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... 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

### In a standard 52 card deck what is the probability of getting 4

**aces**when being dealt a 5 card hand?... 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

... 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

... 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

...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

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

... 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

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

... 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

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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