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  1. ... we will have 3 conditions 1- we have 2 aces in the 5 cards 2- we have 3 aces in the... selecting 1st & 2nd card to be aces is 3/5 * 2/4 so the probability will be [3/5...

    2 Answers · Science & Mathematics · 05/12/2007

  2. ... there's a few possibilities of getting dealt the Aces with 5 cards. If we let: N=non-ace A=Ace...

    4 Answers · Science & Mathematics · 22/09/2009

  3. ... by 52!/13!^4. this is what you divide by next give an ace to each player. this can be done in 4! ways, which is equal to...

    4 Answers · Science & Mathematics · 20/12/2007

  4. ... 13 diamonds in a pack of cards and 4 aces, bearing in mind that one of the aces is...find out what the probability is to draw NOT an ace or diamond as the first card and then draw NOt...

    2 Answers · Science & Mathematics · 30/09/2011

  5. ...hypergeometric distribution to find the solution Let X be the number of aces in a 12 card hand. X has the hypergeometric distribution with the ...

    6 Answers · Science & Mathematics · 30/07/2008

  6. P(drawing an Ace from a deck of 52 cards) = 4/52 = 1/13 By applying ...) (1/13)^1 (12/13)^4 = 0.27924038 P(drawing at least 2 aces) = 1 - P(drawing 0 Aces) - P(drawing 1 Aces...

    2 Answers · Science & Mathematics · 11/04/2008

  7. ... under the assumption the question means being dealt four aces or four 9s in the first four cards... because obviously, if you deal out...

    4 Answers · Science & Mathematics · 24/05/2010

  8. qa qa P[20th card ace, but not ace of spades] = 3/4 P[21st card ace of spades...

    5 Answers · Science & Mathematics · 06/02/2010

  9. ... left out is you must calculate the chance of each receiving an ace assuming that four aces are left in the deck and multiply that...

    4 Answers · Science & Mathematics · 05/07/2008

  10. Number of groups of 3 aces: 4C3 = (4*3*2)/(3*2*1) = 4. Number of groups of 3 ...

    2 Answers · Science & Mathematics · 07/10/2009

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