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There is a standard deck of cards ($52$) with four suits; hearts, diamonds, clubs and spades. What is the probability that you draw at most one ace in two draws?

I'm not entirely sure if this approach is correct or not. Firstly, the probability of drawing two aces (A) is,

$$P(A\cap A) = \left(A^{2nd}\mid A^{1st}\right) \times P\left(A^{1st}\right)$$

Therefore, the probability of at most one ace $(A)$ is given by,

$$P(\text{at most one}) = 1 - P(\text{at least two})$$

But since we are drawing exactly two cards, the probability of at least two aces is the same as the probability of two aces. So,

$$P(\text{at most one})= 1 - P(\text{at least two}) = 1-\left(\frac{4}{52}\times\frac{3}{51}\right) = \frac{660}{663} = \frac{220}{221}$$

Is the approach correct?

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    $\begingroup$ Seems all right. $\endgroup$ – Parcly Taxel Oct 14 '16 at 8:37
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Your solution is correct.

As a check, the probability of drawing two aces in two draws is the number of ways we can select two of the four aces divided by the number of ways we can select two of the $52$ cards in the deck.
$$P(\text{two aces}) = \frac{\dbinom{4}{2}}{\dbinom{52}{2}} = \frac{\dfrac{4!}{2!2!}}{\dfrac{52!}{2!50!}} = \frac{\dfrac{4 \cdot 3}{2 \cdot 1}}{\dfrac{52 \cdot 51}{2 \cdot 1}} = \frac{4 \cdot 3}{52 \cdot 51} = \frac{1}{13 \cdot 17} = \frac{1}{221}$$ Hence, the probability of obtaining at most one ace is $$P(\text{at most one ace}) = 1 - P(\text{two aces}) = 1 - \frac{1}{221} = \frac{220}{221}$$ as you found.

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