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I've come across a problem in which is necessary to obtain the probability that by drawing 2 cards from a standard deck without replacement, the second card is red.

I understand there are many ways to go about this problem (probably simpler than what I'm asking) but what I'm trying to understand here is not how to obtain the probability, but how to count the number of subsets of the deck that can result in the second card being red.

I think first of all the problem I have is in the book they treat this as a combinations problem. But to me it seems more like a permutations problem since were interested in the event in which the second card is red.

Doesn't this mean that we're interested in all the ordered subsets of S (S containing all cards of the deck) in which the subset contains only 2 elements (cards) and the second element is part of the subset of (Diamonds union Hearts)?

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  • $\begingroup$ Are these cards drawn with replacement or without replacement? $\endgroup$
    – QED
    Oct 15, 2020 at 7:36
  • $\begingroup$ @QED without replacement $\endgroup$
    – Marco Castro
    Oct 15, 2020 at 7:39

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There are $52$ cards in a deck. The first card can be drawn in $52$ ways, and since the cards are drawn without replacement, the second card can be drawn in $51$ ways. This gives us $52\times 51$ ways of drawing any $2$ cards from the deck.

Now we find the number of ways two cards can be drawn from the deck so that the second card is red. Now we know that there are $26$ black cards and $26$ red cards in the deck. There can be two possibilities:

  1. The first card is black and the second card is red - This is possible in $26\times26$ ways. This is because there are $26$ black cards to choose from in the first draw, and there are $26$ red cards to choose from in the second draw.
  2. Both the cards are red - This can be done in $26\times25$ ways. This is because there are $26$ red cards to choose from in the first draw, and once the first card was chosen, there are $25$ red cards to choose the second red card from. (It doesnt matter whteher the red card is a Diamond or a Heart)

Thus in total, the number of ways $2$ cards can be drawn from a deck, so that the second card is red is $26\times26+26\times25=26\times51$.

And the probability that out of the two cards drawn, the second card is red is $\frac{26\times51}{52\times51}=\frac12$.

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  • $\begingroup$ This a great answer, I think it is very intuitive and helped me understand what is going on here. Now I understand that 52x51 are the total ways of drawing any 2 cards (nPr). is there a general form which I can calculate 26x51? $\endgroup$
    – Marco Castro
    Oct 15, 2020 at 8:02
  • $\begingroup$ I am afraid there is no general method for the later, it has to be broken down into cases $\endgroup$
    – QED
    Oct 15, 2020 at 8:15
  • $\begingroup$ I have also noticed that the same can be achieved by choosing a card from the 26 red cards and dividing by the ways 2 cards can be chosen out of 52. that is 26C1/52C2. In this case, why we don't count at all the first draw? In the book they explain that for the multiplication principle, the chronological order doesn't matter. But I have problem wrapping my head around the fact that one should choose 2 out of 52 but 1 out of 51 since the first card has already been drawn. $\endgroup$
    – Marco Castro
    Oct 15, 2020 at 8:23

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