Sort by

- Relevance
- |Time

...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

... 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

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 = 4/1322 Answers · Science & Mathematics · 16/03/2009

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

... 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

...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

**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, tony2 Answers · Science & Mathematics · 26/05/2013

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

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

All the

**aces**are drawn in exactly 7 draws iff 3**aces**are...3 Answers · Science & Mathematics · 17/05/2007