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I am struggling to grasp conditional probability as it is something I never fully understood in school. Could anyone explain to me how to answer the following questions? I know how to do a few of them but still confused by the others.

Two cards are chosen without replacement from a 52 card deck. In each case, find the conditional probability both cards are aces.

(a) If you are told one of the cards is the ace of spades.

(b) If you are told one of the cards is an ace.

(c) If you are told one of the cards is a spade.

(d) if you are told the first card selected is an ace.

(e) If you are told the second card selected is an ace.

Find the probability of getting four of a kind, given you have the ace of hearts and the ace of diamonds.

If anyone could help me in developing an understanding of this topic that would be brilliant

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    $\begingroup$ Which ones do you know how to do (and what do you think the answers are?). This may help people understand where the gaps are in your knowledge $\endgroup$
    – lioness99a
    Feb 24 '20 at 15:27
  • $\begingroup$ I would also recommend Khan Academy as a good resource for learning topics outside of a school setting $\endgroup$
    – lioness99a
    Feb 24 '20 at 15:28
  • $\begingroup$ I can do the first two, I think I got 1/221 for part (a) and 1/33 for part (b) $\endgroup$
    – user731762
    Feb 24 '20 at 15:31
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    $\begingroup$ Why not edit your question to include your efforts? Explain where you got $\frac 1{221}$ for instance. It will make it a lot easier to advise you if we see how you are reasoning. $\endgroup$
    – lulu
    Feb 24 '20 at 15:34
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    $\begingroup$ As for an answer of $\frac{1}{221} = \frac{1}{13\times 17} = \frac{4}{52}\times \frac{3}{51}$... that is the probability of having gotten two aces in general without any conditional statement being considered. If we are told one of the cards is the ace of spades... of course it is more likely that we got two aces. This should be obvious since if we were told something else, such as that one of the cards was a Jack, then we obviously can not have two aces. $\endgroup$
    – JMoravitz
    Feb 24 '20 at 15:57

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