# Probability Question: Drawing balls from an urn

Question:

An urn contains 12 balls where 3 balls colored red, 3 balls colored blue, 3 balls colored green, and 3 balls colored white. If we draw 4 balls in succession without replacement, what are the probabilities that

1. the first is not red, the second is not blue, the third is not green, the fourth is not white and all four balls are different color?

2. the first is not red, the second is not blue, the third is not green, the fourth is not white, and the drawn balls may have same color?

I can answer question no. 1 using derangement and the answer is $!4*\frac{3}{12}*\frac{3}{11}*\frac{3}{10}*\frac{3}{9}=\frac{27}{440}$, but how to answer question no. 2? Any help would be appreciated. Thanks in advance.

For the second problem, we first find the probability of the complement, that the first is red or the second is blue or the third is green or the fourth is white. We use Inclusion/Exclusion.

The probability the first is red is $\frac{3}{12}$, with similar expressions for the others. If we add these $\frac{3}{12}$, we get a first approximation $\binom{4}{1}\cdot\frac{3}{12}$.

This first approximation, double counts, for example, the cases where first is red and second is blue. The probability of this is $\frac{3}{12}\cdot\frac{3}{11}$. Subtract all $\binom{4}{2}$ instances. Our second approximation is $\binom{4}{1}\cdot\frac{3}{12}-\binom{4}{2}\cdot\frac{3}{12}\cdot\frac{3}{11}$.

But we have subtracted too much, for example the cases where first is red, second is blue, and third is green. So we add back $\binom{4}{3}\cdot\frac{3}{12}\cdot\frac{3}{11}\cdot\frac{3}{10}$. Finally, we subtract the probability of red then blue then green then white.

So the probability of the complement is $$\binom{4}{1}\cdot\frac{3}{12}- \binom{4}{2}\cdot\frac{3}{12}\cdot\frac{3}{11}+\binom{4}{3}\cdot\frac{3}{12}\cdot\frac{3}{11}\cdot\frac{3}{10}-\binom{4}{4}\cdot\frac{3}{12}\cdot\frac{3}{11}\cdot\frac{3}{10}\cdot\frac{3}{9}.$$

• So the answer would be 147/440? Can we use derangement for the second problem? – Venus May 20 '14 at 5:54
• Yes, that is what I got. About derangements, I do not know. I used the first tool that came to mind. – André Nicolas May 20 '14 at 5:56
• It's ok. I can understand that the probability the first is red is 3/12 but why must we multiply by C(4,1) again? I don't get it the motivation of it. Would you mind explaining to me, please? – Venus May 20 '14 at 5:59
• I am adding the probability that the first is red, the probability the second is blue, the probability the third is green, the probability the fourth is white. Four terms, each equal to $\frac{3}{12}$. I used $\binom{4}{1}\cdot \frac{3}{12}$ instead of $\frac{3}{12}+\frac{3}{12}+\frac{3}{12}+\frac{3}{12}$ or $(4)(3/12)$ because it structurally fits in better with the next term, and the term after that, and suggests what we would do if we had more colours. – André Nicolas May 20 '14 at 6:04
• You are welcome. The point of my comment with the $5!$ is not the denominators, but the fact that all different colours is structurally a much simpler problem than the not red not blue and so on stuff. And it has a simple answer, no need for Inclusion/Exclusion. – André Nicolas May 20 '14 at 6:25