}\binom{25}{8}-\binom{6}{2}\binom{20}{8}+\binom{6}{3}\binom{15}{8}-\binom{6}{4}\binom{10}{8}}{\binom{30}{8}}=\frac{5,000}{26,013}$.

  • $\begingroup$ it would be nice to calculate the result of the expresion, i put into wolframalpha but did not a result back, can you do it? $\endgroup$ – Nikos M. Aug 15 '14 at 22:17
  • 1
    $\begingroup$ @Nikos M. I added the calculation, as you requested. $\endgroup$ – user84413 Aug 15 '14 at 22:48
  • $\begingroup$ great result is $\frac{5,000}{26,013} = 0,19221158651443509014723407527006$, as other answer $\endgroup$ – Nikos M. Aug 16 '14 at 0:22

Similar problems appear in

S. Ghahramani, Fundamentals of Probability with Stochastic Processes, 3rd ed. 2005. p73


P. J. Nahin, Digital Dice: Computational Solutions to Practical Probability Problems 2008., p. 237

using their methods and Mathematica which is not as succinct as user84413 answer I get $$ \frac{5000}{26013} $$

  • $\begingroup$ is this an answer? $\endgroup$ – Nikos M. Aug 15 '14 at 21:54
  • $\begingroup$ @Nikos yes it is. Both books can provide much background and their methods verify an answer already given. Since there are several different answers a verification is important. $\endgroup$ – bobbym Aug 15 '14 at 22:02
  • $\begingroup$ I'd call it a partial answer. Ideal answers here are self-contained, and at least state the theorems they are using to do these computations. (You get the same result I got, so at least it's a good check to my answer. :)) $\endgroup$ – Thomas Andrews Aug 15 '14 at 22:12
  • $\begingroup$ i were asking this question i would not consider this an answer $\endgroup$ – Nikos M. Aug 15 '14 at 22:12
  • $\begingroup$ I would not provide that as answer but as addition... I am not begging for an upvote here,I am just trying add something useful to the thread. Answers have already been provided but perhaps the OP would like to see the detailed solutions to this problem in those books. In my opinion Thomas' and user84413 are better than the book answer as I understand it. $\endgroup$ – bobbym Aug 15 '14 at 22:27

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