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the Trip Aces game

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by sanju09 » Fri May 07, 2010 5:32 am
The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip?
(A) 1/1440
(B) 1/720
(C) 1/120
(D) 1/56
(E) 1/44
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by liferocks » Fri May 07, 2010 6:07 am
from a deck of 8 cards first card can be chosen in 8C1 or 8 ways

the deck will have 8-1 or 7 cards and the second can be chosen in 7C1 or 7 ways
the deck will have 7-1 or 6 cards and the second can be chosen in 6C1 or 6 ways

so 3 cards can be chosen in 8*7*6 ways

and the 3 aces can be chosen in 3*2*1 ways

so the probability is 3*2*1/8*7*6 or 1/56

Ans option D
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by analyst218 » Fri May 07, 2010 11:12 am
sanju09 wrote:The Full House Casino is running a new promotion. Each person visiting the casino has the opportunity to play the Trip Aces game. In Trip Aces, a player is randomly dealt three cards, without replacement, from a deck of 8 cards. If a player receives 3 aces, they will receive a free trip to one of 10 vacation destinations. If the deck of 8 cards contains 3 aces, what is the probability that a player will win a trip?
(A) 1/1440
(B) 1/720
(C) 1/120
(D) 1/56
(E) 1/44
D.
3C3/8C3
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