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... 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

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

...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

...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

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

(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

... = 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

... = 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

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

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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