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C
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Difficulty:
15%
(low)
Question Stats:
80% (01:31) correct
20%
(01:33)
wrong
based on 186
sessions
History
Date
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Not Attempted Yet
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. 1/336
B. 1/120
C. 1/56
D. 1/720
E. 1/1440
A. 1/336
B. 1/120
C. 1/56
D. 1/720
E. 1/1440
27 Sep 2012, 12:48
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joesamson
P=P(ace in the first draw)*P(ace in the second draw)*P(ace in the third draw)=3/8*2/7*1/6=1/56.
Answer: C.
28 Sep 2012, 13:47
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The probability of an event A occurring is the number of outcomes that result in A divided by the total number of possible outcomes.
There is only one result that results in a win: receiving three aces.
Since the order of arrangement does not matter, the number of possible ways to receive 3 cards is a combination problem.
The number of combinations of n objects taken r at a time is
C(n,r) = n!/(r!(n-r)!)
C(8,3) = 8!/(3!(8-3)!)
C(8,3) = 8!/(3!(5!))
C(8,3) = 40320/(6(120))
C(8,3) = 40320/720
C(8,3) = 56
The number of possible outcomes is 56.
Thus, the probability of being dealt 3 aces is 1/56.
HENCE C
There is only one result that results in a win: receiving three aces.
Since the order of arrangement does not matter, the number of possible ways to receive 3 cards is a combination problem.
The number of combinations of n objects taken r at a time is
C(n,r) = n!/(r!(n-r)!)
C(8,3) = 8!/(3!(8-3)!)
C(8,3) = 8!/(3!(5!))
C(8,3) = 40320/(6(120))
C(8,3) = 40320/720
C(8,3) = 56
The number of possible outcomes is 56.
Thus, the probability of being dealt 3 aces is 1/56.
HENCE C
