Three dice: conditional probability After throwing $3$ dice, we know that on every die there is a different number. What is the probability that there is a $6$ on exactly one dice?

I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then 
$$P(B)=\dfrac{6∗5∗4}{6^3}\text{ and }
P(A\cap B)=\dfrac{1*5*4}{6^3}$$


I figured that P(A) - we get 6 on some dice, P(B) - different number on every dice. Then 
  $$P(B)=\dfrac{6∗5∗4}{6^3}\text{ and }
P(A\cap B)=\dfrac{1*5*4}{6^3}$$

Ah, no.  You are evaluating $A\cap B$ as the event of selecting a '6' on a particular die and two different faces for the other two dice. You want the event of selecting a '6' and two different faces for the three dice.
Let $A$ be the event that at least one of the die shows a '6'.
Let $B$ be the event that each die shows a distinct face.
So $A\cap B$ is the event of selecting a '6' and two values from the set of five remaining, while $B$ is the event of selecting three values from the set of six.
$$\begin{align}\mathsf P(A\mid B)&=\dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}\\[1ex]&=\dfrac{\binom 52}{\binom 63}\\[1ex]&=\dfrac{(5\cdot 4)/2}{(6\cdot 5\cdot 4)/(3\cdot 2)}\\[1ex]&=\dfrac{1}{2}\end{align}$$
