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So, the question is:

Given that a throw with 10 dice produced at least one ace, what is probability p of two or more aces?

It is conditional probability problem. The formula is $$P(A|B) = \frac{P(A\cap B)}{P(B)}$$ P(A) - probability of getting two or more aces in a throw of 10 dice.

P(B) - probability that throw with 10 dice produced at least one ace. $$P(B) = 1 - \frac{5^{10}}{6^{10}}$$

But, i don't know what should i do further. The correct answer is: $$1 - \frac{10*5^{9}}{(6^{10} - 5^{10})}$$

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    $\begingroup$ $P(A|B) = \frac{P(A\cap B)}{P(B)}$ and $P(A \cap B) = P$(2 or more aces) = 1- $P$(No ace) - $P$(1 ace) $\endgroup$ – user265328 Sep 7 '15 at 10:19
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$P(A)$ is the probability that the throw produced at least $2$ aces, meaning that $P(AB)$ is the probability that the throw produced at least one ace and at least two aces, which is the same as the probability that the throw produced at least two aces.

You therefore simply need to calculate $P(AB)$ which is the same as $P(A)$.

$P(A)$ is $1-x-y$ where $x$ is the probability that the throw produced no aces and $y$ is the probability that the throw produced precisely one ace.

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