First I made a mistake of not taking into account that this event is dependent.
So to get two '$6$' the probability would be:
$$ P_5(2) = \frac{5!}{2!(5-2)!} \cdot \left(\frac16\right)^2\cdot\left(\frac56\right)^3 $$
Where
- $p = \dfrac16$
- $q = \dfrac56$
- $N= 5$ (the number of elements of the system)
Then there are $3$ dice left on the table and we want to know the probability that one of them is a '$5$'. We rule out the two dice with the '$6$' face on. So the number of element in the system is now $3$. So the probability would be:
$$ P_3(1) = \frac{3!}{1!(3-1)!} \cdot \left(\frac16\right)^1\cdot\left(\frac56\right)^2 $$
But then the professor told me that $p=\dfrac15$ and not $p=\dfrac16$ That's what I don't get.
Of course at the end we multiply those two probabilities to get the final probability that we want.
Best Answer
I am a beginner so see if this makes sense to you. If not let me know. Here are two ways you can solve this.
Method 1
You can solve this using the joint pmf for a multinomial distribution.
$$P(X_6 = 2, X_5 = 1, X_{other} = 2) = \frac{5!}{2!1!2!} \left( \frac{1}{6} \right)^2 \left( \frac{1}{6} \right) \left( \frac{4}{6} \right)^2 \approx .062$$
Method 2
Alternatively you can solve this using the multiplication rule. Here you get the $1/5$ that your professor talked about because when you condition you are on a reduced sample space.
Let $E$ be the event you get two sixes.
Let $F$ be the event you get one five.
Let $G$ be the event you get two of the others.
$$P(EFG) = P(E)P(F \mid E)P(G \mid EF) $$
$$P(E) = {5 \choose 2}\left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^3 $$
$$P(F \mid E) = {3 \choose 1}\left( \frac{1}{5} \right) \left( \frac{4}{5} \right)^2$$
$$P(G \mid EF) = {2 \choose 2}\left( \frac{4}{4} \right)^2 \left( \frac{0}{4} \right)^0 = 1$$
$$P(EFG) \approx .062$$