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My problem is as follows: Suppose you draw a card from a 52-card deck and see that the card is a spade. You then draw another card without replacement. What is the probability that the next card is an ace?

I tried solving this question using conditional probability, but the issue is that the Spade that I initially drew could have also been an Ace. Here are the events that I defined:

A = Draw an ace

B = Draw a spade

$P(A | B) = P(A \cap B) / P(B)$

I'm confused how to calculate the intersection in the numerator. Thanks for your help!

David Gao
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Display Name
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  • Seems to be answered here: https://math.stackexchange.com/questions/3059683/consider-of-drawing-one-card-from-a-deck-of-52-prove-that-the-events-of-a-spa#:~:text=Let%20A%20be%20the%20event,%2F52%3D1%2F13. – Chickenmancer Mar 30 '24 at 00:38
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    This is not quite the same, @Chickenmancer . Display Name wants the conditional probability when drawing two* cards* that the second card is an Ace when given that the first card was a Spade. This needs the Law of Total Probability, partitioning the first draw on whether that Spade was an Ace or not.$$\mathsf P(A_2\mid S_1)=\mathsf P(A_2, A_1\mid S_1)+\mathsf P(A_2, A_1^\complement\mid S_1)$$ Where $A_1, A_2$ are the events of an Ace being the first and second cards respectively, and $S_1$ the event that the first card is a Spade. – Graham Kemp Mar 31 '24 at 09:03
  • It's a duplicate, but Chickenmancer linked the wrong question : https://math.stackexchange.com/questions/3068080/conditional-probability-with-choosing-two-cards-from-a-deck?rq=1 – D. Thomine Mar 31 '24 at 09:41
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    @GrahamKemp, thank you so much! This is exactly what I was thinking. I tried solving it using the Law of Total Probability, but I didn't realize that you can apply the LoTP on conditional probabilities. I'm used to seeing it in the context of: P(A) = P(A, B) + P(A, B^c). But now I realize that conditional probabilities can be manipulated in much the same way as marginal probabilities. – Display Name Mar 31 '24 at 14:44

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