Multinomial Distribution - Special Case

Question

Suppose you roll 6 ’fair’ dice. Let event A be the case where exactly two dice show a number less than 3 (1 or 2). Let event B be the case where where exactly two dice show a number greater than 4 (5 or 6)) Let event C be the case where every die shows a different number. What is probability of rolling the 6 dice and seeing both the A and B conditions? This seems to me like I should use a Multinomial Distribution but is there a way to do so for a problem like this without needing to calculate and sum the probabilities for every possible permutation that would yield this outcome?

Answer 1

It is indeed a multinomial distribution problem where the options are $A$ ($1$ or $2$), $N$either ($3$ or $4$) or $B$ ($5$ or $6$). Each has a $\frac13$ chance of occuring for each die, and we want to find $P(N=A=B=2)$: $$\binom6{2,2,2}(1/3)^2(1/3)^2(1/3)^2=\frac{90}{3^6}=\frac{10}{81}$$ Note that the problem's descriptions of events $A$ and $B$ specify exactly two dice, not "at least one". Thus only a single multinomial probability needs to be found.