Probability of drawing at least 1 red, 1 blue, 1 green, 1 white, 1 black, and 1 grey when drawing 8 balls from a pool of 30?

Question

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.

Answer 1

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 $0.1922$.

Answer 2

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.

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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.

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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.

Answer 3

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

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

$=\binom{6}{1}\binom{25}{8}-\binom{6}{2}\binom{20}{8}+\binom{6}{3}\binom{15}{8}-\binom{6}{4}\binom{10}{8}$,

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

Answer 4

Similar problems appear in

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

and

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 $$ \frac{5000}{26013} $$