# Are these two events independent and if they are, do they have the same probability?

This is the question for context: You have 12 marbles: 5 are vermillion, 4 are emerald, and 3 are ultramarine. You divide them at random in four groups of 3. Let X be the number of groups that contain 3 marbles of all distinct colors (one marble of each color). Find Var(X).

The way I go about solving this:

$$Var(X) = E(X^2) - (E(X))^2$$

Let $$X_i$$ be an indicator variable that is $$1$$ if the $$i^{th}$$ group has four distinct marbles and $$0$$ if not. Then,

$$X = \sum_{i=1}^{4}{X_i}$$

Then,

$$E(X) = E\left(\sum_{i=1}^{4}{X_i}\right) = \sum_{i=1}^{4}{E(X_i)} = \sum_{i=1}^{4}{P(\text{group i having all distinct marbles})}$$

Let's define $$A_i$$ to be the event that group $$i$$ has 3 distinct marbles. Then, I need $$P(A_i)$$.

I found this probability to be $$\frac{5 \times 4 \times 3}{12 \times 11 \times 10} \frac{1}{3!}$$, so the expected value was $$0.3030$$.

To find the other term in the equation for variance, I did the following:

$$E(X^2) = E\left( \sum_{i = 1}^{4}{X_{i}^{2}} + \sum_{i \ne j}^{4}{X_{i} X_{j}} \right) = E\left( \sum_{i = 1}^{4}{X_{i}^{2}}\right) + E\left(\sum_{i \ne j}^{4}{X_{i} X_{j}} \right) = \sum_{i = 1}^{4}{E(X_{i}^{2})} + \sum_{i \ne j}^{4}{E(X_{i} X_{j})} = \sum_{i = 1}^{4}{P(A_{i})} + \sum_{i \ne j}^{4}{P(A_{i} \cap A_{j})}$$

Now, my main question is if the events $$A_i$$ and $$A_j$$ are independent and have the same probability. My thoughts are yes they are independent, since these marbles are not "drawn one at a time", but rather are simply randomly split into groups. And wouldn't the probability of one group having 3 distinct marbles be the same no matter the group? However, I am having second guesses come into play because if one group did indeed have 3 distinct marbles, doesn't that say something about the probability of one of the other groups having 3 distinct marble?

• They have the same probability distribution, but they are not independent: If $A_1 = A_2 = A_3 = 1$, then $A_4 = 0$ necessarily. Commented Apr 10, 2020 at 1:53

No, the event $$A_i$$ and $$A_j$$ are not independent. You can see this trivially by noting that if there is only one marble of the third color, it is never possible for there to be more than one group of marbles to contain all three colors. So if $$X_i = 1$$ for group $$i$$, we must have $$X_j = 0$$ for all $$j \ne i$$. The fact that this dependence results from a nontrivial choice of distribution of marble colors indicates that it exists in the given distribution of colors stated in the question.
• So in order to finish the calculation of the variance, I would need to find $P(A_i | A_j)$, correct? And this could be found as $\frac{4 \times 3 \times 2}{9 \times 8 \times 7} \frac{1}{3!}$ since given $A_{j}$, there are 3 less marbles total and 1 marble less from each color? Commented Apr 10, 2020 at 2:14