probability of obtaining at least one 6 if it is known that all three dice showed different faces Three dice are rolled. What is the probability of obtaining at least one 6 if it is known that all three dice showed different faces? The answer is 0.5. Could you give a hint? 
 A: To justify the answer formally you could use what you know about conditional probability. In particular:
Let $S$ be the event that at least one six occurs, and $D$ that all three dice show different faces. The outcome of our experiment is a triple $(a,b,c)$, where $a$, $b$, $c$ are the faces showed by the first, the second and the third dice respectively. There are $N = 6^3$ such triples. In addition, there are $6 \cdot 5 \cdot 4$ ways of choosing $(a,b,c)$ such that no face is repeated therein, so $P(D) = \frac {6 \cdot 5 \cdot 4}{N}$. Let us now consider the event $S \cap D$ (i.e. one dice shows six, and all of them have different outcomes); there are 3 ways to choose which dice shows six, and $5 \cdot 4$ to select the faces of the other two, in total: $3 \cdot 5 \cdot 4$ cases, which yields $P(S \cap D) = \frac{3 \cdot 5 \cdot 4}{N} > 0$. Finally, we obtain the probability  in question:
$$P(S|D) = \frac {P(S \cap D)}{P(D)} = \frac {3 \cdot 5 \cdot 4}{6 \cdot 5 \cdot 4} = 0.5.$$
A: Hint: if the three dice show different faces after being rolled, then surely three different numbers are obtained. What is the probability that these include the number $6$?
A: If all three dice are different, imagine the 6 numbers from 1 to 6. Three of them have been rolled and three have not. So each number has a 50% chance of being one of the three rolled since no number has an inherent advantage over the others.
A: To be unnecessarily thourough.  There are $6*5*4=120$ ways for the faces to be different.  (An arbitrary first die can have any face, the arbitrary second can have any of the five remaining, etc.)
There are $3(1*5*4)=60$ ways for one of the faces to be a $6$. (The face that is a 6, must be a six, the second can be any of 5 and the third any of 4, and there are 3 chooses for which die is $6$.  So Probability is $60/120 = 1/2$.
2)  The probability of the first die being a six is $1/6$.  The probability of the second die being a six, and the first die not being a six, given the dies are different is $5/6*1/5 = 1/6$.  So probability of one of the first two dice is six is $1/6 + 1/6 = 1/3$  The probability of the third die being six and neither of the first two, given that all there are different is $2/3*1/4 = 1/6$.  So the probability of one the faces being six given they are all different is $1/3 + 1/6 = 1/2$.
3)All combinations of different numbers are equally likely.  $3$ numbers appear.  $3$ do not.  Each is equally likely so that a six appears (or not) is 1/2.
