Expected Value and Variance of Marble Combinations You have 12 marbles: 5 are vermillion, 4 are emerald, and 3 are ultramarine. You divide them at random in four groups of 3 to give to your four best friends. Let $X$ be the number of friends that receive 3 marbles of all distinct colors (one marble of each color). Find $E[X]$ and $Var[X]$.
I'm having trouble with my approach. I started by defining an indicator random variable $$X_i = \begin{cases}1 &\text{if $i-th$ friend gets all 3 distinct marbles}\\0 &\text{otherwise}\end{cases}$$
Let $A_i$ denote the event that friend $i$ receives 3 distinct marbles. Then $E[X_i]=P(A_i)$ and $E[X]=\sum_1^4E[X_i]=\sum_1^4 P(A_i)$. The problem is 2-fold. I'm first having trouble computing $P(A_i)$. I thought it would involve the multinomial, namely: $$P(A_1)=\frac{3!}{\binom{12}{3,3,3,3}}$$ since the total number of arrangements comes from dividing 12 objects into 4 groups of 3 and there are 3! ways to arrange 3 distinct marbles. But I'm assuming the denominator is incorrect since the objects are not all distinct.
Then the issue is computing subsequent probabilities since they are not independent. The event that the first friend gets 3 distinct marbles should certainly influence the probability that the second friend receives 3 distinct marbles. Not sure how to change my approach.
 A: When you calculate a probability as the quotient of the number of favourable cases to the total number of cases, you need to make sure you’re counting the same sort of things in the numerator and the denominator. In the numerator, you’re counting the different orders of the marbles given to one friend, whereas in the denominator you’re counting the different selections of marbles given to all friends, ignoring the order within each friend’s gift – those are two quite different counts, so their ratio doesn’t have any significance and isn’t the probability you want.
You don’t need to distinguish outcomes according to the order within each friend’s gift (all outcomes have the same number of such orders, $3!^4$, so this factor would just cancel in the quotient if you included it); and neither do you need to distinguish the ways to divide the remaining $9$ marbles among the remaining $3$ friends (again because the count of these ways is the same no matter which marbles you give to friend $i$, so it would just cancel in the quotient). All you need is the ratio between the favourable selections for friend $i$ and the total number of selections for friend $i$.
There are $5\cdot4\cdot3$ different triples of marbles with distinct colours (since you can independently choose one marble of each colour), and a total of $\binom{12}3$ triples of marbles selected from the $12$ marbles, so
$$
E[X_i]=P(A_i)=\frac{5\cdot4\cdot3}{\binom{12}3}=\frac3{11}.
$$
I’m not sure what you mean by “subsequent probabilities” that are not independent. This is enough to find $E[X]$; to find $\operatorname{Var}[X]$ you also need $E\left[X_i^2\right]=E[X_i]$ (since $X_i^2=X_i$) and $E[X_iX_j]=P(A_i\cap A_j)$ for $i\ne j$, which you need to determine separately similarly to how I determined $E[X_i]$ above.
