<s>[INST] We throw two dices distinc. What is the probability that there will be at least a 6 given that the dice show different results? We throw two dices distinc. 
What is the probability that there will be at least a 6 given that the dice show different results?
My work
Let 
$A=$We have a six, 
$B=$Different result
Then $P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{1}{6}$
This because: $P(B)=\frac{30}{36}$ and $P(A\cap B)=\frac{5}{36}$
Is good this?
 [/INST] $A\cap B =$ one 6 and one non-6 = $\{(1,6), (2, 6), (3,6), (4,6), (5,6), (6,1), (6,2),(6,3), (6,4), (6,5)\}$.
So $|A\cap B| = 10, P(A\cap B)=\frac{10}{36}, P(A|B) = \frac{10}{30} = \frac{1}{3}$.
(Incidentally, $|A| = 11$, not $12$, because $A = (A \cap B) \cup \{(6,6)\}$.)
</s>