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  1. ... from a deck in 52C3 = 22,100 ways. P(at least 2 aces ) = P( 2 ace , 1 non-aces) + P( 3 aces , 0 non-aces...

    2 Answers · Science & Mathematics · 29/07/2014

  2. by two aces, i take it that it means exactly 2 aces. n[full house with 2 aces...4c2 *12c1*4c3 = A n[2 aces] = 4c2*48c3 = B P[full house | 2 aces] = A/B 48/48c3 = 3/1081 <------- your AD...

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

  3. ...10 and the 1 3 = 4 Now, to YOUR problem. Choose 1 ace from 4 cards, and 4 "others" from 48 cards = 4 ...

    5 Answers · Science & Mathematics · 11/04/2016

  4. ...you're only drawing four times... The probability you get an ace on the first draw is 4/54. Next, there's one less ace...

    6 Answers · Science & Mathematics · 08/01/2012

  5. P(draw ace) = 4/52*48/51 + 48/52*4/51 = 384/2,652 P(draw 2 aces) = 4/52*3/51 = 12/2,652 P(draw 1 or 2 aces) = P(draw ace) + P(draw 2 aces) = 384/2,652 + 12/2,652 = 396/2...

    5 Answers · Science & Mathematics · 29/01/2010

  6. (1/4) of the time the first card will be the spade Ace and the probability of the next card being a spade is (12/51) (3/4) of...

    4 Answers · Science & Mathematics · 06/07/2011

  7. ... = 0) = 4/52 = 1/13 (since it's P(first card is Ace)) P(X1 = 1) = 48/52 (no ace first) * 4 / 51 (Ace ...

    1 Answers · Science & Mathematics · 01/03/2013

  8. ... = 52!/(47!5!) Total combinations that include exactly 1 ace: (4!/(3!1!))*(48!/(44!4!) So ((4!/(3!1!))*(48!/(44!4!))/(52!/(47!5...

    7 Answers · Science & Mathematics · 21/04/2007

  9. just multiply the number of ways of choosing 3 aces from 4 and 10 cards from the remaining 52 C(4, 3)*C...

    2 Answers · Science & Mathematics · 13/08/2010

  10. First, find the probability that the first card is an Ace and the 2nd to 5th are not: (4/52) * (48/51) * (47/50) * (46/49...

    6 Answers · Science & Mathematics · 13/01/2008

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