If I have 2 balls and 8 bins and I define the following indicator variable: $X_i = 1$ if bucket $X_i$ is empty and 0 otherwise. I'm trying to find the following quantity: $E[X_1X_2]$ How would I go about doing this since the two indicator variables are not independent.
I understand that $E[X_1 X_2] = Pr[X_1 = 1, \ X_2 = 1]$ but I'm struggling quite a lot calculating that probability. I've reached the following:
$Pr[X_1 = 1 \cap \ X_2 = 1] = Pr[X=1] + Pr[X=2] - Pr[X_1 = 1 \cup \ X_2 = 1]$
Where $Pr[X=1] = Pr[X=2] = \frac{7.6}{8^3}$ and I'm struggling to calculate the final quantity. Is this the correct approach, or is there a much simpler way?
If the placement of each ball is independent and uniformly at random among all $8$ bins, then the outcome $X_1 X_2 = 1$ means both bins $1$ and $2$ are empty. This occurs with probability $(6/8)^2$.
If multiple balls are not permitted to be placed in the same bin, but again the placement is uniformly at random among all empty bins--that is to say, any pair of bins is equally likely to be selected for the two balls--then the probability that both bins $1$ and $2$ are empty is simply $\binom{6}{2}/\binom{8}{2}$.