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 
$$P(B)=\dfrac{6∗5∗4}{6^3}\text{ and }
P(A\cap B)=\dfrac{1*5*4}{6^3}$$

The required probability is $$\frac {\binom 3 1 \times 5 \times 4} {6 \times 5 \times 4} = \frac 3 6 = \frac 1 2.$$
