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12 cards are removed from a fair pack of 52 cards. None are aces. What is the probability that if two more cards are removed that only 1 out of the 2 cards is an ace?

I can't seem to get this right and it is driving me crazy. Can any help in expaning to me pls?

2 Answers 2

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After removing 12 cards, we're left with a deck of 40 cards, of which 4 are aces. There are two ways to exactly one ace:

  • we draw an ace first (4 out of 40 chance), and then 'not an ace' (36 out of 39 chance)
  • we draw 'not an ace' first (36 out of 40 chance), and then an ace (4 out of 39 chance).

So the answer is $\frac{4}{40} \cdot \frac{36}{39} + \frac{36}{40} \cdot \frac{4}{39} = \frac{6}{65} +\frac{6}{65} = \frac{12}{65}.$

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We left with 40 cards. Out of which 4 are aces.

So we are picking 2 cards. One is ace and one other.

Ace = $\binom{4}{1}$

Other card = $\binom{36}{1}$

Total = $\binom{40}{2}$

Probability = $\frac{\binom{4}{1} \cdot \binom{36}{1}}{\binom{40}{2}}$

= $ \frac{ \frac{4!}{1!.3!} \cdot \frac{36!}{1!.35!}}{\frac{40.39.38!}{38!.2!}}$

= $ \frac{ \frac41 . \frac{36}{1}}{\frac{40.39}{2.1}} = \frac{12}{65}$