Conditional probability of rolling three dice

Question

Three dice are all unbiased. We roll them at once. What's the probability of obtaining

at least one 6 supposing that all three outcomes differ from each other?

Let $B$ denote the event that all outcomes are different from each other!

Let $A$ denote the event that at least one 6 is obtained!

The relevant formula is: $P(A|B) = \frac{P(AB)}{P(B)}$.

$$P(B) = \frac{6\cdot 5 \cdot 4}{6 \cdot 6 \cdot 6}$$.

$$P(AB) = \frac{(5\cdot 4) + (6\cdot 4) + (6\cdot 5)}{6 \cdot 6 \cdot 6}$$.

Question: Is my computation correct?

Answer 1

No. The answer is 1/2. Your computation of $P(B)$ is correct. Now if you have a six, there are three possibilities for the die giving the 6 and then $5\times 4$ possibilities for the two other dice. So $P(A \cap B)/P(B) = 60/120$.

Answer 2

Alternative route:

Set $\{1,2,3,4,5,6\}$ has $\binom63=20$ (equiprobable) subsets that have cardinality $3$ and $\binom52=10$ of those subsets contains a $6$.

So the probability is $\frac{10}{20}=\frac12$.