Conditional probability questions 
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*Three dice are rolled. If no two show the same face, what is the probability that one is an ace?

*Given that a throw with ten dice produced at least one ace, what is the probability p of two or more aces?


Answers provided by the Author William Feller is


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*$1-\frac{5*4*3}{6*5*4}=\frac12$
2.$p=1-\frac{10*5^9}{6^{10}-5^{10}}=0.6147724$
I don't understand how did the author calculate these answers?
If any member knows may answer these questions.
 A: Explanation of Question 1
The desired probability is: $ P(\text{an ace} | \text{each face is different}) $
So, one of the first things that we should calculate is how many different ways we can roll three die with the faces all being different (the given in the question). This number becomes the size of our sample space (also, the question implicitly assumes that order does not matter. We then want to calculate how many of those possible rolls (in the sample space) do not have an ace. The difference is our desired probability. 
Worked out solution:
Ways to roll three die with all different faces: 6*5*4


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*First dice in the roll can be any of the 6 faces

*Second dice in roll can have any face other than that on the first (5)

*Third dice in roll can have any face other than first two die (4)


Ways to roll three die with all different faces and no ace: 5*4*3


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*First dice in the roll can be any of the 5 non-ace faces

*Second dice in roll can have any face other than the first or an ace (4)

*Third dice in roll can have any face other than first two die or an ace (3)


Putting these together:
$ P(\text{no ace} | \text{each face is different}) = \frac{5*4*3}{6*5*4} $
$ P(\text{an ace} | \text{each face is different}) = 1 - P(\text{no ace} | \text{each face is different}) $
$ P(\text{an ace} | \text{each face is different}) = 1 - \frac{5*4*3}{6*5*4} = \frac{1}{2} $
