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  1. ...replace the card each time, then each time, you have a 4/52 chance of drawing an ace... so, since each time is independent of the others, it is (4/52...

    4 Answers · Science & Mathematics · 01/08/2008

  2. ... the probability that the two cards drawn are non-aces = 48C2/52C2 = 1128 / 1326 = 0.8507 or 85.07% <&...

    4 Answers · Science & Mathematics · 17/02/2011

  3. P(Ace) is 4/52 = 1/13 P(Spade) is 13/52 = 1/4 P(Ace and Space) = 1/52 therefore P(Ace or Spade) = 1/13 +1/4 - 1/52 = 4/13

    22 Answers · Science & Mathematics · 16/03/2009

  4. 4C2 * 12C3 * 4C1^3 = 84,480 choosing 2 aces from 4 aces choosing 3 ranks from the remaining 12...

    5 Answers · Science & Mathematics · 16/11/2011

  5. ... of 5 cards from 52 (which includes any of four aces) Subtract from that the number of permutations of 5 ...

    5 Answers · Science & Mathematics · 01/09/2010

  6. ...be any one of the 52 cards. 4/52 = 1/13 of those cards are aces. The probability is 1/13. 2. The second card dealt...

    2 Answers · Science & Mathematics · 10/12/2012

  7. Ace car rental: cost = .20 m + $30 Bell car rental cost = .30 m + 15 If the costs are to be the same, then clearly, they must be equal; so: .20m + 30 = .30m + 15 always, tony

    2 Answers · Science & Mathematics · 26/05/2013

  8. your AD ---------- 4 aces, 48 non-aces P(x) = [ P(4,1)*P(48,x-2) ] *P(3...

    2 Answers · Science & Mathematics · 13/05/2013

  9. You give 4 aces to the hand, you are left with 9 cards. The one king can be...to be chosen from the remaining (52 - 8) = 44 cards (as the aces and kings are gone) which can be done in 44C8...

    1 Answers · Science & Mathematics · 25/09/2009

  10. All the aces are drawn in exactly 7 draws iff 3 aces are...

    3 Answers · Science & Mathematics · 17/05/2007

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