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Bill draws two cards from a standard deck of 52 cards without replacement. What is the probability the first card he draws is the ace of hearts and the second card he draws is a king?

What I did: A regular card deck has 52 cards, therefore, the chances of getting the ace of hearts as the first card is :

$\frac{1}{52}$

Then, since one card is used we are left with 51 cards. There are four king cards in the set so I can represent that as :

$\frac{4}{51}$

Then I get $\frac{1}{52}$$*$$\frac{4}{51}$ which simplifies to $\frac{1}{156}$.

I have a feeling I am wrong because we draw 4 cards so it could be $\frac{4}{48}$. Any help on the question?

Edit: The cards are not replaced after the ace is drawn

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    $\begingroup$ Your argument is correct as is. $\endgroup$ – Ethan Bolker Feb 19 '18 at 0:57
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    $\begingroup$ Why do you say we draw $4$ cards? You only draw two. $\endgroup$ – Ross Millikan Feb 19 '18 at 0:59
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    $\begingroup$ You're good. This is related to conditional probability: $P(AH, King) = P(AH)P(King|AH)$. Where there's only 1 direction in which this can happen. Where are you getting "another 4" though? $\endgroup$ – EDZ Feb 19 '18 at 0:59
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    $\begingroup$ @altrey Yes, but you're not drawing "another 4". Big difference. You're only drawing another 1 card. That card can take 1 of 51 values and we're interested in 4 of those values. Does that make sense? There are 51 possible events, 4 of which we care about, but the process will only reveal 1 outcome. $\endgroup$ – EDZ Feb 19 '18 at 1:38
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    $\begingroup$ $\frac{1}{52}*\frac{4}{51}=\frac{1}{156}$????? $\endgroup$ – Adam Bailey Feb 19 '18 at 10:50

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