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Math Help - Card Probability

  1. #1
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    Card Probability

    I'm not sure if I'm approaching this problem correctly:

    A single playing card is drawn at random from each of six well-shuffled decks of playing cards. Let A be the event that all six cards drawn are different.
    (a) Find P(A).
    (b) Find the probability that at least tow of the drawn cards match.

    I think for (b) is will simply be 1 - P(A), but I need to find P(A) first. Here is the way I think I should solve it: The first deck will always be a "different" card, so it has probability of 1. The second deck has a probability of 51/52 of being a different card. The third would be 50/52, etc. So it comes down to P(A) = 52*51*...*47 / (52^6). Is this correct?
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  2. #2
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    Quote Originally Posted by uberbandgeek6 View Post
    I'm not sure if I'm approaching this problem correctly:

    A single playing card is drawn at random from each of six well-shuffled decks of playing cards. Let A be the event that all six cards drawn are different.
    (a) Find P(A).
    (b) Find the probability that at least tow of the drawn cards match.

    I think for (b) is will simply be 1 - P(A), but I need to find P(A) first. Here is the way I think I should solve it: The first deck will always be a "different" card, so it has probability of 1. The second deck has a probability of 51/52 of being a different card. The third would be 50/52, etc. So it comes down to P(A) = 52*51*...*47 / (52^6). Is this correct?
    Funny how they tell you the decks are well shuffled. If the cards are chosen randomly, this detail really doesn't matter!

    Yes, you are correct for both (a) and (b). Good job!
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  3. #3
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    Quote Originally Posted by uberbandgeek6 View Post
    So it comes down to P(A) = 52*51*...*47 / (52^6). Is this correct?
    51*50*...*47 / 52^5 ; same, but shorter...
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